What's wrong with this proof by contradiction?

Question

Problem Consider the following proposition. What’s wrong with the following proof of the proposition?
Proposition: For every real number $$, $^2≥0$.
Proof: Suppose not. Then for every real number $$, $^2<0$. In particular, plugging in $x=3$ we would get $9<0$, which is clearly false. This contradiction shows that for every number $$, $^2≥0$.

My Answer I'm thinking that this doesn't work because you're only showing that one case contradicts the hypothesis, not every case. Is this all I need to say to answer this question fully?

Your negation is wrong. It should be " There exists a number a such that a^2<0" — kareemmatheson, Feb 21 '16 at 03:08
When you say something is true for everything, it is false if it is false for any one thing. — Saikat, Feb 21 '16 at 03:34

Carl Mummert · Accepted Answer · 2016-02-21T03:16:10.540

The error is that, when the proof writer took the negation of the "for every" quantifier, they did not switch it to "for some". The correct negation would be: "then there exists a real $x$ for which $x^2 < 0$".

To disprove a "for all" quantifier, it is only necessary to give one counterexample. So the error you have identified is not actually an error - in the sense that if the part of the proof before that had been correct then this part would also have been correct. The initial error is earlier, with the bad negation.

To disprove a "there exists" quantifier, it is necessary to show there is no example, not just to show that a particular example doesn't work.

That makes perfect sense. Thanks for the help. – dibdub Feb 21 '16 at 03:12 — dibdub, Feb 21 '16 at 03:12

What's wrong with this proof by contradiction?

1 Answers1