You can't divide by ze

>if there's 2 y-values for the same x-value then it's not a function.
Correct.

>0 and 1 both yield 1.
Those are two x values giving the same y value, brainlet.

>whats your point?
The point is you can't take the factorial of both sides of an equation like that.
>I can also make up arbitrary equations
You're retarded. Coming up with one example where the results are still equal doesn't nullify all the other examples where the results aren't still equal. If you wrongly try to claim multiplication and addition are interchangeable and someone disproves you with the example of 3 + 3 = 6 and 3 * 3 = 9, it isn't a valid counterargument to point out 2 * 2 = 4 and 2 + 2 = 4.

>doesn't necessarily mean the pic is wrong tho
What? That's exactly what it means.

but i showed how it works sometimes, so maybe it works in case 1/0 too

>Go swallow some glass.

other user is b8ing

>it works sometimes
Broken watch theorem

the fact that you are getting 2 values of x should tell you right off the bat that you are not dealing with a well defined function.

sqrt(9) = +3
sqrt(9) = -3

and square roots are not a well defined function, that's why we use principle square roots in applications, so your point?

then im just gonna use principle factorials

>it works sometimes
No it doesn't. 1!/1! being equal to 2!/2! isn't because factorials of numerators and denominators preserve equality, they're just a coincidental case where the factorials of numerators and denominators are equal, like how 2+2 coincidentally equals 2*2 even though multiplication and addition aren't actually the same thing.
>I was just pretending to be retarded.
Fuck off.

>it's doesn't fit my agenda so im just gonna call it "coincidence"
ok
>multiplication and addition aren't actually the same thing.
but multiplication is just repeated addition

The mistake here is in line 1 where you assume that 1/0 is defined by equating it to x and then operating on x. Of course you got a nonsensical result. Therefore, 1/0 is undefined.

You're close
[eqn]
\frac{1}{0} = x \\
1 = 0x \\
\frac{1}{x} = 0 \\
\frac{1}{x} + 1 = 0+1 \\
\frac{1+x}{x} = 1 \\
1+x = x \\
1+x -x = x-x \\
1 = 0
[/eqn]
If we then plug that into the original equation
[eqn]
\frac{1}{0} = x \\
0 = 1 \\
\frac{1}{1} = x \\
1 = x
[/eqn]
And that's our final answer

>Treating factorial as if !/! Is equal to 1
>Thinking factorials can be distributed across the numerator and denominator

Sure Is brainlet season

I like how math went downhill like 400 years ago and never recovered. Now its just proofs for arbitrary things that everyone already accepted as defacto, or shit like x=y via unintelligble nonsense.

That post was unintelligible nonsense.

Retard

>y=x^2 is not a function cause (-1)^2=1^2

>[math]\left(\frac{n}{k}\right)! = \frac{n!}{k!}[/math]
jesus christ

>mfw people on this board can't see great math when it's right in front of them

nice trips but lemme explain why OP's pic is absolutely wrong :
at the second line, a supposition is made, that the gamma function (because OP also admits implicitly that x is real, and well defined in the domain of the gamma function, extension of the factorial to most reals) preserves quotients, as in [math](\frac{x}{y})!=\frac{x!}{y!}[/math], if that was true in the domain of gamma, that would mean the following :
[math](\frac{x+1}{x})!=\frac{(x+1)!}{x!}=x+1 \\
\Longleftrightarrow x+1=(1+\frac{1}{x})![/math]
a relation which clearly doesn't respect the continuity of the function, so nope, the argument made in this pic is doubly wrong

Jesus christ you guys are retards...
obviously factorial is a group homomorphism on the division group of Q...

f(x) = x!
There you go, now factorial is a function.

Kek, that's gotta be embarrassing.

what's the domain and codomain?

If you actualy want to use factorial in a serious way you could at least use the gamma function so it works on reals and complex numbers too.