Prove me wrong

...

when I fucked your mom I I gave her a great orgasm. somehow your dad accomplished the same thing. therefore I am your dad. prove me wrong.

why do chinks do math tests with markers
Here watch this
>0! is 1 by definition
>you can't apply any transitive property of equality by definition
QED and get rekt

>dividing by !
>

>

You can't divide by ! LOL.

kys

log(9) =log(10-1)=log(10)/log(1)=1/0=infinity

Your logical fallacy is located in your last step. You seem to use the following inference rule:
[eqn]\frac{x\,k\,=\,y\,k}{x\,=\,y}[/eqn]
which is true if and only if [math]k[/math] is simplifiable. The problem is that [math]![/math] is not simplifiable, for it is not a number. In fact, [math]![/math] is an operator denoting a function called the factorial. The factorial is defined, for any natural integer [math]n[/math], as [math]n!\,=\,\prod_{i\,=\,1}^n i[/math]. Using this definition, we may immediately find that [math]0!\,=\,\prod_{i\,=\,1}^0 i\,=\,1[/math] as an empty product and that [math]1!\,=\,\prod_{i\,=\,1}^1 i\,=\,1[/math] as a non-empty product with only one term. However, since [math]1\,=\,s\left(0\right)[/math] for the succession function [math]s[/math] postulated by Peano's arithmetic, we have [math]1\,\neq\,0[/math], which shows that the factorial function is not injective. Therefore, you cannot use that inference rule.

I hope my explanation has been useful to you. :)

Cool. Now how about EVERY anonymous Veeky Forums poster goes to their Universities and finally demands mathematics to be taken seriously and replace 'percentage' based bullshit with 'logarithms' and geometry.

Well done.

Class dismissed.
//
酷。 現在怎麼樣EVERY匿名/ SCI /海報去他們的大學,最後要求要認真對待和替換“百分比”基於與胡說“對數”和幾何數學。

幹得好。

下課。

>You can't divide by !
Not with that attitude, user.

Consider a successor function s(n) (injective) as defined by Peano's axioms. If 0 equals 1, then s(0) = s(1). but s(0) = 1 and s(1) = 2, ergo, 0 != 1.
kys.

well then a guess peano is a stupid faggot. probably why his name is retarded.

nice ad hominem faggot, enjoy your non-countable sets.
i bet you are a finitist brainlet.

>non-countable sets
There is no such thing as a non-countable set

exactly

you've just proved that the output of the factorial is the same, it does not follow that the arguments are equal
sin(pi/2) = 1, sin(5pi/2)=1, therefore pi/2 = 5pi/2, 0 = 2pi

! Is not a number, idiot.

Oh really? Is that what they teach you in biology class nowadays?

That's not how factorials work. 0! is not operation.

1! is also not... Factorial. it's just 1.
Factorial can apply for numbers, which have n-1 > 1

fuck off ching chong

kek. poster just got BTFO. i'm not even going to ask if he'll ever recover, it's clear he will not

Behave motherfucker, dudes invented gunpowder and I think it should be used on you.

馬鈴薯階乘

The fucking retard probably got taught that numbers are just a social construct in his "quantitative relations in the united states class"

anime website

So what you're saying in a long winded way is that 0! is defined as 1 and you're not allowed to infer that 0!=1! . This is so we can do useful things and we just need to roll with it.

I'm not disagreeing, I'm just trying to parse your moonlanguage

Also 0^0 is 1

you forgot the
!

>n^0 = 1
>m^0 =1
>m^0 = n^0
>m = n
QED

numbers are a lie

This.

factorial is non-math. Fucking worthless operation.
>hurr 1! = 1*0! = 0! but 0! is not defined XD
how is this shit even allowed, really.

>what is the gamma function?

Of course you can, otherwise you could not do "1! = 1, 1=2/2, therefore 1! = 2/2"

It's the whole point of transitivity. Of course, from stupid definitions come stupid results, but nobody's stopping you from working in a space where 0 = 1 if you define it to be true (and for bigger n like, especially if n is prime, 0 = n is not even uncommon and even has practical applications: modulo rings)

So, the result is 1 = 2 = 3 = ... (and, because s(n) is injective, therefore also 0 = 1)

Where's the contradiction?

But the factorial functions is defined for all natural numbers, including 0?

0! := 1
(n+1)! = (n+1) * n!

Lol wtf you can't "divide by !" because it's a mathematical function
Imagine it being a function like f(x)

f(0)=1
f(1)=1

f has the same value for 0 as it does for 1
but that doesn't mean that 0 and 1 are the same

>2^2=4
>(-2)^2=4
>=>(-2)=2
>=>4=0
btfo

It's a restriction of the gamma function you brainlet, you can't just invert it and get only what you started with.

That's like saying since 2^2 = (-2)^2 it must mean 2 = -2.

But !: R -> R is not a one-to-one function, i.e.

~ (Ax)(Ay) !(x) = !(y) => (x = y)

Suppose dividing with [math]![/math] was possible (which is where your proof falls short). Given then that [math]n! = n(n-1)!\\[/math] we would get:[eqn]
\dfrac{n!}{!} = n = n(n-1)
[/eqn]
It is immediately evident that this is not true for all [math]n \in \mathbb{N}_{>0}[/math], neither is it true for [math]n = 1[/math] since [math]1 \ne 1(1-1)[/math].
[math]\text{Q.E.D.}[/math]

[math]\dfrac{n!}{n}=(n-1)! \implies 1!=\left(1-1\right)!=0![/math]

f(x)=x2 - x + 1

f(0)=1

f(1)=1

0=1 prove me wrong