How can I prove by contradiction that there is no real number such that $q^2 = -1$

Question

How can I prove by contradiction that there is no real number such that $q^2 = -1$.

You would have to assume that there exists a $q$ that satisfies $q^2 = -1$.

But I can´t understand how I am supposed to prove this.

Do I have to first assume that $q$ is positive, and make it such that $q = -1/q >0$.

I cannot see how this contradicts my assumption that there exists a real number which satisfies $q^2 = -1$

The square of every natural, rational, real number is $\ge 0$. — Mauro ALLEGRANZA, Sep 11 '19 at 09:31
Why not distinguishing the three cases $q>0$ , $q<0$ , $q=0$ to show $q^2\ge 0$ for every real $q$ which is the desired contradiction ? — Peter, Sep 11 '19 at 09:33
A "good" proof requires some work... You have to start from the Axioms for the real numbers. Then prove trichotomy and then prove that "if $a \ne 0$, then $a^2 > 0$". — Mauro ALLEGRANZA, Sep 11 '19 at 09:50
@Peter That's also a way to prove, but the problem is to prove it by using the method of CONTRADICTION — MafPrivate, Sep 11 '19 at 09:52
@IsaacYIUMathStudio So we are only allowed to use $q^2=-1$ ? In this case, I suggest that we conclude that this implies $q=-i$ or $q=i$ which contradicts the assumption that $q$ can be real ? Or isn't that "proof by contradiction" either ? — Peter, Sep 11 '19 at 10:30

MafPrivate · Accepted Answer · 2019-09-11T09:35:30.337

0

Suppose $q>0$ (obviously, $q\ne0$), then $q=\dfrac{-1}{q}<0$. Contradiction.

Similarly, suppose $q<0$, then $q=\dfrac{-1}{q}>0$. Contradiction.

Therefore, there are no such real numbers $q$ that $q^2=-1$.

edited Sep 11 '19 at 09:35

answered Sep 11 '19 at 09:32

MafPrivate

1

You are halfway there. Now suppose $q \lt 0 \ldots$ – Ross Millikan Sep 11 '19 at 09:33
Thank you very much. Problem for me is understanding why this is a contradiction, why is "q = -1/q < 0" a contradiction? – Bob Pen Sep 11 '19 at 09:35
Because you have supposed $q>0$ @dondeman – FFjet Sep 11 '19 at 09:36
At first, we mention that $q>0$, however, $\dfrac{-1}{q}<0$, this makes a contradiction. – MafPrivate Sep 11 '19 at 09:36

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