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

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:

1. Let $p$ be "$x^2 + x - 2 = 0$" and $q$ be "$x \neq 0$".
2. Assume that $x^2 + x - 2 = 0$ and assume $x = 0$.
3. (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 6:52
• Substitute assumption $x=0$ into $x^2+x-2=0.$ Then... – M. Strochyk Oct 10 '13 at 6:54
• 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 6:57
• Since the statement $"-2=0"$ is false, then the assumption $"x=0"$ must be false. – M. Strochyk Oct 10 '13 at 7:20

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 $x \neq 0$.

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 $\mathbb{R}$.)