1. ## Re: Interesting mathematical statements

This isn't exactly an interesting maths statement but more or less something I found humourous.

2. ## Re: Interesting mathematical statements

Originally Posted by leehuan
This isn't exactly an interesting maths statement but more or less something I found humourous.
that is rather amusing.

I have seen people do the same for exponential limits.

3. ## Re: Interesting mathematical statements

$\noindent Not so much surprising a fact as it's own consequences, but... \\ Let F(x) be the integral of the invertible function f(x) and f^{-1}(x) be the inversion of f(x).\\Then the following statement is always true: \\ \int f^{-1}(x) dx = xf^{-1}(x)-F\left(f^{-1}(x)\right) +C\\Due to the arbitrariness of f, the relationship is completely bijective. This means that if a function and it's inverse are expressible in elementary terms, and only one of them has an obvious integral in elementary form, then the other function is guaranteed to have an integral expressible in elementary terms.$

4. ## Re: Interesting mathematical statements

Carmichael's theorem:

$\\ When n>12, the nth Fibonacci number has at least one prime divisor that DOES NOT divide any earlier Fibonacci number.$

5. ## Re: Interesting mathematical statements

I don't know if anyone's posted this yet:
$If a theory \varphi can prove that \varphi is consistent, then \varphi is inconsistent.$

6. ## Re: Interesting mathematical statements

Originally Posted by leehuan
Carmichael's theorem:

$\\ When n>12, the nth Fibonacci number has at least one prime divisor that DOES NOT divide any Fibonacci number.$
Any EARLIER Fibonacci number.

It wouldn't make any sense without that word.

7. ## Re: Interesting mathematical statements

Originally Posted by KingOfActing
I don't know if anyone's posted this yet:
$If a theory \varphi can prove that \varphi is consistent, then \varphi is inconsistent.$
Then the computational extension of the theorem is as follows:

If a (turing complete) machine can decide whether or not any given input will halt or not, then said machine cannot decide the input comprising itself.

By the law of the excluded middle, this is a contradiction, and as such, any such machine cannot possibly exist.

8. ## Re: Interesting mathematical statements

Then the computational extension of the theorem is as follows:

If a (turing complete) machine can decide whether or not any given input will halt or not, then said machine cannot decide the input comprising itself.

By the law of the excluded middle, this is a contradiction, and as such, any such machine cannot possibly exist.
I don't think this statement is quite true. Specifically, the "input comprising itself" bit - I don't know of a proof that shows that this is the case. If you're alluding to the diagonalisation argument, the result is slightly more complex.

Suppose that A is a Turing machine that given any machine T and input I, decides whether B halts on I.

Let C be a machine that, when given I, calls A with machine I and input I and halts iff A concludes that I doesn't halt on input I.

If A is called with machine C and input C and halts, then that means that A must halt on C when given C. Inversely if A is called with (C, C) and doesn't halt, then that means A must halt on (C, C). Here lies the contradiction. (unless I've made a mistake)

So C is the input that breaks T, not T. Unless I've misunderstood what you've said.

But the more important question is - who are you? What kind of high school kid has this kind of background in maths? Where did you learn this stuff?

9. ## Re: Interesting mathematical statements

Originally Posted by GoldyOrNugget
I don't think this statement is quite true. Specifically, the "input comprising itself" bit - I don't know of a proof that shows that this is the case. If you're alluding to the diagonalisation argument, the result is slightly more complex.

Suppose that A is a Turing machine that given any machine T and input I, decides whether B halts on I.

Let C be a machine that, when given I, calls A with machine I and input I and halts iff A concludes that I doesn't halt on input I.

If A is called with machine C and input C and halts, then that means that A must halt on C when given C. Inversely if A is called with (C, C) and doesn't halt, then that means A must halt on (C, C). Here lies the contradiction. (unless I've made a mistake)

So C is the input that breaks T, not T. Unless I've misunderstood what you've said.

But the more important question is - who are you? What kind of high school kid has this kind of background in maths? Where did you learn this stuff?
He's more gay for maths than leehuan

10. ## Re: Interesting mathematical statements

Originally Posted by KingOfActing
He's more gay for maths than leehuan
Oi!

11. ## Re: Interesting mathematical statements

Originally Posted by GoldyOrNugget
I don't think this statement is quite true. Specifically, the "input comprising itself" bit - I don't know of a proof that shows that this is the case. If you're alluding to the diagonalisation argument, the result is slightly more complex.

Suppose that A is a Turing machine that given any machine T and input I, decides whether B halts on I.

Let C be a machine that, when given I, calls A with machine I and input I and halts iff A concludes that I doesn't halt on input I.

If A is called with machine C and input C and halts, then that means that A must halt on C when given C. Inversely if A is called with (C, C) and doesn't halt, then that means A must halt on (C, C). Here lies the contradiction. (unless I've made a mistake)

So C is the input that breaks T, not T. Unless I've misunderstood what you've said.

But the more important question is - who are you? What kind of high school kid has this kind of background in maths? Where did you learn this stuff?
That's why I said comprising. There are a few gaps that I wasn't bothered filling in, because I know the input requires two machines in order to break the supposed machine.

12. ## Re: Interesting mathematical statements

That's why I said comprising. There are a few gaps that I wasn't bothered filling in, because I know the input requires two machines in order to break the supposed machine.
You're an IMO guy/girl aren't you?

13. ## Re: Interesting mathematical statements

Originally Posted by GoldyOrNugget
You're an IMO guy/girl aren't you?
If only I were...

14. ## Re: Interesting mathematical statements

Originally Posted by turntaker
-1 x -1 = 2
actually = 1, but anyway

16. ## Re: Interesting mathematical statements

$\lim_{m \to \infty}\frac{(2m)!/m!}{2^{m+\frac{1}{2}}m^me^{-m}}=1$

17. ## Re: Interesting mathematical statements

Originally Posted by leehuan
$\lim_{m \to \infty}\frac{(2m)!/m!}{2^{m+\frac{1}{2}}m^me^{-m}}=1$
that's not very interesting.

it's a fairly trivial consequence of Stirling's Approximation... and Limits...

18. ## Re: Interesting mathematical statements

Too bad, cause even simple things like the power series for exp are interesting when you first see it.

(Also it was not a consequence in my homework. It was an intermediate step to GET to Stirling's approximation)

19. ## Re: Interesting mathematical statements

Originally Posted by leehuan
Too bad, cause even simple things like the power series for exp are interesting when you first see it.

(Also it was not a consequence in my homework. It was an intermediate step to GET to Stirling's approximation)
I know. But I'm not the one doing it, so ¯\_(ツ)_/¯

20. ## Re: Interesting mathematical statements

$\\ Almost all real numbers have a (roughly; aka in the limit) equal proportion of 0s and 1s in their binary expansion.$

$\\ To see this, note that if \ \{B_n\}_{n\geq 1} \ are a sequence of iid Bernoulli(0.5) trials, then the random variable \ Z_{0.5} = \sum_{n=1}^{\infty}\frac{B_n}{2^n} \sim U(0,1), \ is uniformly distributed$

$\\ Moreover, whenever we do the same thing with Bernoulli(p) trials instead, to obtain \ Z_p \ it turns out that Z_p is not absolutely continuous (i.e. it has no density), so that all of its probability mass accumulates on a set of measure 0$

21. ## Re: Interesting mathematical statements

Originally Posted by Sy123
$\\ Almost all real numbers have a (roughly; aka in the limit) equal proportion of 0s and 1s in their binary expansion.$

$\\ To see this, note that if \ \{B_n\}_{n\geq 1} \ are a sequence of iid Bernoulli(0.5) trials, then the random variable \ Z_{0.5} = \sum_{n=1}^{\infty}\frac{B_n}{2^n} \sim U(0,1), \ is uniformly distributed$

$\\ Moreover, whenever we do the same thing with Bernoulli(p) trials instead, to obtain \ Z_p \ it turns out that Z_p is not absolutely continuous (i.e. it has no density), so that all of its probability mass accumulates on a set of measure 0$
It is as if you're trying to make our boys Uri and Daniel proud at the same time

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