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My proof by contradiction ends with $x=-x$ when x is a positive integer. What is the correct terminology for why this is a contradiction?

Right now it says "This yields a contradiction since a positive integer cannot equal the negative value of itself", but I'm sure there's a theorum or something for this.

Thank you!

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  • $\begingroup$ You should probably give some context as to what the proof is, and what you tried. $\endgroup$ – Mark Pineau Apr 14 '17 at 11:17
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    $\begingroup$ What does the $/$ mean? $\endgroup$ – kingW3 Apr 14 '17 at 11:17
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    $\begingroup$ Did you mean $x\ne -x$? If you did then that isn't necessarily a contradiction unless $x=0$. $\endgroup$ – Laars Helenius Apr 14 '17 at 11:19
  • $\begingroup$ @LaarsHelenius: You mean "unless $x\ne 0$", right? $\endgroup$ – Henning Makholm Apr 14 '17 at 11:22
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    $\begingroup$ In some cases, $x = -x$ and not just for 0. In a field of characteristic 2 for example. $\endgroup$ – badjohn Apr 14 '17 at 11:25
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Trichotomy Law: If $a,b\in\mathbb{R}$ then exactly one of the following holds $$(i)a>b\quad (ii)a=b\quad (iii)a<b$$

In your case, we have $-x<x$ since $x$ is positive integer. Thus, by the above law, $-x\neq x$. This is in contradiction to the one you obtained $-x=x$.

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You can do another proof by contradiction for that:

Assume $x\ne0$ and $x=-x$. Since $x$ is not $0$, we may divide by $x$ to achieve: $1=-1$. This is a falsity. Therefore we conclude $x=0$, which is not a positive number.

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Just note that $x=-x$ is equivalent to $x+x = 0$, equivalent to $x = 0$.

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