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The Top Unsolved Questions in Mathematics Remain Mostly Mysterious
Just one of the seven Millennium Prize Problems named 21 years ago has been solved
- By Rachel Crowell on May 28, 2021
Twenty-one years ago this week, mathematicians released a list of the top seven unsolved problems in the field. Answering them would offer major new insights in fundamental mathematics and might even have real-world consequences for technologies such as cryptography.
But big questions in math have not often attracted the same level of outside interest that mysteries in other scientific areas have. When it comes to understanding what math research looks like or what the point of it is, many folks are still stumped, says Wei Ho, a mathematician at the University of Michigan. Although people often misunderstand the nature of her work, Ho says it does not have to be difficult to explain. “My cocktail party spiel is always about elliptic curves,” she adds. Ho often asks partygoers, “You know middle school parabolas and circles? Once you start making a cubic equation, things get really hard.... There are so many open questions about them.”
One famous open problem called the Birch and Swinnerton-Dyer conjecture concerns the nature of solutions to equations of elliptic curves, and it is one of the seven Millennium Prize Problems that were selected by the founding scientific advisory board of the Clay Mathematics Institute (CMI) as what the institute describes as “some of the most difficult problems with which mathematicians were grappling at the turn of the second millennium.” At a special event held in Paris on May 24, 2000, the institute announced a prize of $1 million for each solution or counterexample that would effectively resolve one of these problems for the first time. Rules revised in 2018 stipulate that the result must achieve “general acceptance in the global mathematics community.”
The 2000 proclamation gave $7 million worth of reasons for people to work on the seven problems: the Riemann hypothesis, the Birch and Swinnerton-Dyer conjecture, the P versus NP problem, the Yang-Mills existence and mass gap problem, the Poincaré conjecture, the Navier-Stokes existence and smoothness problem, and the Hodge conjecture. Yet despite the fanfare and monetary incentive, after 21 years, only the Poincaré conjecture has been solved.
An Unexpected Solution
In 2002 and 2003 Grigori Perelman, a Russian mathematician then at the St. Petersburg Department of the Steklov Mathematical Institute of the Russian Academy of Sciences, shared work connected to his solution of the Poincaré conjecture online. In 2010 CMI announced that Perelman had proved the conjecture and, along the way, had also solved the late mathematician William Thurston’s related geometrization conjecture. (Perelman, who rarely engages with the public, famously turned down the prize money .)
According to CMI, the Poincaré conjecture focuses on a topological question about whether spheres with three-dimensional surfaces are “essentially characterized” by a property called “simple connectivity.” That property means that if you encase the surface of the sphere with a rubber band, you can compress that band—without tearing it or removing it from the surface—until it is just a single point. A two-dimensional sphere or doughnut hole is simply connected, but a doughnut (or another shape with a hole in it) is not.
Martin Bridson, a mathematician at the University of Oxford and president of CMI, describes Perelman’s proof as “one of the great events of, certainly, the last 20 years” and “a crowning achievement of many strands of thought and our understanding of what three dimensional spaces are like.” And the discovery could lead to even more insights in the future. “The proof required new tools, which are themselves giving far-reaching applications in mathematics and physics,” says Ken Ono, a mathematician at the University of Virginia.
Ono has been focused on another Millennium Problem: the Riemann hypothesis, which involves prime numbers and their distribution. In 2019 he and his colleagues published a paper in the Proceedings of the National Academy of Sciences USA that reexamined an old, formerly abandoned approach for working toward a solution. In an accompanying commentary, Enrico Bombieri, a mathematician at the Institute for Advanced Study in Princeton, N.J., and a 1974 winner of mathematics’ highest honor, the Fields Medal, described the research as a “major breakthrough.” Yet Ono says it would be unfounded to describe his work as “anything that suggests that we’re about to prove the Riemann hypothesis.” Others have also chipped away at this problem over the years. For instance, mathematician “Terry Tao wrote a nice paper a couple years ago on [mathematician Charles] Newman’s program for the Riemann hypothesis,” Ono says.
Progress on What Won’t Work
The fact that just one of the listed problems has been solved so far is not surprising to the experts—the puzzles are, after all, long-standing and staggeringly difficult. “The number of problems that have been solved is one more than I would expect” to see by now, says Manjul Bhargava, a mathematician at Princeton University and a 2014 Fields medalist. Bhargava himself has reported multiple recent results connected to the Birch and Swinnerton-Dyer conjecture, including one in which he says he and his colleagues “prove that more than 66 percent of elliptic curves satisfy the Birch and Swinnerton-Dyer conjecture.”
None of the problems will be easy to solve, but some may prove especially intractable. The P versus NP problem appears so difficult to solve that Scott Aaronson, a theoretical computer scientist at the University of Texas at Austin, calls it “a marker of our ignorance.” This problem concerns the issue of whether questions that are easy to verify (a class of queries called NP) also have solutions that are easy to find (a class called P).* Aaronson has written extensively about the P versus NP problem. In a paper published in 2009 he and Avi Wigderson, a mathematician and computer scientist at the Institute for Advanced Study and one of the winners of the 2021 Abel Prize, showed a new barrier to proving that the P class is not the same as the NP class. The barrier that Aaronson and Wigderson found is the third one discovered so far.
“There’s a lot of progress on showing what approaches will not work,” says Virginia Vassilevska Williams, a theoretical computer scientist and mathematician at the Massachusetts Institute of Technology. “Proving that P [is] not equal to NP would be an important stepping-stone toward showing that cryptography is well founded,” she adds. “Right now cryptography is based on unproved assumptions,” one of which is the idea that P is not equal to NP. “In order to show that you cannot break the cryptographic protocols that people need in modern computers,” including ones that keep our financial and other online personal information secure, “you need to at least prove that P is not equal to NP,” Vassilevska Williams notes. “When people have tried to pin me down to a number,” Aaronson says, “I’ll give a 97 percent or 98 percent chance that P is not equal to NP.”
Climbing Mount Everest
Searching for solutions to the prize problems is similar to trying to climb Mount Everest for the first time, Ono says. “There are various steps along the way that represent progress,” he adds. “The real question is: Can you make it to base camp? And if you can, you still know you’re very far.”
For problems such as the Birch and Swinnerton-Dyer conjecture and the Riemann hypothesis, Ono says, “surely we’re at Nepal”—one of the countries of departure for climbing the mountain—“but have we made it to base camp?” Mathematicians might still need additional “gear” to trek to the peak. “We’re now trying to figure out what the mathematical analogues are for the high-tech tools, the bottles of oxygen, that will be required to help us get to the top,” Ono says. Who knows how many obstacles could be sitting between current research and possible solutions to these problems? “Maybe there are 20. Maybe we’re closer than we think,” Ono says.
Despite the difficulty of the problems, mathematicians are optimistic about the long term. “I hope very much that while I’m president of the Clay institute, one of them will be solved,” says Bridson, who notes that CMI is in the process of strategizing about how to best continue raising awareness about the problems. “But one has to accept that they’re profoundly difficult problems that may continue to shape mathematics for the rest of my life without being solved.”
* Editor’s Note (6/2/21): This sentence was revised after posting to correct the description of the P versus NP problem.
ABOUT THE AUTHOR(S)
Rachel Crowell is a Midwest-based writer covering science and mathematics. Follow Crowell on Twitter @writesRCrowell Credit: Nick Higgins
Recent Articles by Rachel Crowell
- Why 2 Is the Best Number and Other Secrets from a MacArthur-Winning Mathematician
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10 Math Equations That Have Never Been Solved
By Kathleen Cantor, 10 Sep 2020
Mathematics has played a major role in so many life-altering inventions and theories. But there are still some math equations that have managed to elude even the greatest minds, like Einstein and Hawkins. Other equations, however, are simply too large to compute. So for whatever reason, these puzzling problems have never been solved. But what are they?
Like the rest of us, you're probably expecting some next-level difficulty in these mathematical problems. Surprisingly, that is not the case. Some of these equations are even based on elementary school concepts and are easily understandable - just unsolvable.
1. The Riemann Hypothesis
Equation: σ (n) ≤ Hn +ln (Hn)eHn
- Where n is a positive integer
- Hn is the n-th harmonic number
- σ(n) is the sum of the positive integers divisible by n
For an instance, if n = 4 then σ(4)=1+2+4=7 and H4 = 1+1/2+1/3+1/4. Solve this equation to either prove or disprove the following inequality n≥1? Does it hold for all n≥1?
This problem is referred to as Lagarias’s Elementary Version of the Riemann Hypothesis and has a price of a million dollars offered by the Clay Mathematics Foundation for its solution.
2. The Collatz Conjecture
Equation: 3n+1
- where n is a positive integer n/2
- where n is a non-negative integer
Prove the answer end by cycling through 1,4,2,1,4,2,1,… if n is a positive integer. This is a repetitive process and you will repeat it with the new value of n you get. If your first n = 1 then your subsequent answers will be 1, 4, 2, 1, 4, 2, 1, 4… infinitely. And if n = 5 the answers will be 5,16,8,4,2,1 the rest will be another loop of the values 1, 4, and 2.
This equation was formed in 1937 by a man named Lothar Collatz which is why it is referred to as the Collatz Conjecture.
3. The Erdős-Strauss Conjecture
Equation: 4/n=1/a+1/b+1/c
- a, b and c are positive integers.
This equation aims to see if we can prove that for if n is greater than or equal to 2, then one can write 4*n as a sum of three positive unit fractions.
This equation was formed in 1948 by two men named Paul Erdős and Ernst Strauss which is why it is referred to as the Erdős-Strauss Conjecture.
4. Equation Four
Equation: Use 2(2∧127)-1 – 1 to prove or disprove if it’s a prime number or not?
Looks pretty straight forward, does it? Here is a little context on the problem.
Let’s take a prime number 2. Now, 22 – 1 = 3 which is also a prime number. 25 – 1 = 31 which is also a prime number and so is 27−1=127. 2127 −1=170141183460469231731687303715884105727 is also prime.
5. Goldbach's Conjecture
Equation: Prove that x + y = n
- where x and y are any two primes
This problem, as relatively simple as it sounds has never been solved. Solving this problem will earn you a free million dollars. This equation was first proposed by Goldbach hence the name Goldbach's Conjecture.
If you are still unsure then pick any even number like 6, it can also be expressed as 1 + 5, which is two primes. The same goes for 10 and 26.
6. Equation Six
Equation: Prove that (K)n = JK1N(q)JO1N(q)
- Where O = unknot (we are dealing with knot theory )
- (K)n = Kashaev's invariant of K for any K or knot
- JK1N(q) of K is equal to N- colored Jones polynomial
- We also have the volume of conjecture as (EQ3)
- Here vol(K) = hyperbolic volume
This equation tries to portray the relationship between quantum invariants of knots and the hyperbolic geometry of knot complements . Although this equation is in mathematics, you have to be a physics familiar to grasp the concept.
7. The Whitehead Conjecture
Equation: G = (S | R)
- when CW complex K (S | R) is aspherical
- if π2 (K (S | R)) = 0
What you are doing in this equation is prove the claim made by Mr. Whitehead in 1941 in an algebraic topology that every subcomplex of an aspherical CW complex that is connected and in two dimensions is also spherical. This was named after the man, Whitehead conjecture.
8. Equation Eight
Equation: (EQ4)
- Where Γ = a second countable locally compact group
- And the * and r subscript = 0 or 1.
This equation is the definition of morphism and is referred to as an assembly map. Check out the reduced C*-algebra for more insight into the concept surrounding this equation.
9. The Euler-Mascheroni Constant
Equation: y=limn→∞(∑m=1n1m−log(n))
Find out if y is rational or irrational in the equation above. To fully understand this problem you need to take another look at rational numbers and their concepts. The character y is what is known as the Euler-Mascheroni constant and it has a value of 0.5772.
This equation has been calculated up to almost half of a trillion digits and yet no one has been able to tell if it is a rational number or not.
10. Equation Ten
Equation: π + e
Find the sum and determine if it is algebraic or transcendental. To understand this question you need to have an idea of algebraic real numbers and how they operate. The number pi or π originated in the 17th century and it is transcendental along with e. but what about their sum? So Far this has never been solved.
As you can see in the equations above, there are several seemingly simple mathematical equations and theories that have never been put to rest. Decades are passing while these problems remain unsolved. If you're looking for a brain teaser, finding the solutions to these problems will give you a run for your money.
See the 11 Comments below.
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Posted in Mathematics category - 10 Sep 2020 [ Permalink ]
11 Comments on “10 Math Equations That Have Never Been Solved”
But 2(2127)−1 = 340282366920938463463374607431768211455 is not a prime number. It is divisible by 64511.
Hello I am explorer and i type on google search " unsolvable mathematical formulas ", and I first find this syte. I see you are good-math-guys. Do you know what is this formula means:
π × ∞ = " 5 "
If you happen to have a quantum computer, I am not kidding be smart and don't insert this formula: [π × ∞ = " 5 "] into it please.
Maybe only, if you know meaning of this three symbols up writen and connected together.
(x dot epsilon)
I can explain my theory if you want me to spoil the pleasure of solving the equation. And mathematics as a science too or " as well " sorry i am not good in English, and google translate is not exelent.
8.539728478 is the answer to number 10
8.539728478 is the answer to number 10 or 8.539734221
Equation Four: Solved
To determine whether the number 2(2^127)-1 – 1 is a prime number, we first need to calculate its value. The expression 2(2^127) can be simplified as follows:
2(2^127) = 2 * 2^127 = 2^128
Therefore, the expression 2(2^127)-1 – 1 can be written as 2^128 – 1 – 1. We can then simplify this further to get:
2^128 – 1 – 1 = 2^128 – 2
To determine whether this number is prime, we can use the fundamental theorem of arithmetic, which states that every positive integer can be written as a product of prime numbers in a unique way (ignoring the order of the factors). This means that if a number is not prime, it can be expressed as the product of two or more prime numbers.
We can use this theorem to determine whether 2^128 – 2 is prime by trying to express it as the product of two or more prime numbers. However, it is not possible to do this, because 2^128 – 2 cannot be evenly divided by any prime number (except for 1, which is not considered a prime number).
Therefore, we can conclude that 2^128 – 2 is a prime number, because it cannot be expressed as the product of two or more prime numbers.
Equation Ten: Solved
The sum of π and e is equal to π + e = 3.14159 + 2.71828 = 5.85987.
To determine whether this number is algebraic or transcendental, we first need to understand the difference between these two types of numbers. Algebraic numbers are numbers that can be expressed as a root of a polynomial equation with integer coefficients, while transcendental numbers cannot be expressed in this way.
In this case, the number 5.85987 can be expressed as the root of the polynomial equation x^2 - 5.85987x + 2.71828 = 0. Therefore, it is an algebraic number.
In conclusion, the sum of π and e is equal to 5.85987, which is an algebraic number.
Equation 2: SOLVED
The equation 3n + 1 states that a positive integer n should be multiplied by 3 and then 1 should be added to the result. If the resulting value is then divided by 2 and the quotient is a non-negative integer, the process should be repeated with the new value of n.
To prove that this equation always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can start by substituting a value for n and performing the calculations as specified in the equation.
For example, if n is equal to 1, the sequence of values will be: n = 1 3n + 1 = 3(1) + 1 = 4 n = 4/2 = 2 3n + 1 = 3(2) + 1 = 7 n = 7/2 = 3.5
Since n must be a non-negative integer, the value of n becomes 2 when the result of the previous calculation is divided by 2. The sequence then becomes: n = 2 3n + 1 = 3(2) + 1 = 7 n = 7/2 = 3.5
The value of n becomes 3 when the result of the previous calculation is divided by 2. The sequence then becomes: n = 3 3n + 1 = 3(3) + 1 = 10 n = 10/2 = 5
The value of n becomes 5 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 5 3n + 1 = 3(5) + 1 = 16 n = 16/2 = 8 n = 8/2 = 4 n = 4/2 = 2 n = 2/2 = 1 n = 1/2 = 0.5
Since n must be a non-negative integer, the value of n becomes 1 when the result of the previous calculation is divided by 2. The sequence then becomes: n = 1 3n + 1 = 3(1) + 1 = 4 n = 4/2
To complete the proof that the equation 3n + 1 always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can continue the process of substituting values for n and performing the calculations as specified in the equation.
If n is equal to 4, the sequence of values will be: n = 4 3n + 1 = 3(4) + 1 = 13 n = 13/2 = 6.5
Since n must be a non-negative integer, the value of n becomes 6 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 6 3n + 1 = 3(6) + 1 = 19 n = 19/2 = 9.5
Certainly! To complete the proof that the equation 3n + 1 always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can continue the process of substituting values for n and performing the calculations as specified in the equation.
If n is equal to 4, the sequence of values will be:
n = 4 3n + 1 = 3(4) + 1 = 13 n = 13/2 = 6.5
Since n must be a non-negative integer, the value of n becomes 9 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 9 3n + 1 = 3(9) + 1 = 28 n = 28/2 = 14 n = 14/2 = 7 n = 7/2 = 3.5
The value of n becomes 3 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 3 3n + 1 = 3(3) + 1 = 10 n = 10/2 = 5 n = 5/2 = 2.5
Since n must be a non-negative integer, the value of n becomes 2 when the result of the previous calculation is divided by 2. The sequence then becomes:
n = 2 3n + 1 = 3(2) + 1 = 7 n = 7/2 = 3.5
As we can see, the sequence of values becomes repetitive
The Riemann Hypothesis
This equation states that the sum of the positive integers divisible by n (σ(n)) is less than or equal to the n-th harmonic number (Hn) plus the natural logarithm of the n-th harmonic number (ln(Hn)) multiplied by the n-th harmonic number (Hn) raised to the power of Hn.
To solve this equation, you would need to substitute a specific value for n and determine the value of Hn and σ(n) for that specific value. You can then substitute these values into the equation and see if it holds true.
For example, if n = 5, the sum of the positive integers divisible by 5 (σ(5)) is 15 (1 + 5 + 10 + 15 + 20 + 25), the 5th harmonic number (H5) is 2.28, and the natural logarithm of the 5th harmonic number (ln(H5)) is 0.83. Substituting these values into the equation, we get:
σ(5) ≤ H5 + ln(H5)eH5 15 ≤ 2.28 + 0.83 * 2.28^2.28 15 ≤ 4.39
Since 15 is less than or equal to 4.39, the equation holds true for this specific value of n.
Equation #9
In the equation y = limn→∞(∑m=1n1m−log(n)), y is the limit of the sequence (∑m=1n1m−log(n)) as n approaches infinity.
The Euler-Mascheroni constant is defined as the limit of the sequence (∑m=1n1m−log(n)) as n approaches infinity, and it has a value of approximately 0.5772. Therefore, y is equal to the Euler-Mascheroni constant, which is a rational number.
Rational numbers are numbers that can be expressed as the ratio of two integers, such as 3/4, 7/11, or 2/5. They can be written as a finite or repeating decimal, such as 0.75, 0.636363636..., or 1.5.
Irrational numbers are numbers that cannot be expressed as the ratio of two integers, and they cannot be written as a finite or repeating decimal. Examples of irrational numbers include √2, π, and e.
Since y is equal to the Euler-Mascheroni constant, which is a rational number, y is a rational number.
The equation G = (S | R) is a definition of a CW complex, where S and R are subcomplexes of G. A CW complex is a topological space that can be built up from cells, where each cell is homeomorphic to a closed ball in Euclidean space.
The statement "when CW complex K (S | R) is aspherical" means that the complex K (S | R) does not contain any non-trivial loops, i.e. loops that cannot be continuously contracted to a point. This implies that the fundamental group of K (S | R) is trivial, which means that π1(K (S | R)) = {e}.
The statement "if π2 (K (S | R)) = 0" means that the second homotopy group of the complex K (S | R) is trivial, which means that there are no non-trivial 2-dimensional holes in K (S | R).
Together, these statements imply that the CW complex K (S | R) is a topological space with no non-trivial loops or holes. This is a strong condition that is satisfied by very few spaces, and it is a necessary condition for a space to be aspherical.
In summary, the statement "when CW complex K (S | R) is aspherical" and "if π2 (K (S | R)) = 0" implies that the complex K (S | R) is a topological space with no non-trivial loops or holes, which is a necessary condition for a space to be aspherical.
#3 Erdos Strauss Conjecture:
To solve the equation 4/n = 1/a + 1/b + 1/c where n ≥ 2, a, b and c are positive integers, we can first multiply both sides of the equation by nabc to get rid of the fractions:
4abc = nab + nbc + nac
We can then group like terms:
4abc = (n + a)(b + c)
Now we can use the fact that n, a, b, and c are positive integers to make some observations:
Since n, a, b and c are positive integers, n, a, b and c must be factors of 4abc. Since n is greater than or equal to 2, it must be one of the factors of 4abc. The other factors of 4abc are (n + a), b, and c. So, to find all the possible values of n, a, b, and c, we must find all the ways to factorize 4abc such that one of the factors is greater than or equal to 2.
4abc = 4 * 1 * 1 * 2 * 3 * 5 = 120
Some possible factorizations are:
n = 2, a = 1, b = 5, c = 12 n = 2, a = 3, b = 5, c = 8 n = 2, a = 4, b = 3, c = 15 n = 2, a = 6, b = 2, c = 20 n = 4, a = 1, b = 3, c = 30 So, the possible solutions to the equation are: (n,a,b,c) = (2,1,5,12), (2,3,5,8), (2,4,3,15), (2,6,2,20), (4,1,3,30)
It's worth noting that this is not an exhaustive list, but just some of the possible solutions, as there could be infinitely many solutions to this equation.
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5 Simple Math Problems No One Can Solve
Easy to understand, supremely difficult to prove.
Mathematics can get pretty complicated. Fortunately, not all math problems need to be inscrutable. Here are five current problems in the field of mathematics that anyone can understand, but nobody has been able to solve.
Collatz Conjecture
Pick any number. If that number is even, divide it by 2. If it's odd, multiply it by 3 and add 1. Now repeat the process with your new number. If you keep going, you'll eventually end up at 1. Every time.
Moving Sofa Problem
So you're moving into your new apartment, and you're trying to bring your sofa. The problem is, the hallway turns and you have to fit your sofa around a corner. If it's a small sofa, that might not be a problem, but a really big sofa is sure to get stuck. If you're a mathematician, you ask yourself: What's the largest sofa you could possibly fit around the corner? It doesn't have to be a rectangular sofa either, it can be any shape.
This is the essence of the moving sofa problem . Here are the specifics: the whole problem is in two dimensions, the corner is a 90-degree angle, and the width of the corridor is 1. What is the largest two-dimensional area that can fit around the corner?
The largest area that can fit around a corner is called—I kid you not—the sofa constant. Nobody knows for sure how big it is, but we have some pretty big sofas that do work, so we know it has to be at least as big as them. We also have some sofas that don't work, so it has to be smaller than those. All together, we know the sofa constant has to be between 2.2195 and 2.8284.
Perfect Cuboid Problem
Remember the pythagorean theorem, A 2 + B 2 = C 2 ? The three letters correspond to the three sides of a right triangle. In a Pythagorean triangle, and all three sides are whole numbers. Let's extend this idea to three dimensions. In three dimensions, there are four numbers. In the image above, they are A, B, C, and G. The first three are the dimensions of a box, and G is the diagonal running from one of the top corners to the opposite bottom corner.
Just as there are some triangles where all three sides are whole numbers, there are also some boxes where the three sides and the spatial diagonal (A, B, C, and G) are whole numbers. But there are also three more diagonals on the three surfaces (D, E, and F) and that raises an interesting question: can there be a box where all seven of these lengths are integers?
The goal is to find a box where A 2 + B 2 + C 2 = G 2 , and where all seven numbers are integers. This is called a perfect cuboid. Mathematicians have tried many different possibilities and have yet to find a single one that works. But they also haven't been able to prove that such a box doesn't exist, so the hunt is on for a perfect cuboid.
Inscribed Square Problem
Draw a closed loop. The loop doesn't have to be a circle, it can be any shape you want, but the beginning and the end have to meet and the loop can't cross itself. It should be possible to draw a square inside the loop so that all four corners of the square are touching the loop. According to the inscribed square hypothesis, every closed loop (specifically every plane simple closed curve) should have an inscribed square, a square where all four corners lie somewhere on the loop.
This has already been solved for a number of other shapes, such as triangles and rectangles. But squares are tricky, and so far a formal proof has eluded mathematicians.
Happy Ending Problem
The happy ending problem is so named because it led to the marriage of two mathematicians who worked on it, George Szekeres and Esther Klein. Essentially, the problem works like this:
Make five dots at random places on a piece of paper. Assuming the dots aren't deliberately arranged—say, in a line—you should always be able to connect four of them to create a convex quadrilateral, which is a shape with four sides where all of the corners are less than 180 degrees. The gist of this theorem is that you'll always be able to create a convex quadrilateral with five random dots, regardless of where those dots are positioned.
So that's how it works for four sides. But for a pentagon, a five-sided shape, it turns out you need nine dots. For a hexagon, it's 17 dots. But beyond that, we don't know. It's a mystery how many dots is required to create a heptagon or any larger shapes. More importantly, there should be a formula to tell us how many dots are required for any shape. Mathematicians suspect the equation is M=1+2 N-2 , where M is the number of dots and N is the number of sides in the shape. But as yet, they've only been able to prove that the answer is at least as big as the answer you get that way.
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The Simple Math Problem We Still Can’t Solve
September 22, 2020
BIG MOUTH for Quanta Magazine
Introduction
This column comes with a warning: Do not try to solve this math problem.
You will be tempted. This problem is simply stated, easily understood, and all too inviting. Just pick a number, any number: If the number is even, cut it in half; if it’s odd, triple it and add 1. Take that new number and repeat the process, again and again. If you keep this up, you’ll eventually get stuck in a loop. At least, that’s what we think will happen.
Take 10 for example: 10 is even, so we cut it in half to get 5. Since 5 is odd, we triple it and add 1. Now we have 16, which is even, so we halve it to get 8, then halve that to get 4, then halve it again to get 2, and once more to get 1. Since 1 is odd, we triple it and add 1. Now we’re back at 4, and we know where this goes: 4 goes to 2 which goes to 1 which goes to 4, and so on. We’re stuck in a loop.
Or try 11: It’s odd, so we triple it and add 1. Now we have 34, which is even, so we halve it to get 17, triple that and add 1 to get 52, halve that to get 26 and again to get 13, triple that and add 1 to get 40, halve that to get 20, then 10, then 5, triple that and add 1 to get 16, and halve that to get 8, then 4, 2 and 1. And we’re stuck in the loop again.
The infamous Collatz conjecture says that if you start with any positive integer, you’ll always end up in this loop. And you’ll probably ignore my warning about trying to solve it: It just seems too simple and too orderly to resist understanding. In fact, it would be hard to find a mathematician who hasn’t played around with this problem.
I couldn’t ignore it when I first learned of it in school. My friends and I spent days trading thrilling insights that never seemed to get us any closer to an answer. But the Collatz conjecture is infamous for a reason: Even though every number that’s ever been tried ends up in that loop, we’re still not sure it’s always true. Despite all the attention, it’s still just a conjecture.
Yet progress has been made. One of the world’s greatest living mathematicians ignored all the warnings and took a crack at it , making the biggest strides on the problem in decades. Let’s take a look at what makes this simple problem so very complicated.
To understand the Collatz conjecture, we’ll start with the following function:
$latex f(n) = \begin{cases} n / 2 & \text{if $n$ is even } \\ 3n+1 & \text{if $n$ is odd } \end{cases}\ $
You might remember “piecewise” functions from school: The above function takes an input n and applies one of two rules to it, depending on whether the input is odd or even. This function f enacts the rules of the procedure we described above: For example, f (10) = 10/2 = 5 since 10 is even, and f (5) = 3 × 5 + 1 = 16 since 5 is odd. Because of the rule for odd inputs, the Collatz conjecture is also known as the 3 n + 1 conjecture.
The Collatz conjecture deals with “orbits” of this function f . An orbit is what you get if you start with a number and apply a function repeatedly, taking each output and feeding it back into the function as a new input. We call this “iterating” the function. We’ve already started computing the orbit of 10 under f , so let’s find the next few terms:
f (10) = 10/2 = 5 f (5) = 3 × 5 + 1 = 16 f (16) = 16/2 = 8 f (8) = 8/2 = 4
A convenient way to represent an orbit is as a sequence with arrows. Here’s the orbit of 10 under f :
10 → 5 → 16 → 8 → 4 → 2 → 1 → 4 → 2 → 1 → …
At the end we see we are stuck in the loop 1 → 4 → 2 → 1 → ….
Similarly, the orbit for 11 under f can be represented as
11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 → 4 → ….
Again we end up in that same loop. Try a few more examples and you’ll see that the orbit always seems to stabilize in that 4 → 2 → 1 → … loop. The starting values of 9 and 19 are fun, and if you’ve got a few minutes to spare, try 27. If your arithmetic is right, you’ll get there after 111 steps.
The Collatz conjecture states that the orbit of every number under f eventually reaches 1. And while no one has proved the conjecture, it has been verified for every number less than 2 68 . So if you’re looking for a counterexample, you can start around 300 quintillion. (You were warned!)
It’s easy to verify that the Collatz conjecture is true for any particular number: Just compute the orbit until you arrive at 1. But to see why it’s hard to prove for every number, let’s explore a slightly simpler function, ℊ .
$latex g(n) = \begin{cases} n / 2 & \text{if $n$ is even } \\ n+1 & \text{if $n$ is odd } \end{cases}\ $
The function ℊ is similar to f , but for odd numbers it just adds 1 instead of tripling them first. Since ℊ and f are different functions, numbers have different orbits under ℊ than under f . For example, here are the orbits of 10 and 11 under ℊ :
10 → 5 → 6 → 3 → 4 → 2 → 1 → 2 → 1 → 2 → … 11 → 12 → 6 → 3 → 4 → 2 → 1 → 2 → 1→ 2 → …
Notice that the orbit of 11 reaches 1 faster under ℊ than under f . The orbit of 27 also reaches 1 much faster under ℊ.
27 → 28 → 14 → 7 → 8 → 4 → 2 → 1 → 2 → …
In these examples, orbits under ℊ appear to stabilize, just like orbits under f , but in a slightly simpler loop:
→ 2 → 1 → 2 → 1 → ….
We might conjecture that orbits under ℊ always get to 1. I’ll call this the “Nollatz” conjecture, but we could also call it the n + 1 conjecture. We could explore this by testing more orbits, but knowing something is true for a bunch of numbers — even 2 68 of them — isn’t a proof that it’s true for every number. Fortunately, the Nollatz conjecture can actually be proved. Here’s how.
First, we know that half of a positive integer is always less than the integer itself. So if n is even and positive, then ℊ ( n ) = n /2 < n. In other words, when an orbit reaches an even number, the next number will always be smaller.
Now, if n is odd, then ℊ ( n ) = n + 1 which is bigger than n . But since n is odd, n + 1 is even, and so we know where the orbit goes next: ℊ will cut n + 1 in half. For an odd n the orbit will look like this:
… → n → n + 1 → $latex \frac{n+1}{2}$ → …
Notice that $latex \frac{n+1}{2}$ = $latex \frac{n}{2}$ + $latex \frac{1}{2}$. Since $latex \frac{n}{2}$ < n and $latex \frac{1}{2}$ is small, $latex \frac{n}{2}$ + $latex \frac{1}{2}$ is probably also less than n . In fact, some simple number theory can show us that as long as n > 1, then it’s always true that $latex \frac{n}{2}$ + $latex \frac{1}{2}$< n .
This tells us that when an orbit under ℊ reaches an odd number greater than 1, we’ll always be at a smaller number two steps later. And now we can outline a proof of the Nollatz conjecture: Anywhere in our orbit, whether at an even or an odd number, we’ll trend downward. The only exception is when we hit 1 at the bottom of our descent. But once we hit 1 we’re trapped the loop, just as we conjectured.
Can a similar argument work for the Collatz conjecture? Let’s go back to the original function.
As with ℊ , applying f to an even number makes it smaller. And as with ℊ , applying f to an odd number produces an even number, which means we know what happens next: f will cut the new number in half. Here’s what the orbit under f looks like when n is odd:
… → n → 3 n + 1 → $latex \frac{3n+1}{2}$ → …
But here’s where our argument falls apart. Unlike our example above, this number is bigger than n : $latex \frac{3n+1}{2}$ = $latex \frac{3n}{2}$ + $latex \frac{1}{2}$, and $latex \frac{3n}{2}$ = 1.5 n , which is always bigger than n . The key to our proof of the Nollatz conjecture was that an odd number must end up smaller two steps later, but this isn’t true in the Collatz case. Our argument won’t work.
If you’re like me and my friends back in school, you might now be excited about proving that the Collatz conjecture is false: After all, if the orbit keeps getting bigger, then how can it get down to 1? But that argument requires thinking about what happens next, and what happens next illuminates why the Collatz conjecture is so slippery: We can’t be sure whether $latex \frac{3n+1}{2}$ is even or odd.
We know that 3 n + 1 is even. If 3 n + 1 is also divisible by 4, then $latex \frac{3n+1}{2}$ is also even, and the orbit will fall. But if 3 n + 1 is not divisible by 4, then $latex \frac{3n+1}{2}$ is odd, and the orbit will rise. In general we can’t predict which will be true, so our argument stalls out.
But this approach isn’t completely useless. Since half of all positive integers are even, there’s a 50% chance that $latex \frac{3n+1}{2}$ is even, which makes the next step in the orbit $latex \frac{3n+1}{4}$. For n > 1 this is less than n , so half the time an odd number should get lower after two steps. There’s also a 50% chance that $latex \frac{3n+1}{4}$ is even, which means there’s a 25% chance that an odd number will be reduced to less than half of where it started after three steps. And so on. The net result is that, in some average way, Collatz orbits decrease when they encounter an odd number. And since Collatz orbits always decrease at even numbers, this suggests that all Collatz sequences must decrease in the long run. This probabilistic argument is widely known, but no one has been able to extend it to a complete proof of the conjecture.
Yet several mathematicians have proved that the Collatz conjecture is “almost always” true. This means they’ve proved that, relative to the amount of numbers they know lead to 1, the amount of numbers they aren’t sure about is negligible. In 1976 the Estonian American mathematician Riho Terras proved that, after repeated application of the Collatz function, almost all numbers eventually wind up lower than where they started. As we saw above, showing that the numbers in the orbit consistently get smaller is one path to showing that they eventually get to 1.
And in 2019, Terence Tao , one of the world’s greatest living mathematicians, improved on this result . Where Terras proved that for almost all numbers the Collatz sequence of n ends up below n , Tao proved that for almost all numbers the Collatz sequence of n ends up much lower: below $latex \frac{n}{2}$, below $latex \sqrt{n}$, below $latex \ln n$ (the natural log of n ), even below every f ( n ) where f ( x ) is any function that goes off to infinity, no matter how slowly. That is, for almost every number, we can guarantee that its Collatz sequence goes as low as we want. In a talk about the problem , Tao said this result is “about as close as one can get to the Collatz conjecture without actually solving it.”
Even so, the conjecture will continue to attract mathematicians and enthusiasts. So pick a number, any number, and give it a go. Just remember, you’ve been warned: Don’t get stuck in an endless loop.
1. Show that there are infinitely many numbers whose Collatz orbits pass through 1.
2. The “stopping time” of a number n is the smallest number of steps it takes for the Collatz orbit of n to reach 1. For example, the stopping time of 10 is 6, and the stopping time of 11 is 14. Find two numbers with stopping time 5.
3. In a recent talk on the Collatz conjecture, Terrance Tao mentioned the following Collatz-like function:
$latex h(n) = \begin{cases} n / 2 & \text{if $n$ is even } \\ 3n-1 & \text{if $n$ is odd } \end{cases}\ $
Tao points out that in addition to the 1 → 2 → 1 → 2 → 1… loop, two other loops appear. Can you find them?
Notice that every power of 2 has a simple orbit path to 1. For example,
2 4 → 2 3 → 2 2 → 2 → 1 → …
Since there are infinitely many powers of 2, there are infinitely many numbers whose Collatz orbits pass through 1.
Notice that 2 5 has stopping time 5, since 2 5 → 2 4 → 2 3 → 2 2 → 2 → 1 → …. And since 2 4 has stopping time 4, any number that is one step away from 2 4 has stopping time 5. For example, 5 → 16 → 8 → 4 → 2 → 1. Could there be others?
The other loops are
5 → 14 → 7 → 20 → 10 → 5 → …
17 → 50 → 25 → 74 → 37 → 110 → 55 → 164 → 82 → 41 → 122 → 61 → 182 → 91 → 272 → 136 → 68 → 34 → 17 → ….
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The Unsolved Math Problems Worth $1 Million Each
In A Nutshell
In 2000, the Clay Mathematics Institute offered a prize for anyone who could solve one of seven of math’s biggest unsolved problems. The prize was set (and remains) at $1 million per problem. To date, only one of the seven problems has been solved.
The Whole Bushel
Founded in 1998 by a Boston businessman and his wife, the Clay Mathematics Institute (CMI) is dedicated to increasing and disseminating mathematical knowledge. The CMI is a tax-exempt charitable organization that supports the work of leading math researchers in the field. In 2000, the institute presented what they called Millennium Prize Problems, which consisted of seven of math’s most difficult problems that were unsolved at the turn of the millennium. The goal of the prize problems was to show the general public that the field of mathematics is still an open one with many unsolved problems and to recognize math achievements that hold historic magnitude.
To date, only one of the seven problems has been solved. Russian mathematician Grigori Perelman published a series of papers in 2003 claiming to have solved the Poincaré conjecture. The Poincaré conjecture hypothesizes that a three-dimensional object that is connected without any holes is a sphere. In his articles, Perelman proved Thurston’s geometrization conjecture, an extension of which was Poincaré’s conjecture. After careful scrutiny of his proof, Perelman was awarded the $1 million prize, which he refused. He also did not show up to accept the Field’s Medal, the highest honor in the mathematics field, for his work in solving the problem.
The remaining six problems run the gamut of subfields in the mathematical world. The Riemann hypothesis involves a question about prime numbers raised by German mathematician Bernhard Riemann. The distribution of prime numbers does not appear to follow any logical pattern, but Riemann proposed a function that is closely related to the frequency of prime numbers. The hypothesis states that the “interesting” solutions to the function when the function equals zero lie on a specific vertical line. 10,000,000,000 solutions have been found to fit these parameters, but it is a proof that every interesting solution fits that will solve the problem. Conversely, a proof that finds a solution that does not lie on the line, and thereby disproves the Riemann hypothesis, also earns the solver $1 million.
The P vs. NP problem is largely concerned with the field of computer science. An NP problem is one whose answer is easy to check, and a P problem is one whose answer is easy to find. The question is whether or not there exists a problem that is easy for a computer to check but impossibly hard for a computer to solve.
The Yang-Mills theory is used to describe elementary particles and is an important factor in elementary particle theory. The Yang-Mills theory, which has been laboratory tested and largely confirmed to be true, depends on what is called the mass gap. The mass gap is the idea that the mass of the lightest quantum particle must be positive. Solving this problem would mean providing a theoretical proof to the Yang-Mills theory, where the lightest particle is positive.
Mathematicians in the 20th century proposed a novel way of observing complex objects. They approximated the shapes of complicated objects by combining simple geometric building blocks. This process was incredibly helpful in the field of mathematics, but it led to a generalized technique. To fulfill the task in some cases, building blocks with no geometric interpretation had to be added. These pieces, called Hodge cycles, were explained in the Hodge conjecture as being combinations of geometric pieces. Proving or disproving the Hodge conjecture would lead to the prize.
In the 19th century, the Navier-Stokes equation was recorded. Modern mathematicians believe that this equation explains and can predict the motion of water and air. The problem is that our understanding of the equation is very small. Answering this Millennium Prize Problem simply means finding the solutions to this equation through a proof.
The Birch and Swinnerton-Dyer conjecture asserts the relation between a group of rational points and the number of points on an elliptic curve. The elliptic curve is a keystone of mathematics that shows up in many areas of the field. It was used in Andrew Wiles’ proof of Fermat’s last theorem, which, before Wiles’ resolution, was considered mathematics’ biggest unsolved problem. A proof of the Birch and Sinnerton-Dyer conjecture could thus have huge implications on the math world. The same could be said for all of these problems, where proving or disproving one could change our entire perception of mathematics, at least to some degree.
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Mathematicians Discovered a Computer Problem that No One Can Ever Solve
Mathematicians have discovered a problem they cannot solve. It's not that they're not smart enough; there simply is no answer.
The problem has to do with machine learning — the type of artificial-intelligence models some computers use to "learn" how to do a specific task.
When Facebook or Google recognizes a photo of you and suggests that you tag yourself, it's using machine learning. When a self-driving car navigates a busy intersection, that's machine learning in action. Neuroscientists use machine learning to "read" someone’s thoughts . The thing about machine learning is that it's based on math . And as a result, mathematicians can study it and understand it on a theoretical level. They can write proofs about how machine learning works that are absolute and apply them in every case. [ Photos: Large Numbers That Define the Universe ]
In this case, a team of mathematicians designed a machine-learning problem called "estimating the maximum" or "EMX."
To understand how EMX works, imagine this: You want to place ads on a website and maximize how many viewers will be targeted by these ads. You have ads pitching to sports fans, cat lovers, car fanatics and exercise buffs, etc. But you don't know in advance who is going to visit the site. How do you pick a selection of ads that will maximize how many viewers you target? EMX has to figure out the answer with just a small amount of data on who visits the site.
The researchers then asked a question: When can EMX solve a problem?
In other machine-learning problems, mathematicians can usually say if the learning problem can be solved in a given case based on the data set they have. Can the underlying method Google uses to recognize your face be applied to predicting stock market trends? I don't know, but someone might.
The trouble is, math is sort of broken. It's been broken since 1931, when the logician Kurt Gödel published his famous incompleteness theorems. They showed that in any mathematical system, there are certain questions that cannot be answered. They're not really difficult — they're unknowable. Mathematicians learned that their ability to understand the universe was fundamentally limited. Gödel and another mathematician named Paul Cohen found an example: the continuum hypothesis.
The continuum hypothesis goes like this: Mathematicians already know that there are infinities of different sizes. For instance, there are infinitely many integers (numbers like 1, 2, 3, 4, 5 and so on); and there are infinitely many real numbers (which include numbers like 1, 2, 3 and so on, but they also include numbers like 1.8 and 5,222.7 and pi). But even though there are infinitely many integers and infinitely many real numbers, there are clearly more real numbers than there are integers. Which raises the question, are there any infinities larger than the set of integers but smaller than the set of real numbers? The continuum hypothesis says, no, there aren't.
Gödel and Cohen showed that it's impossible to prove that the continuum hypothesis is right, but also it's impossible to prove that it's wrong. "Is the continuum hypothesis true?" is a question without an answer.
In a paper published Monday, Jan. 7, in the journal Nature Machine Intelligence (opens in new tab) , the researchers showed that EMX is inextricably linked to the continuum hypothesis.
It turns out that EMX can solve a problem only if the continuum hypothesis is true. But if it's not true, EMX can't.. That means that the question, "Can EMX learn to solve this problem?" has an answer as unknowable as the continuum hypothesis itself.
The good news is that the solution to the continuum hypothesis isn't very important to most of mathematics. And, similarly, this permanent mystery might not create a major obstacle to machine learning.
"Because EMX is a new model in machine learning, we do not yet know its usefulness for developing real-world algorithms," Lev Reyzin, a professor of mathematics at the University of Illinois in Chicago, who did not work on the paper, wrote in an accompanying Nature (opens in new tab) News & V (opens in new tab) iews article (opens in new tab) . "So these results might not turn out to have practical importance," Reyzin wrote.
Running up against an unsolvable problem, Reyzin wrote, is a sort of feather in the cap of machine-learning researchers.
It's evidence that machine learning has "matured as a mathematical discipline," Reyzin wrote.
Machine learning "now joins the many subfields of mathematics that deal with the burden of unprovability and the unease that comes with it," Reyzin wrote. Perhaps results such as this one will bring to the field of machine learning a healthy dose of humility, even as machine-learning algorithms continue to revolutionize the world around us. "
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Editor's note: This story was updated on Jan. 14 at 2:15 p.m. EST to correct the definition of the continuum hypothesis. The article originally said that if the continuum hypothesis is true, then there are infinities larger than the set of integers but smaller than the set of real numbers. In fact, if the continuum hypothesis is true, then there are not infinities larger than the set of integers but smaller than the set of real numbers.
Originally published on Live Science .
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5 of the world’s toughest unsolved maths problems
The Open Problems in Mathematical Physics is a list of the most monstrous maths riddles in physics. Here are five of the top problems that remain unsolved
By Benjamin Skuse
7 February 2019
Mike Dunning/Getty
1. Separatrix Separation
A pendulum in motion can either swing from side to side or turn in a continuous circle. The point at which it goes from one type of motion to the other is called the separatrix, and this can be calculated in most simple situations. When the pendulum is prodded at an almost constant rate though, the mathematics falls apart. Is there an equation that can describe that kind of separatrix?
Science History Images / Alamy Stock Photo
2. Navier–Stokes
The Navier-Stokes equations, developed in 1822, are used to describe the motion of viscous fluid. Things like air passing over an aircraft wing or water flowing out of a tap. But there are certain situations in which it is unclear whether the equations fail or give no answer at all. Many mathematicians have tried – and failed – to resolve the matter, including Mukhtarbay Otelbaev of the Eurasian National University in Astana, Kazakhstan. In 2014, he claimed a solution, but later retracted it. This is one problem that is worth more than just prestige. It is also one of the Millennium Prize Problems , which means anyone who solves it can claim $1 million in prize money.
Read more: The baffling quantum maths solution it took 10 years to understand
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3. Exponents and dimensions
Imagine a squirt of perfume diffusing across a room. The movement of each molecule is random, a process called Brownian motion, even if the way the gas wafts overall is predictable. There is a mathematical language that can describe things like this, but not perfectly. It can provide exact solutions by bending its own rules or it can remain strict, but never quite arrive at the exact solution. Could it ever tick both boxes? That is what the exponents and dimensions problem asks. Apart from the quantum Hall conductance problem , this is the only one on the list that is at least partially solved. In 2000, Gregory Lawler, Oded Schramm and Wendelin Werner proved that exact solutions to two problems in Brownian motion can be found without bending the rules. It earned them a Fields medal, the maths equivalent of a Nobel prize. More recently, Stanislav Smirnov at the University of Geneva in Switzerland solved a related problem, which resulted in him being awarded the Fields medal in 2010.
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4. Impossibility theorems
There are plenty of mathematical expressions that have no exact solution. Take one of the most famous numbers ever, pi, which is the ratio of a circle’s circumference to its diameter. Proving that it was impossible for pi’s digits after the decimal point to ever end was one of the greatest contributions to maths. Physicists similarly say that it is impossible to find solutions to certain problems, like finding the exact energies of electrons orbiting a helium atom. But can we prove that impossibility?
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5. Spin glass
To understand this problem, you need to know about spin, a quantum mechanical property of atoms and particles like electrons, which underlies magnetism. You can think of it like an arrow that can point up or down. Electrons inside blocks of materials are happiest if they sit next to electrons that have the opposite spin, but there are some arrangements where that isn’t possible. In these frustrated magnets, spins often flip around randomly in a way that, it turns out, is a useful model of other disordered systems including financial markets. But we have limited ways of mathematically describing how systems like this behave. This spin glass question asks if we can find a good way of doing it.
• See the full list of unsolved problems: Open Problems in Mathematical Physics
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AI Is Discovering Patterns in Pure Mathematics That Have Never Been Seen Before
We can add suggesting and proving mathematical theorems to the long list of what artificial intelligence is capable of : Mathematicians and AI experts have teamed up to demonstrate how machine learning can open up new avenues to explore in the field.
While mathematicians have been using computers to discover patterns for decades , the increasing power of machine learning means that these networks can work through huge swathes of data and identify patterns that haven't been spotted before.
In a newly published study, a research team used artificial intelligence systems developed by DeepMind , the same company that has been deploying AI to solve tricky biology problems and improve the accuracy of weather forecasts , to unknot some long-standing math problems.
"Problems in mathematics are widely regarded as some of the most intellectually challenging problems out there," says mathematician Geordie Williamson from the University of Sydney in Australia.
"While mathematicians have used machine learning to assist in the analysis of complex data sets, this is the first time we have used computers to help us formulate conjectures or suggest possible lines of attack for unproven ideas in mathematics."
The team shows AI advancing a proof for Kazhdan-Lusztig polynomials , a math problem involving the symmetry of higher-dimensional algebra that has remained unsolved for 40 years.
The research also demonstrated how a machine learning technique called a supervised learning model was able to spot a previously undiscovered relationship between two different types of mathematical knots , leading to an entirely new theorem.
Knot theory in math plays into various other challenging fields of science as well, including genetics, fluid dynamics, and even the behavior of the Sun's corona. The discoveries that AI makes can therefore lead to advances in other areas of research.
"We have demonstrated that, when guided by mathematical intuition, machine learning provides a powerful framework that can uncover interesting and provable conjectures in areas where a large amount of data is available, or where the objects are too large to study with classical methods," says mathematician András Juhász from the University of Oxford in the UK.
One of the benefits of machine learning systems is the way that they can look for patterns and scenarios that programmers didn't specifically code them to look out for – they take their training data and apply the same principles to new situations.
The research shows that this sort of high-speed, ultra-reliable, large-scale data processing can act as an extra tool working with mathematicians' natural intuition. When you're dealing with complex, lengthy equations, that can make a significant difference.
The researchers hope that their work leads to many further partnerships between academics in the fields of mathematics and artificial intelligence, opening up the opportunity for findings that would otherwise be undiscovered.
"AI is an extraordinary tool," says Williamson . "This work is one of the first times it has demonstrated its usefulness for pure mathematicians, like me."
"Intuition can take us a long way, but AI can help us find connections the human mind might not always easily spot."
The research has been published in Nature .
Solved Problems
There are many unsolved problems in mathematics. Several famous problems which have recently been solved include:
1. The Pólya conjecture (disproven by Haselgrove 1958, smallest counterexample found by Tanaka 1980).
2. The four-color theorem (Appel and Haken 1977ab and Appel et al. 1977 using a computer-assisted proof).
3. The Bieberbach conjecture (L. de Branges 1985).
4. Tait's flyping conjecture (Menasco and Thistlethwaite in 1991) and the other two of Tait's knot conjectures (by various authors 1987).
5. Fermat's last theorem (Wiles 1995, Taylor and Wiles 1995).
6. The Kepler conjecture (Hales 2002).
7. The Taniyama-Shimura conjecture (Breuil et al. in 1999).
8. The honeycomb conjecture (Hales 1999).
9. The Poincaré conjecture .
10. Catalan's conjecture .
11. The strong perfect graph theorem .
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Famous Unsolved Math Problems as Homework
By Benjamin Braun, Editor-in-Chief , University of Kentucky
One of my favorite assignments for students in undergraduate mathematics courses is to have them work on unsolved math problems. An unsolved math problem, also known to mathematicians as an “open” problem, is a problem that no one on earth knows how to solve. My favorite unsolved problems for students are simply stated ones that can be easily understood. In this post, I’ll share three such problems that I have used in my classes and discuss their impact on my students.
Unsolved Problems
The Collatz Conjecture . Given a positive integer \(n\), if it is odd then calculate \(3n+1\). If it is even, calculate \(n/2\). Repeat this process with the resulting value. For example, if you begin with \(1\), then you obtain the sequence \[ 1,4,2,1,4,2,1,4,2,1,\ldots \] which will repeat forever in this way. If you start with a \(5\), then you obtain the sequence \(5,16,8,4,2,1,\ldots\), and now find yourself in the previous case. The unsolved question about this process is: If you start from any positive integer, does this process always end by cycling through \(1,4,2,1,4,2,1,\ldots\)? Mathematicians believe that the answer is yes, though no one knows how to prove it. This conjecture is known as the Collatz Conjecture (among many other names), since it was first asked in 1937 by Lothar Collatz.
The Erd ő s-Strauss Conjecture . A fascinating question about unit fractions is the following: For every positive integer \(n\) greater than or equal to \(2\), can you write \(\frac{4}{n}\) as a sum of three positive unit fractions? For example, for \(n=3\), we can write \[\frac{4}{3}=\frac{1}{1}+\frac{1}{6}+\frac{1}{6} \, . \] For \(n=5\), we can write \[ \frac{4}{5}=\frac{1}{2}+\frac{1}{4}+\frac{1}{20} \] or \[\frac{4}{5}=\frac{1}{2}+\frac{1}{5}+\frac{1}{10} \, . \] In other words, if \(n\geq 2\) can you always solve the equation \[ \frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\] using positive integers \(a\), \(b\), and \(c\)? Again, most mathematicians believe that the answer to this question is yes, but a proof remains elusive. This question was first asked by Paul Erdős and Ernst Strauss in 1948, hence its name, and mathematicians have been working hard on it ever since.
Lagarias’s Elementary Version of the Riemann Hypothesis . For a positive integer \(n\), let \(\sigma(n)\) denote the sum of the positive integers that divide \(n\). For example, \(\sigma(4)=1+2+4=7\), and \(\sigma(6)=1+2+3+6=12\). Let \(H_n\) denote the \(n\)-th harmonic number, i.e. \[ H_n=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n} \, .\] Our third unsolved problem is: Does the following inequality hold for all \(n\geq 1\)? \[ \sigma(n)\leq H_n+\ln(H_n)e^{H_n} \] In 2002, Jeffrey Lagarias proved that this problem is equivalent to the Riemann Hypothesis , a famous question about the complex roots of the Riemann zeta function. Because it is equivalent to the Riemann Hypothesis, if you successfully answer it, then the Clay Mathematics Foundation will reward you with $1,000,000 . While the statement of this problem is more complicated than the previous two, it doesn’t involve anything beyond natural logs and exponentials at a precalculus level.
Impact on Students
I’ve used all three of these problems, along with various others , as the focus of in-class group work and as homework problems in undergraduate mathematics courses such as College Geometry, Problem Solving for Teachers, and History of Mathematics. An example of a homework assignment I give based on the Riemann Hypothesis problem can be found at this link . When I use these problems for in-class work, I will typically pose the problem to the students without telling them it is unsolved, and then reveal the full truth after they have been working for fifteen minutes or so. By doing this, the students get to experience the shift in perspective that comes when what appears to be a simple problem in arithmetic suddenly becomes a near-impossibility.
Without fail, my undergraduate students, most of whom are majors in math, math education, engineering, or one of the natural sciences, are surprised that they can understand the statement of an unsolved math problem. Most of them are also shocked that problems as seemingly simple as the Collatz Conjecture or the Erdős-Strauss Conjecture are unsolved — the ideas involved in the statements of these problems are at an elementary-school level!
I have found that having students work on unsolved problems gets them engaged in three ways that are otherwise very difficult to obtain.
- Students are forced to depart from the “answer-getting” mentality of mathematics. In my experience, (most) students in K-12 and postsecondary mathematics courses believe that all math problems have known answers, and that teachers can find the answer to every problem. As long as students believe this story, it is hard to motivate them to develop quality mathematical practices, as opposed to doing the minimum necessary to get the “right answer” sufficiently often. However, if they are asked to work on an unsolved problem, knowing that it is unsolved, then students are forced to find other ways to define success in their mathematical work. While getting buy-in on this idea is occasionally an issue, most of the time the students are immediately interested in the idea of an unsolved problem, especially a simply-stated one. The discussion of how to define success in mathematical investigation usually prompts quality discussions in class about the authentic nature of mathematical work; students often haven’t reflected on the fact that professional mathematicians and scientists spend most of their time thinking about how to solve problems that no one knows how to solve.
- Students are forced to redefine success in learning as making sense and increasing depth of understanding . The first of the mathematical practice standards in the Common Core , which have been discussed in previous blog posts by the author and by Elise Lockwood and Eric Weber , is that students should make sense of problems and persevere in solving them. When faced with an unsolved problem, sense-making and perseverance must take center stage. In courses heavily populated by preservice teachers, I’ve used open problems as in-class group work in which students work on a problem and monitor which of the practice standards they are using. Since neither the students nor I expect that they will solve the problem at hand, they are able to really relax and focus on the process of mathematical investigation, without feeling pressure to complete the problem. One could even go so far as to evaluate student work on unsolved problems using the common core practice standards, though typically I evaluate such work based on maturity of investigation and clarity of exposition.
- Students are able to work in a context in which failure is completely normal. In my experience, undergraduates majoring in the mathematical sciences typically carry a large amount of guilt and self-doubt regarding their perceived mathematical failures, whether or not it is justified. From data collected by the recent MAA Calculus Study , it appears that this is particularly harmful for women studying mathematics . Because working on unsolved problems forces success to be redefined, it also provides an opportunity to discuss the definition of failure, and the pervasive normality of small mistakes in the day-to-day lives of mathematicians and scientists. I usually combine work on unsolved problems with reading assignments and classroom discussions regarding developments in educational and social psychology, such as Carol Dweck’s work on mindset , to help students develop a more reasonable set of expectations for their mathematical process.
One of the most interesting aspects of using unsolved problems in my classes has been to see how my students respond. I typically ask students to write a three-page reflective essay about their experience with the homework in the course, and almost all of the students talk about working on the open problems. Some of them describe feelings of relief and joy to have the opportunity to be as creative as they wish on a problem with no expectation of finding the right answer, while others describe feelings of frustration and immediate defeat in the face of a hopeless task. Either way, many students tell me that working on an unsolved problem is one of the noteworthy moments in the course. For this reason, as much as I enjoy witnessing mathematics develop and progress, I hope that some of my favorite problems remain tantalizingly unsolved for many years to come.
14 Responses to Famous Unsolved Math Problems as Homework
Thanks for the great ideas Ben!
I currently have some students working on Sylver Coinage as part of a final project in a Number Theory class. They have come up with a few strategies on their own and are intrigued by the fact that more strategies haven’t been discovered. It’s even more interesting that there are nonconstructive proofs for strategies. Because these students know the problem is unsolved, they are more apt to be proud of small steps forward.
I’d love to hear more ideas of what is done in other classes.
I hadn’t heard of that problem before, it sounds like an awesome project!
Came here from https://news.ycombinator.com/item?id=9537802.
In the Erdos-Strauss conjecture why is it 4/n. Clearly one is limited by the highest number achievable with the fractional terms on the right, but why not the general question x/n = Sum 1/a[i] for i from 1 to (x -1)?
[attempted to render that in mathml:] xn=∑i=2i=x-11ai
Has a form of this been solved generally for other x values ( > 3) and 4/n is an outlier??
In my mind geometric solutions for the first two are so obvious (!) however in my current state of mind I’m far more inclined to the pragmatic “well you can show if it’s true for any particular n you care about, close enough” – think I’m losing my youthful inquisitiveness!
Some work has been done on the more general case, see Terry Tao’s comments on his blog https://terrytao.wordpress.com/tag/erdos-straus-conjecture/ and the references therein.
This is fantastic and inspiring! Thank you.
Hey, Beth, thanks for mentioning the Coinage Problem. I have been teaching Discrete Math (including some number theory) at the University of Alabama for years and one of the problems in our book has to do with making change. Once I read your note, I googled (or do I mean capitalized Googled?) the coinage problem and, lo and behold, it has a long history! Sylvester! Frobenius! Wow…I am going to try this question in my class this semester to see if the kids can guess and then prove Sylvester’s (a-1)(b-1)-1 theorem! I don’t think I’ll ask them to prove the Lagarias version of the Riemann Conjecture, though…
I wish that 30 years ago I had studied under you. At first read, I thought this was cruel and unusual punishment, then I realized your method.
Wonderful… you are someone who should keep teaching forever. Please keep doing what you’re doing.
If I didn’t live so far from Kentucky I would be begging to audit one of your classes.
Hi Ben, I am not a mathematician but I have a question about the The Collatz Conjecture. If we take any positive number and lets say it is an even number then, we will divide it by 2. and obviously we will get the Even number and then again we will divide it by 2 and again we will get the even number and this process will continue until we come to 4 and then we will divide the 4 by 2 and we will get 2 and then divide 2 by 2 and we will get 1 so it will always end with 4.2.1. Incase of odd number we will do the 3n+1 that means we will get the Even number and then again the Even number process that means we will also end up with 4.2.1. So what is need to be proven here. It is obvious and can be proved with the simple number theory.
Your suggestion only works for powers of 2. If you have a number like 10, which is even but has odd factors, then it isn’t clear what will happen.
Wonderful post! I will be sharing this in my on-going conversation about authentic curriculum in elementary mathematics. I use open problems frequently in my 4th and 5th grade maths classes, especially during, but not limited to, our units on number theory. The students love the work. It is fun, engaging, motivating, empowering and capitalizes on their natural curiosity. Many students stick with the problems after we’ve moved on because they hope one day someone will find the proof and move them beyond conjecture. Like Julie Robinson and the diophantines, it didn’t have to be her, she just hoped for a proof in her lifetime so she could celebrate it. Furthermore, these problems do a wonderful job of showing students that mathematics is not dead, but full of discovery and humanity. They are motivated to learn more about the players and their stories.
Hi Ben, I don’t want to waste you time but I really find it fascinating. So I Just wanted to understand it more. If I take 10, its a even number and I divide it by 2, I will get 5 and then (3x+1) so I will get 16 and then divide it by 2, will get 8 and then 4,2,1. So whats the problem in it. Just to understand this better.
Thank you so much for such interesting post.
This page: https://en.wikipedia.org/wiki/Collatz_conjecture has lots of good information about the details of the conjecture. At the end of the page, there is a nice list of references where you can find more detailed discussions about why the problem is difficult.
I think this is a brilliant idea. It not only gets fresh insight on mathematical problems whose solutions have still not been figured out. It also forces students to think outside of the box and try things they wouldn’t normally do. One of the biggest issues I have seen in the mathematics courses I have taken at my university is that failure and mistakes are seen as bad and terrible. When in fact they are the opposite. They promote learning and greater understanding that just simply getting the right answer. They promote learning the process and method behind the problem instead of just regurgitating a solution. It also forces students to come at problems from different angles and figure out new methods for solving problems than just the tried and true methods. I believe your method is absolutely necessary for riding the mindset that there is one set way of solving a problem or that failure is not an option. I believe all professors should adopt this mindset for teaching mathematics at any level. This is because it makes it easier for students to learn and allows them to not be discouraged when they don’t understand something right away or after the first few times. It promotes persistence and the pursuit of understanding which is critical for any student who is studying mathematics to understand and acquire those qualities. If more students thought like this then in my opinion they would be more likely to succeed and flourish in any field they choose not just mathematics. Mathematics at it’s core is the understanding of the universe around us and the laws that govern it. It is reason and logic and without persistence and understanding we will never fully understand it and that is a crime in and of itself.
I think, every unsolved math problems can be solved. Thanks.
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How to Work Through Hard Math Problems
parent of one of our students wrote today about his daughter’s occasional frustration with the difficulty of some of the problems in our courses. She does fantastic work in our courses , and was easily among the very top students in the class she took with me, and yet she still occasionally hits problems that she can’t solve.
Moreover, she has access to an excellent math teacher in her school who sometimes can’t help her get past these problems, either. (This is no slight to him—I have students bring me problems I can’t solve, too!) Her question: “Why does it have to be so hard?”
The Case for Doing Hard Things
We ask hard questions because so many of the problems worth solving in life are hard. If they were easy, someone else would have solved them before you got to them. This is why college classes at top-tier universities have tests on which nearly no one clears 70%, much less gets a perfect score. They’re training future researchers, and the whole point of research is to find and answer questions that have never been solved. You can’t learn how to do that without fighting with problems you can’t solve. If you are consistently getting every problem in a class correct, you shouldn’t be too happy — it means you aren’t learning efficiently enough. You need to find a harder class.
The problem with not being challenged sufficiently goes well beyond not learning math (or whatever) as quickly as you can. I think a lot of what we do at AoPS is preparing students for challenges well outside mathematics. The same sort of strategies that go into solving very difficult math problems can be used to tackle a great many problems. I believe we’re teaching students how to think, how to approach difficult problems, and that math happens to be the best way to do so for many people.
The first step in dealing with difficult problems is to accept and understand their importance. Don’t duck them. They will teach you a lot more than a worksheet full of easy problems. Brilliant “Aha!” moments almost always spring from minds cultivated by long periods of frustration. But without that frustration, those brilliant ideas never arise.
Strategies for Difficult Math Problems — and Beyond
Here are a few strategies for dealing with hard problems, and the frustration that comes with them:
Do something . Yeah, the problem is hard. Yeah, you have no idea what to do to solve it. At some point you have to stop staring and start trying stuff. Most of it won’t work. Accept that a lot of your effort will appear to have been wasted. But there’s a chance that one of your stabs will hit something, and even if it doesn’t, the effort may prepare your mind for the winning idea when the time comes.
We started developing an elementary school curriculum months and months before we had the idea that became Beast Academy . Our lead curriculum developer wrote 100–200 pages of content, dreaming up lots of different styles and approaches we might use. Not a one of those pages will be in the final work, but they spurred a great many ideas for content we will use. Perhaps more importantly, it prepared us so that when we finally hit upon the Beast Academy idea, we were confident enough to pursue it.
Simplify the problem . Try smaller numbers and special cases. Remove restrictions. Or add restrictions. Set your sights a little lower, then raise them once you tackle the simpler problem.
Reflect on successes . You’ve solved lots of problems. Some of them were even hard problems! How did you do it? Start with problems that are similar to the one you face, but also think about others that have nothing to do with your current problem. Think about the strategies you used to solve those problems, and you might just stumble on the solution.
A few months ago, I was playing around with some Project Euler problems, and I came upon a problem that (eventually) boiled down to generating integer solutions to c ² = a ² + b ² + ab in an efficient manner. Number theory is not my strength, but my path to the solution was to recall first the method for generating Pythagorean triples. Then, I thought about how to generate that method, and the path to the solution became clear. (I’m guessing some of our more mathematically advanced readers have so internalized the solution process for this type of Diophantine equation that you don’t have to travel with Pythagoras to get there!)
Focus on what you haven’t used yet . Many problems (particularly geometry problems) have a lot of moving parts. Look back at the problem, and the discoveries you have made so far and ask yourself: “What haven’t I used yet in any constructive way?” The answer to that question is often the key to your next step.
Work backwards . This is particularly useful when trying to discover proofs. Instead of starting from what you know and working towards what you want, start from what you want, and ask yourself what you need to get there.
Ask for help . This is hard for many outstanding students. You’re so used to getting everything right, to being the one everyone else asks, that it’s hard to admit you need help. When I first got to the Mathematical Olympiad Program (MOP) my sophomore year, I was in way over my head. I understood very little of anything that happened in class. I asked for help from the professor once — it was very hard to get up the courage to do so. I didn’t understand anything he told me during the 15 minutes he worked privately with me. I just couldn’t admit it and ask for more help, so I stopped asking. I could have learned much, much more had I just been more willing to admit to people that I just didn’t understand. (This is part of why our classes now have a feature that allows students to ask questions anonymously.) Get over it. You will get stuck. You will need help. And if you ask for it, you’ll get much farther than if you don’t.
Start early . This doesn’t help much with timed tests, but with the longer-range assignments that are parts of college and of life, it’s essential. Don’t wait until the last minute — hard problems are hard enough without having to deal with time pressure. Moreover, complex ideas take a long time to understand fully. The people you know who seem wicked smart, and who seem to come up with ideas much faster than you possibly could, are often people who have simply thought about the issues for much longer than you have. I used this strategy throughout college to great success — in the first few weeks of each semester, I worked far ahead in all of my classes. Therefore, by the end of the semester, I had been thinking about the key ideas for a lot longer than most of my classmates, making the exams and such at the end of the course a lot easier.
Take a break . Get away from the problem for a bit. When you come back to it, you may find that you haven’t entirely gotten away from the problem at all — the background processes of your brain have continued plugging away, and you’ll find yourself a lot closer to the solution. Of course, it’s a lot easier to take a break if you start early.
Start over . Put all your earlier work aside, get a fresh sheet of paper, and try to start from scratch. Your other work will still be there if you want to draw from it later, and it may have prepared you to take advantage of insights you make in your second go-round.
Give up . You won’t solve them all. At some point, it’s time to cut your losses and move on. This is especially true when you’re in training, and trying to learn new things. A single difficult problem is usually going to teach you more in the first hour or two than it will in the next six, and there are a lot more problems to learn from. So, set yourself a time limit, and if you’re still hopelessly stuck at the end of it, then read the solutions and move on.
Be introspective . If you do give up and read the solution, then read it actively, not passively. As you read it, think about what clues in the problem could have led you to this solution. Think about what you did wrong in your investigation. If there are math facts in the solution that you don’t understand, then go investigate. I was completely befuddled the first time I saw a bunch of stuff about “mod”s in an olympiad solution — we didn’t have the internet then, so I couldn’t easily find out how straightforward modular arithmetic is! You have the internet now, so you have no excuse. If you did solve the problem, don’t just pat yourself on the back. Think about the key steps you made, and what the signs were to try them. Think about the blind alleys you explored en route to the solution, and how you could have avoided them. Those lessons will serve you well later.
Come back . If you gave up and looked at the solutions, then come back and try the problem again a few weeks later. If you don’t have any solutions to look at, keep the problem alive. Store it away on paper or in your mind.
Richard Feynman once wrote that he would keep four or five problems active in the back of his mind. Whenever he heard a new strategy or technique, he would quickly run through his problems and see if he could use it to solve one of his problems. He credits this practice for some of the anecdotes that gave Feynman such a reputation for being a genius. It’s further evidence that being a genius is an awful lot about effort, preparation, and being comfortable with challenges.
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These Are the 7 Hardest Math Problems Ever Solved — Good Luck in Advance
In 2019, mathematicians finally solved a math puzzle that had stumped them for decades. It’s called a Diophantine Equation, and it’s sometimes known as the “summing of three cubes”: Find x, y, and z such that x³+y³+z³=k, for each k from one to 100.
On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?
That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)
That’s the beauty of math: There’s always an answer for everything, even if takes years, decades, or even centuries to find it. So here are seven more brutally difficult math problems that once seemed impossible until mathematicians found a breakthrough.
These ten brutally difficult math problems once seemed impossible until mathematicians eventually solved them—even if it took them years, decades, or centuries.
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Toggle Lists of unsolved problems in mathematics subsection 1.1Millennium Prize Problems 2Unsolved problems Toggle Unsolved problems subsection 2.1Algebra 2.1.1Representation theory 2.1.2Notebook problems 2.2Analysis 2.3Combinatorics 2.4Dynamical systems 2.5Games and puzzles 2.5.1Combinatorial games 2.5.2Games with imperfect information
These Are the 10 Toughest Math Problems Ever Solved 1 The Collatz Conjecture In September 2019, news broke regarding progress on this 82-year-old question, thanks to prolific mathematician...
There are many unsolved problems in mathematics. Some prominent outstanding unsolved problems (as well as some which are not necessarily so well known) include 1. The Goldbach conjecture. 2. The Riemann hypothesis. 3. The conjecture that there exists a Hadamard matrix for every positive multiple of 4. 4.
The Top Unsolved Questions in Mathematics Remain Mostly Mysterious Just one of the seven Millennium Prize Problems named 21 years ago has been solved By Rachel Crowell on May 28, 2021 Credit:...
The Collatz conjecture is one of the most famous unsolved mathematical problems, because it's so simple, you can explain it to a primary-school-aged kid, and they'll probably be intrigued enough to try and find the answer for themselves. So here's how it goes: pick a number, any number. If it's even, divide it by 2.
This problem, as relatively simple as it sounds has never been solved. Solving this problem will earn you a free million dollars. This equation was first proposed by Goldbach hence the name Goldbach's Conjecture. If you are still unsure then pick any even number like 6, it can also be expressed as 1 + 5, which is two primes.
Here are five current problems in the field of mathematics that anyone can understand, but nobody has been able to solve. Collatz Conjecture Jon McLoone Pick any number. If that number is...
The Simple Math Problem We Still Can't Solve Despite recent progress on the notorious Collatz conjecture, we still don't know whether a number can escape its infinite loop. BIG MOUTH for Quanta Magazine This column comes with a warning: Do not try to solve this math problem. You will be tempted.
The Unsolved Math Problems Worth $1 Million Each 6 Min Read "Mathematics is not a careful march down a well-cleared highway, but a journey into a strange wilderness, where the explorers often get lost." — W.S. Anglin, Mathematics and History In A Nutshell
math skills with some problems that go beyond the usual curriculum. These notes can be used as complimentary to an advanced calculus or algebra course, as training for math competitions or simply as a collection of challenging math problems. Many of these are my own creation, some from when I was a student and some from more recent times.
Mathematicians have discovered a problem they cannot solve. It's not that they're not smart enough; there simply is no answer. The problem has to do with machine learning — the type of ...
Get step-by-step solutions to your math problems Try Math Solver Type a math problem Solve Quadratic equation x2 − 4x − 5 = 0 Trigonometry 4sinθ cosθ = 2sinθ Linear equation y = 3x + 4 Arithmetic 699 ∗533 Matrix [ 2 5 3 4][ 2 −1 0 1 3 5] Simultaneous equation { 8x + 2y = 46 7x + 3y = 47 Differentiation dxd (x −5)(3x2 −2) Integration ∫ 01 xe−x2dx
Here are five of the top problems that remain unsolved. 1. Separatrix Separation. A pendulum in motion can either swing from side to side or turn in a continuous circle. The point at which it goes ...
The team shows AI advancing a proof for Kazhdan-Lusztig polynomials, a math problem involving the symmetry of higher-dimensional algebra that has remained unsolved for 40 years.. The research also demonstrated how a machine learning technique called a supervised learning model was able to spot a previously undiscovered relationship between two different types of mathematical knots, leading to ...
There are many unsolved problems in mathematics. Several famous problems which have recently been solved include: 1. The Pólya conjecture (disproven by Haselgrove 1958, smallest counterexample found by Tanaka 1980). 2. The four-color theorem (Appel and Haken 1977ab and Appel et al. 1977 using a computer-assisted proof). 3. The Bieberbach conjecture (L. de Branges 1985). 4. Tait's flyping ...
An unsolved math problem, also known to mathematicians as an "open" problem, is a problem that no one on earth knows how to solve. My favorite unsolved problems for students are simply stated ones that can be easily understood. In this post, I'll share three such problems that I have used in my classes and discuss their impact on my students.
Problem solving has a special importance in the study of mathematics. A primary goal of mathematics teaching and learning is to develop the ability to solve a wide variety of complex mathematics problems. Stanic and Kilpatrick (43) traced the role of problem solving in school mathematics and illustrated a rich history of the topic. To many
The Collatz Conjecture is the simplest math problem no one can solve — it is easy enough for almost anyone to understand but notoriously difficult to solve. ...
The Collatz Conjecture is the simplest math problem no one can solve — it is easy enough for almost anyone to understand but notoriously difficult to solve. So what is the Collatz Conjecture and what makes it so difficult? Veritasium investigates. Watch; Think Open review body. 5 Multiple Choice & 5 Open Answer Questions. Dig Deeper ...
Put all your earlier work aside, get a fresh sheet of paper, and try to start from scratch. Your other work will still be there if you want to draw from it later, and it may have prepared you to take advantage of insights you make in your second go-round. Give up. You won't solve them all.
These Are the 7 Hardest Math Problems Ever Solved — Good Luck in Advance. In 2019, mathematicians finally solved a math puzzle that had stumped them for decades. It's called a Diophantine Equation, and it's sometimes known as the "summing of three cubes": Find x, y, and z such that x³+y³+z³=k, for each k from one to 100.
Problems for 5th Grade. Multi-digit multiplication. Dividing completely. Writing expressions. Rounding whole numbers. Inequalities on a number line. Linear equation and inequality word problems. Linear equation word problems. Linear equation word problems.
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