# What's wrong with this proof by contradiction (Session 3.3 of How to prove it)?

I am reading How to prove it and doing the exercise in session 3.3.

Theorem: For every real number $$x$$, $$x^2 \geq 0$$.

What's wrong with the following proof of the theorem?

Proof. Suppose not. Then for every real number $$x$$, $$x^2 \lt 0$$. In particular, plugging in $$x = 3$$ we would get $$9 \lt 0$$, which is clearly false. This contradiction shows that for every number $$x$$, $$x^2 \geq 0$$.

This proof makes sense to me but why it is incorrect?

• The negation of an existential quantifier is a universal quantifier. – Hyperion Feb 9 at 3:27
• You CAN'T put $x = 3$. The negation of the theorem is: there exists a real number $x$ such that $x^{2} <0$. – Seewoo Lee Feb 9 at 3:27
• To construct a proof by contradiction, you need to prove the absurdity of the statement "There exists some $x \in \mathbb{R}$ such that $x^2 < 0$. – Hyperion Feb 9 at 3:30
• I think you make a logic error in line 3, which might have been said this way instead:: "Proof. Suppose not. Then there is at least one number x, such that..." – eSurfsnake Feb 9 at 3:36
• As an example of why this is logic is wrong, note that by removing the squaring everywhere, the exact same "proof" could be used to show "for every real number $x, x \ge 0$". This is a useful method for understanding where you went wrong: see if you can use the same technique to "prove" something you know is false, and then figure out where it stopped making sense. – Paul Sinclair Feb 20 at 17:16

## 3 Answers

Since The negation of $$\forall x(x^2 \geq 0)$$ is $$\exists x(x^2 \lt0)$$, the sentence " Then for every real number $$x$$, $$x^2 \lt 0$$" is therefore incorrect.

As mentioned in the comments, the negation of an existential quantifier is a universal quantifier. Hence, $$\neg(\forall x: p(x))$$ is not $$\forall x:\neg p(x)$$ but $$\exists x:\neg p(x)$$.

If your theorem is false, then it is not the case that for every real number, $$x, x^2\ge 0$$.

It doesn't follow that if the theorem is false, all squares are negative. For the theorem to be false, we need only one counter-example. We don't need $$x$$ to be less than zero for all real numbers. Your contradiction says too much. So checking one square and showing it is positive, won't help you. You need to show that they all are positive. Perhaps a proof by induction?

What I've said isn't anything not already in the answers or comments.