Is there any unsolvable math questions?

Question

I was just wondering,Im so rubbish a Maths lol

EVERY DAYS A SCHOOL DAY! x

Answer 1

Hi Stacey.
In theory, everything should be soluble. i think the biggest problem currently unresolved is, an equation to turn a square of a certain area into a circle of a certain area (and vice versa). Some smarty pants may crack it one day, though.
Cheers, Steve.

Answer 2

yeah if you don't have enough info, you can't solve things.



oh and 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2

= 806

Answer 3

2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2

Answer 4

Yes and no. There have been questions, or rather theorems and postulates that have been posed that appear to be true but have not been proved. As the field of Mathematics advances some problems are solved.

For example, in the mid 1600's Fermat pose a theorem, called "Fermat's Last Theorem" which was unproven until the 1990's. The type of mathematics needed to prove the theorem were not developed or discovered until the mid 1900's.

http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html

The Riemann hypothesis is still unproven, but will, more than likely one day be proven.
http://en.wikipedia.org/wiki/Riemann_hypothesis

Now there are some problems that are impossible. For example, the integral of e ^ (x^2) :

∫ exp( x^2) dx

has no closed form solution. there is no function whose derivative is exp(x^2). So if you were asked to find the closed form solution to the integral, then yes this is an unsolvable math problem.

You can always construct unsolvable probelms as well.

Answer 5

There exist within maths problems which are undecidable. This was actually proved by Godel. There also exist problems as yet unsolved. Kepler's theorem for the closest packing of regular spheres as face-centered cubic is yet unproved.
Golbach's conjecture that every even number is the sum of two primes is unproved.
The Reimann hypothesis that the R/z function has non-trivial poles with a real part=1/2. ($1M prize!)
(Erdos showed that in any partion n>2n there must exist a prime, which is a consequence of the Goldbach conjecture, so I would guess that this is the track to prove GP and perhaps even Rz)
The problem with Godels proof is that it does not specify all problems which are not provable.So 1,2 and 3 could be by definition intractable.

Answer 6

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 twin prime conjecture (i.e., the conjecture that there are an infinite number of twin primes).

5. Determination of whether NP-problems are actually P-problems.

6. The Collatz problem.

7. Proof that the 196-algorithm does not terminate when applied to the number 196.

8. Proof that 10 is a solitary number.

9. Finding a formula for the probability that two elements chosen at random generate the symmetric group S_n.

10. Solving the happy end problem for arbitrary n.

11. Finding an Euler brick whose space diagonal is also an integer.

12. Proving which numbers can be represented as a sum of three or four (positive or negative) cubic numbers.

13. Lehmer's Mahler measure problem and Lehmer's totient problem on the existence of composite numbers n such that phi(n)|(n-1), where phi(n) is the totient function.

14. Determining if the Euler-Mascheroni constant is irrational.

15. Deriving an analytic form for the square site percolation threshold.

16. Determining if any odd perfect numbers exist.

The Clay Mathematics Institute (http://www.claymath.org/millennium/) of Cambridge, Massachusetts (CMI) has named seven "Millennium Prize Problems," selected by focusing on important classic questions in mathematics that have resisted solution over the years. A $7 million prize fund has been established for the solution to these problems, with $1 million allocated to each. The problems consist of the Riemann hypothesis, Poincaré conjecture, Hodge conjecture, Swinnerton-Dyer Conjecture, solution of the Navier-Stokes equations, formulation of Yang-Mills theory, and determination of whether NP-problems are actually P-problems.

In 1900, David Hilbert proposed a list of 23 outstanding problems in mathematics (Hilbert's problems, a number of which have now been solved, but some of which remain open. In 1912, Landau proposed four simply stated problems, now known as Landau's problems, which continue to defy attack even today. One hundred years after Hilbert, Smale (2000) proposed a list of 18 outstanding problems.

K. S. Brown, D. Eppstein, S. Finch, and C. Kimberling maintain webpages of unsolved problems in mathematics. Classic texts on unsolved problems in various areas of mathematics are Croft et al. (1991), in geometry, and Guy (1994), in number theory.

SEE ALSO: Beal's Conjecture, Catalan's Conjecture, Fermat's Last Theorem, Hilbert's Problems, Kepler Conjecture, Landau's Problems, Mathematics Contests, Mathematics Prizes, Poincaré Conjecture, Problem, Solved Problems, Szemerédi's Theorem, Twin Primes.

Answer 7

This is taken from a longer article about Gregory Chaitin's work.

Number theory is the foundation of pure mathematics. It describes how to deal with concepts such as counting, adding, and multiplying. Chaitin's search for Omega in number theory started with "Diophantine equations"--which involve only the simple concepts of addition, multiplication and exponentiation (raising one number to the power of another) of whole numbers.

Chaitin formulated a Diophantine equation that was 200 pages long and had 17,000 variables. Given an equation like this, mathematicians would normally search for its solutions. There could be any number of answers: perhaps 10, 20, or even an infinite number of them. But Chaitin didn't look for specific solutions, he simply looked to see whether there was a finite or an infinite number of them.

He did this because he knew it was the key to unearthing Omega. Mathematicians James Jones of the University of Calgary and Yuri Matijasevic of the Steklov Institute of Mathematics in St Petersburg had shown how to translate the operation of Turing's computer into a Diophantine equation. They found that there is a relationship between the solutions to the equation and the halting problem for the machine's program. Specifically, if a particular program doesn't ever halt, a particular Diophantine equation will have no solution. In effect, the equations provide a bridge linking Turing's halting problem--and thus Chaitin's halting probability--with simple mathematical operations, such as the addition and multiplication of whole numbers.

Chaitin had arranged his equation so that there was one particular variable, a parameter which he called N, that provided the key to finding Omega. When he substituted numbers for N, analysis of the equation would provide the digits of Omega in binary. When he put 1 in place of N, he would ask whether there was a finite or infinite number of whole number solutions to the resulting equation. The answer gives the first digit of Omega: a finite number of solutions would make this digit 0, an infinite number of solutions would make it 1. Substituting 2 for N and asking the same question about the equation's solutions would give the second digit of Omega. Chaitin could, in theory, continue forever. "My equation is constructed so that asking whether it has finitely or infinitely many solutions as you vary the parameter is the same as determining the bits of Omega," he says.

But Chaitin already knew that each digit of Omega is random and independent. This could only mean one thing. Because finding out whether a Diophantine equation has a finite or infinite number of solutions generates these digits, each answer to the equation must therefore be unknowable and independent of every other answer. In other words, the randomness of the digits of Omega imposes limits on what can be known from number theory--the most elementary of mathematical fields. "If randomness is even in something as basic as number theory, where else is it?" asks Chaitin. He thinks he knows the answer. "My hunch is it's everywhere," he says. "Randomness is the true foundation of mathematics."

The fact that randomness is everywhere has deep consequences, says John Casti, a mathematician at the Santa Fe Institute in New Mexico and the Vienna University of Technology. It means that a few bits of maths may follow from each other, but for most mathematical situations those connections won't exist. And if you can't make connections, you can't solve or prove things. All a mathematician can do is aim to find the little bits of maths that do tie together.

"Chaitin's work shows that solvable problems are like a small island in a vast sea of undecidable propositions," Casti says.

Answer 8

There are a lot of problems. Some of them are solved using mathematics, but the most of them are unsolvable, especially if the hypothesis of the problem is not correct.Even the hypothesis is correct the problem maybe unsolvable using mathematics, but we can solve it using a suitable software programme in the computer.

Answer 9

Sure add your phone number plus the number 2. Subtract 3 times your age. Add 2 plus 3 and subtract the original by the number of people who died in the civil war.



Oh and to this
"Yes there is....

3 soldies arrive at a motel, the clerk charges them $30.00 for the night, each soldier gives them $10.00. They go up to the room. The clerk felt bad for charging full price to the soldiers so he had the bellboy take them $5.00... The bellboy before arriving kept $2.00 and gave the soldiers back $1.00.. so they payed $9.00 each for the room. if ....:

3x9=27
27+2=29

where is the other $1.00?"


The answer is they paid 25 add the bell boys 2 dollars devide that by 3 would be 9 each this means now they paid 27 dollars.. and the bell boy gave them 3 dollars back which makes 30 :).

Answer 10

Try the Riemann's Hipothesys, it has been unsolved by mathematicians for 150-200 years.