Terminology for $x=-x$ when x is a positive integer

Question

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!

You should probably give some context as to what the proof is, and what you tried. — Mark Pineau, Apr 14 '17 at 11:17
Did you mean $x\ne -x$? If you did then that isn't necessarily a contradiction unless $x=0$. — Laars Helenius, Apr 14 '17 at 11:19
In some cases, $x = -x$ and not just for 0. In a field of characteristic 2 for example. — badjohn, Apr 14 '17 at 11:25
@badjohn: But the integers are clearly not a field of characteristic $2$. Indeed, in a field of characteristic $2$ (or of any positive characteristic) you cannot define an order, and therefore the term "positive" does not make sense. Indeed, wherever "positive" makes sense, the sum of two positive elements is again a positive element, which excludes $x+x=0$ for any positive $x$, as $0$ by definition is not positive. — celtschk, Apr 14 '17 at 11:42
If $x$ is a positive integer then $-x<x$ and because you obtained $-x=x$, we get a contradiction according to Trichotomy Law. — Juniven Acapulco, Apr 14 '17 at 11:42
@ΘΣΦGenSan So how would I write that in the "yields a contradiction because..." form? — GiantDuck, Apr 14 '17 at 11:46
@celtschk Yes but at the time of my post, there was less information in the question. I don't think that it mentioned "positive integer" at that point. If it did then I apologise for having read it too quickly. — badjohn, Apr 14 '17 at 11:47
You already said that you end up with $-x=x$ where $x$ is a positive integer, right? This is not possible because $-x<x$. See and google Trichotomy Law — Juniven Acapulco, Apr 14 '17 at 11:47
@ΘΣΦGenSan Gotcha. If you want to post as an answer I'll accept it -- thanks! — GiantDuck, Apr 14 '17 at 11:48
@GiantDuck Thanks. I am an old man now and my knowledge is quite rusty. I am here to try to polish it up. It's good to know that I am not going senile and just missing half of your question. — badjohn, Apr 14 '17 at 11:49
@HenningMakholm Of course! I need to read what I type before I hit send more often! — Laars Helenius, Apr 14 '17 at 12:08

Thorgott · Answer 1 · 2017-09-30T13:49:13.630

2

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.

edited Sep 30 '17 at 13:49

answered Apr 14 '17 at 11:53

Thorgott

11,682

score 2 · Accepted Answer · answered Apr 14 '17 at 11:59

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 $-x=x$.

Yes · Answer 3 · 2017-04-14T12:31:42.537

1

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

edited Apr 14 '17 at 12:31

answered Apr 14 '17 at 11:51

Yes

20,719

Terminology for $x=-x$ when x is a positive integer

3 Answers3