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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 $x^2 + 3 > 0$?

ming
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  • Proving the negation of that statement would be gibberish, because the statement is true. – Steve B Dec 08 '18 at 02:02
  • What do you mean, “proving the contradiction”? Do you mean a proof by contradiction? – Arturo Magidin Dec 08 '18 at 02:03
  • 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 08 '18 at 02:03
  • The contradiction of a statement and the negation of a statement are the same thing. Therefore proving them would be the same thing. In practice, I doubt that anyone sets out to prove the negation of a statement. – Steve B Dec 08 '18 at 02:09
  • Yes @ArturoMagidin – ming Dec 08 '18 at 02:17
  • Wait, but people set out to prove the prove/disprove the contradiction.. but they don't set out to prove/disprove the negation? Then how are they the same thing? @SteveB – ming Dec 08 '18 at 02:18
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    @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 08 '18 at 02:51
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    Stop using the wrong words, you may stop getting confusing answers. In the meantime, perhaps see https://math.stackexchange.com/questions/112774/proof-by-contradiction-circular-reasoning – Arturo Magidin Dec 08 '18 at 02:52
  • @SteveB: “the contradiction of a statement” is not something people talk about in this context; your replies are likely to generate confusion rather than clarify. – Arturo Magidin Dec 08 '18 at 03:02
  • Also see Are proofs by contradiction really logical? – user21820 Jan 02 '19 at 08:11

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What you are actually asking about, according to the comments, is “proof by contradiction” (which is not “prove the contradiction”).

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.

Arturo Magidin
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