abstract algebra - how do I prove that $1 > 0$ in an ordered field? - Mathematics Stack Exchange
algebra precalculus - Prove $0! = 1$ from first principles - Mathematics Stack Exchange
definition - Why is 1 greater than 0? Show the proof. - Mathematics Stack Exchange
Proof That 1 = 0 (It's Legit)
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We need $0!$ to be defined as $1$ so that many mathematical formulae work. For example we would like $$n! = n \times (n-1)!$$ to work when $n=1,$ ie $1! = 1 \times 0!.$ Also we require that the formula for the number of ways of choosing $k$ objects from $n$ is valid for $k=n.$ ie $${n \choose k} = \frac{n!}{k!(n-k)!}$$ is valid when $k=n.$
Things need to work when we extend our definition of the factorial via the gamma function.
$$\Gamma(z) = \int\limits_0^\infty t^{z-1} e^{-t} \,\mathrm{d}t,\qquad \Re(z)>0.$$
The above gives $\Gamma(n)=(n-1)!$ and so we require $0!=1,$ since $\Gamma(1)=1.$
I'm not sure that there is anything to prove. I think it follows directly from the definition of factorial:
$$ n! := \prod_{k = 1}^n k$$
So if $n=0$ the right hand side is the empty product, which is $1$ by convention.
The usual proof relies on $a^2> 0$ for any nonzero real number $a$. If you can prove that as a lemma from the basic axioms, and you note that 1 is the multiplicative identity, you will be done.
It does very much depends on what "basic axioms" you use. Believe it or not not all mathematicians use the same. YOu should list which ones you have.
From an algebra perspective, we can use the ordered field axioms.
There's a lot of things to prove.
Intermediate and nescessary things to prove but our big goal is:
a) For all $a\ne 0$ we know $a^2 > 0$.
Then we are done, as $1 = (1)^2 > 0$.
To prove a) we usually have an axioms $a > b; and x > 0$ then $ax > bx$. We also have if $a > b$ then $a + x > b+x$ for all x$.
From there we can prove $x > 0 \iff 0 > -x$ by noting $x > 0 \implies x-x > 0 -x$ so $0 > -x$.
We need to prove tedious little things such as $(-x)(y) = x(-y) = -xy$ but using the axiom $a(b+c) = ab + bc$ and that for all $a$ there is a unique $-a$ so that $a+(-a) = 0$. Then we need to prove $(-x)(-y) = xy$.
This is all to prove that if $a > b$ and $x < 0$ then $ax < ab$ (because $x < 0 \implies -x > 0 \implies a(-x) > b(-x) \implies -ax > -bx \implies -ax +ax + bx > -bx +ax +bx \implies $bx > ax$).
So with all that in mind we can prove: If $x > 0$ then $x^2 =x*x > 0*x = 0$. And if $x < 0$ then $x*x > 0*x$ so $x^2 > 0$. So as long as $x \ne 0$ we have $x^2 > 0$!
So $1 = 1^2 > 0$!