Proof by contradiction: If $x$ is a real number and $x^2 + x - 2 = 0$, then $x \neq 0$

Question

Here's the problem again:

Use a proof by contradiction to prove the following universal statement:

If $x$ is a real number and $x^2 + x - 2 = 0$, then $x \neq 0$

Here's my attempt at it:

Let $p$ be "$x^2 + x - 2 = 0$" and $q$ be "$x \neq 0$".
Assume that $x^2 + x - 2 = 0$ and assume $x = 0$.
(This is where I got stuck)

I'm not sure how to proceed, I've worked with statements like, "If $3n+2$ is odd, then $n$ is odd" but I've never worked with equations like this. How would I proceed with this proof by contradiction?

Conclude immediately by substitution that $-2=0$. This is false. The end. — André Nicolas, Oct 10 '13 at 06:52
So after I substitute it into the proof, I get that -2 = 0 as Andre Nicolas said earlier, I can just say that I proved that x != 0? So it's just a really short proof? — Camille, Oct 10 '13 at 06:57
Since the statement $"-2=0"$ is false, then the assumption $"x=0"$ must be false. — M. Strochyk, Oct 10 '13 at 07:20

score 1 · Answer 1 · answered Oct 10 '13 at 07:18

The key to this proof is that for $x \in \mathbb{R}$, either $x=0$ or $x\neq 0$.

Proposition:

If $x \in \mathbb{R}$ and $x^{2} + x -2 = 0$, then $x \neq 0$.

Proof by Contradiction:

Suppose that if $x \in \mathbb{R}$ and $x^{2} + x -2 = 0$, then $x=0$.

This implies that $0^{2} + 0 - 2 = 0$, i.e. $-2 = 0$. This claim is absurd, and we're done.

(The claim is absurd because $0$, the additive identity, is unique in $\mathbb{R}$.)

Proof by contradiction: If $x$ is a real number and $x^2 + x - 2 = 0$, then $x \neq 0$

1 Answers1