# Proving the contradiction/negation of a statement

Is proving the contradiction and proving the negation the same thing? Proving it true is proving the original statement false and proving it's false proves the statement is true?

So if a statement is an implication we just assume that the hypothesis is false? What if there's a quantifier in front?

What would proving the negation of this statement look like?:

$$\forall x \in \Bbb R$$, if $$x > 2$$, then $$x^2 + 3 > 0$$?

• Proving the negation of that statement would be gibberish, because the statement is true. – Steve B Dec 8 '18 at 2:02
• What do you mean, “proving the contradiction”? Do you mean a proof by contradiction? – Arturo Magidin Dec 8 '18 at 2:03
• The negation of the statement you write is “there exists $x\in\mathbb{R}$ such that $x\gt 2$, and $x^2+3\leq 0$.” But that statement is false. – Arturo Magidin Dec 8 '18 at 2:03
• @ming: You are saying things in a confusing manner, which is why you are getting confusing answers. The statement “prove the contradiction” is incorrect. A “proof by contradiction” is a method by which you prove a given statement; it is not what you are proving. By contrast, the negation of a statement is itself a statement (which is obtained by negating the given statement). They are different things. Proving the negation shows that the given statement is false. A proof by contradiction establishes the truth of the given statement. – Arturo Magidin Dec 8 '18 at 2:51
• Stop using the wrong words, you may stop getting confusing answers. In the meantime, perhaps see math.stackexchange.com/questions/112774/… – Arturo Magidin Dec 8 '18 at 2:52

A proof by contradiction is a method of proof in which one assumes the negation of what you want to prove, and deduce a statement that is impossible. In classical logic, this means that the original statement must be true, because of the law of the excluded middle: it cannot be false (because if it were false that would lead to a contradiction), and if it is cannot be false, then it must be true.

In order to do a proof by contradiction, you must know the negation of the statement; but you are not trying to prove that the negation is true. You are assuming that the negation is true, and trying to deduce a statement known to be false/impossible.

In the case you give, the statement you want to prove is $$\forall x\in\mathbb{R}\Bigl( x\gt 2 \rightarrow x^2+3\gt 0\Bigr)$$

The negation of this statement is \begin{align*} &\neg\Biggl( \forall x\in\mathbb{R}\Bigl(x\gt 2 \rightarrow x^2+3\gt 0\Bigr)\Biggr)\\ &\exists x\in\mathbb{R}\Biggl(\neg\Bigl( x\gt 2\rightarrow x^2+3\gt 0\Bigr)\Biggr)\\ &\exists x\in\mathbb{R}\Biggl( x\gt 2 \text{ and } \neg(x^2+3\gt 0)\Biggr)\\ &\exists x\in\mathbb{R}\Bigl( x\gt 2 \text{ and }x^2+3\leq 0\Bigr) \end{align*} So, to do a proof by contradiction, you would start by assuming that there is a real number $$x$$ that is both greater than 2, and also has the property that $$x^2+3\leq 0$$. From this, you would want to deduce something utterly impossible.

If you were trying to prove the negation, you would be trying to prove that there is a real number $$x$$ that is both greater than 2 and also has the property that $$x^2+3\leq 0$$.

Different things entirely.

If you do a proof by contradiction successfully, you have established that the given statement is true.

If you successfully prove the negation, you have established that the given statement is false.