If prime, 2017 is the year Riemmen's Hipothesis is proven right or wrong

If prime, 2017 is the year Riemmen's Hipothesis is proven right or wrong.

Math general.

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5*11*158029
What's another year.

Is it possible to have two functions defined on the same domain (for the sake of argument let's just say R^n, otherwise the answer would be trivial), that is equal at every x, yet we can't show them too be using straightforward algebraic manipulation?

well?

nevermind, such an a does not exist

This is a good question. Conventionally, two functions are identified precisely when they evaluate identically for all arguments. It reduces to the Yoneda lemma: if the composites coincide for choices of maps, then they are identical. The identity of indiscernibles. This is why we use the univalence axiom.

That's way above my head.
Let's just consider functions defined everywhere on R first.
Looking at polynomials we could have two which are equal at every x but whose "strings" initially are not equal. In the simplest case it could just be a matter of terms having different position in the string defining them. However in this case we could easily
rearrange them through a finite series of steps preserving equivalency and get the same string for both of them.
The step above in complexity would be comparing two polynomials, one who is in factored form, the other in expanded form. In this case too we would have the same function essentially (if they're equal everywhere) only represented at some instance by differing strings, but we could rearrange them and obtain the same string.
Furthermore we could look use exponents, logs to have two equal functions represented by different strings. However in all examples I can think of we could easily rearrange the strings and obtain the same string.

Would it be possible in theory to have two such functions that evaluate to the same value at every x in R, yet which can't be shown using the simple mechanics of algebra to be equal, ie obtaining the same string representing them.
Almost surely not, as long as you're using well defined operations in defining your functions, right?

>If prime
>babby can't determine whether prime
gtfo babby

I think it depends on the operations allowed. There could well be a pair of power series which both converge to the same values but which cannot be rearranged in a finite number of algebraic steps to look like the other. These would be very odd functions, though. I don't know how to formalize this question, though.

How can he predict the thread's number, you triple nigger

are you kidding? on a slow moving board like Veeky Forums?

this isn't /an/ or /po/

Maybe.

QED

it's still slow enough to manage

That's not how meme magic works.

>8693019

ah well i forgot an arrow

damnit

ha!

For a statement S, could there ever be a proof P1 of that statement such that P1 doesn't prove S directly, but rather it proves that another proof P2 must exist and be valid, and that P2 does prove S?
And P1 is unequal to P2.

Ie could we prove that a proof must exist for some statement, without actually finding that proof?

Suppose P1 exists so that P1 proves P2 that proves S directly.
by P1's existence, S is proven, so P1 is a direct proof of S and thus equivalent to P2?
I don't know, tricky

Yes, of course P2 is a proof of S. But theorems could have several proofs, I think the proof of the pythagorean theorem for example has 370 different proofs.
Thus P1 would be a proof that atleast one other proof which is not P1, must exist.

math.stackexchange.com/questions/1642225/proving-the-existence-of-a-proof-without-actually-giving-a-proof

Thanks user, I'll look through it later when I have the time.

she's got fucking ugly FASD philtrum and upper lip

DELETE THIS NOW

she's got the FAS, deal with it.
just look at the broad nostrils and the comparably long distance between the eyes. probably attention deficit and a slight mental retardation, too.

Those are classic mongoloid features you ignoramus. There are billions of them.
Finns are part mongoloid.

see

Are there any primes ITT?

bigger distance between the eyes is correlated with intelligence

years can't be primes

>that tfw when no math gf

bump

integrating the equation f'(x)=g'(x) gives you
f(x)=g(x)+C, for some constant C,
because a =/= 0 , evaluating at 0 gives you
0=f(0)=g(0)+C=1+C, so C must be -1.
evaluating at point 'a' gives you
a^a=a^a - 1, so 0=1, contradicting the existence of a.

You can think of a function as a (possibly infinite) ordered tuple. Looking at it this way, it's obvious that most functions can't be represented algebraically, let alone compared.

Right here, bby

Of course, program equivalence is undecidable so you only need the domain N, no need for reals.

But I guess you question is whether two elementary function have this property, and this reduces whether you can have a normalform for them