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I must prove the following:

Prop. : let $x \in \Bbb{R}, n \in \Bbb{N}-\{0\}$ then $$x^n=0 \to x=0$$ Proof : by contradiction I have $x \neq 0$, by trichotomy one of the following holds

  1. $x <0 $
  2. $x >0$

1) if $x >0$ then $x^n>0$ therefore $x^n\geq 0$ and $ x^n\neq 0$ and (by hypothesis) $x^n=0$, this is absurd

2) if $x <0$ then:

  • if $n$ is even then $x^n>0$ therefore $x^n\geq 0 $ and $ x^n\neq 0$ and (by hypothesis) $x^n=0$, this is absurd
  • if $n$ is odd then $x^n<0$ therefore $x^n\leq 0 $ and $x^n\neq 0$ and (by hypothesis) $x^n=0$, this is absurd

Is this correct?

mle
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  • What is this wedge product? – Semiclassical Jul 23 '14 at 15:35
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    @Semiclassical It's logical conjunction and disjunction: $\land$ is and $\lor$ is or. Your proof looks correct to me. – James Jul 23 '14 at 15:36
  • I believe he means "or" in logic. Though I think technically $x<0\oplus x>0$ should be in place of $x <0 \vee x >0$. – Jam Jul 23 '14 at 15:36
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    I figured out what OP means. This is a very unfortunately written post. What the first line should read is: Let $x^n = a$ and $a=0$, then $x=0$. This is really unnecessary and should instead be written let $x^n = 0$, then $x=0$. @mle, DO NOT mix "or" and "and" logical symbols with regular math. It is very confusing for the reader. Instead write out your statements in words. Overuse of symbols is bad form. – Cameron Williams Jul 23 '14 at 15:40
  • @CameronWilliams You're probably right. I'd attempted to edit some of the language to make it more understandable but some of it is quite confusing. – Jam Jul 23 '14 at 15:42
  • You should appeal to an axiom in each of the "therefore" statements you make, or otherwise use a different approach. – abnry Jul 23 '14 at 15:46
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    Please stop using $\wedge$ for "and" and $\vee$ for "or"... – Emily Jul 23 '14 at 16:18

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This is the approach I'd take. More generally, I will use only the axioms of a field. I think the basic idea of your proof is good, but you don't justify why $x > 0$ implies $x^n > 0$. As I show below, you can avoid this casework altogether.

As you started, assume $x \neq 0$. We will show that this implies $x^n \neq 0$, which will give us the contradiction we want.

But to show this use contradiction within a contradiction argument. While $x \neq 0$, suppose $x^n = 0$ for some positive integer $n$. We must have $n > 1$ since $x^1 =x \neq 0$. Let $m > 0$ be the smallest such $n$. By the properties of a field, since $x \neq 0$, there is an inverse $x^{-1}$ so that $x^{-1}x^m = x^{-1}0 = 0$. But this means there is a smaller $n=m-1 > 0$ such that $x^n = 0$. This is a contradiction, so we must have $x^n \neq 0.$

But then as already mentioned, this is another contradiction. So the assumption $x \neq 0$ is false.

abnry
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  • My one small edit to this is to change "for some $n$" to "for some positive integer $n$". Then you need to justify the definition of $m$ by noting that every nonempty set of positive integers has a least element by the well-ordering principle. This may beyond the OP's intended level of care, however. – John Hughes Jul 23 '14 at 15:59