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In logic, proof by contradiction is a form of proof that establishes the truth or validity of a proposition by showing that the proposition's being false would imply a contradiction. Since by the law of bivalence a proposition must be either true or false, and its falsity has been shown impossible, the proposition must be true.

In other words, to prove by contradiction that ${\displaystyle P}$, show that ${\displaystyle \lnot P\Rightarrow \perp }$ or its equivalent ${\displaystyle \lnot P\Rightarrow (Q\land \lnot Q)}$ . Then, since ${\displaystyle \lnot P}$ implies a contradiction, conclude ${\displaystyle P}$ .

Proof by contradiction is also known as indirect proof, apagogical argument, reductio ad impossibile. It is a particular kind of the more general form of argument known as reductio ad absurdum.

## Examples

A classic proof by contradiction from mathematics is the proof that the square root of 2 is irrational. If it were rational, it could be expressed as a fraction ${\displaystyle {\frac {a}{b}}}$ in lowest terms, where ${\displaystyle a,b\in \mathbb {Z} }$ , at least one of which is odd. But if ${\displaystyle {\frac {a}{b}}={\sqrt {2}}}$ , then ${\displaystyle a^{2}=2b^{2}}$ . Therefore ${\displaystyle a^{2}}$ must be even. Because the square of an odd number is odd, that in turn implies that ${\displaystyle a}$ is even. This means that ${\displaystyle b}$ must be odd because ${\displaystyle {\frac {a}{b}}}$ is in lowest terms.

On the other hand, if ${\displaystyle a}$ is even, then ${\displaystyle a^{2}}$ is a multiple of 4. If ${\displaystyle a^{2}}$ is a multiple of 4 and ${\displaystyle a^{2}=2b^{2}}$ , then ${\displaystyle 2b^{2}}$ is a multiple of 4, and therefore ${\displaystyle b^{2}}$ is even, and so is ${\displaystyle b}$ .

So ${\displaystyle b}$ is odd and even, a contradiction. Therefore the initial assumption—that ${\displaystyle {\sqrt {2}}}$ can be expressed as a fraction—must be false.

### The length of the hypotenuse is less than the sum of the lengths of the two legs

The method of proof by contradiction has also been used to show that for any non-degenerate Right triangle, the length of the hypotenuse is less than the sum of the lengths of the two remaining sides. The proof relies on the Pythagorean theorem. Letting ${\displaystyle c}$ be the length of the hypotenuse and ${\displaystyle a}$ and ${\displaystyle b}$ the lengths of the legs, the claim is that ${\displaystyle a+b>c}$ .

As usual, we start the proof by negating the claim and assuming that ${\displaystyle a+b\leq c}$ . The next step is to show that this leads to a contradiction. Squaring both sides, we have ${\displaystyle (a+b)^{2}\leq c^{2}}$ or, equivalently, ${\displaystyle a^{2}+2ab+b^{2}\leq c^{2}}$ . A triangle is non-degenerate if each edge has positive length, so we may assume that ${\displaystyle a}$ and ${\displaystyle b}$ are greater than 0. Therefore, ${\displaystyle a^{2}+b^{2} . Taking out the middle term, we have ${\displaystyle a^{2}+b^{2} . We know from the Pythagorean theorem that ${\displaystyle a^{2}+b^{2}=c^{2}}$. We now have a contradiction since strict inequality and equality are mutually exclusive. The latter was a result of the Pythagorean theorem and the former the assumption that ${\displaystyle a+b\leq c}$ . The contradiction means that it is impossible for both to be true and we know that the Pythagorean theorem holds. It follows that our assumption that ${\displaystyle a+b\leq c}$ must be false and hence ${\displaystyle a+b>c}$ , proving the claim.

### In mathematics

Say we wish to disprove proposition ${\displaystyle p}$ . The procedure is to show that assuming ${\displaystyle p}$ leads to a logical contradiction. Thus, according to the law of non-contradiction, ${\displaystyle p}$ must be false.

Say instead we wish to prove proposition ${\displaystyle p}$ . We can proceed by assuming "not ${\displaystyle p}$" (i.e. that ${\displaystyle p}$ is false), and show that it leads to a logical contradiction. Thus, according to the law of non-contradiction, "not ${\displaystyle p}$" must be false, and so, according to the law of the excluded middle, ${\displaystyle p}$ is true.

In symbols:

To disprove ${\displaystyle p}$ : one uses the tautology ${\displaystyle (p\to (R\land \lnot R))\to \lnot p}$ , where ${\displaystyle R}$ is any proposition and the ${\displaystyle \land }$ symbol is taken to mean "and". Assuming ${\displaystyle p}$ , one proves ${\displaystyle R}$ and ${\displaystyle \lnot R}$ , and concludes from this that ${\displaystyle p\to (R\land \lnot R)}$ . This and the tautology together imply ${\displaystyle \lnot p}$ .

To prove p: one uses the tautology (¬p → (R ∧ ¬R)) → p where R is any proposition. Assuming ¬p, one proves R and ¬R, and concludes from this that ¬p → (R ∧ ¬R). This and the tautology together imply p.

For a simple example of the first kind, consider the proposition, ¬p: "there is no smallest rational number greater than 0". In a proof by contradiction, we start by assuming the opposite, p: that there is a smallest rational number, say, r0.

Now let x = r0/2. Then x is a rational number greater than 0 and less than r0. (In the above symbolic argument, "x is the smallest rational number" would be R and "r0 (which is different from x) is the smallest rational number" would be ¬R.) But that contradicts our initial assumption, p, that r0 was the smallest rational number. So we can conclude that the original proposition, ¬p, must be true — "there is no smallest rational number greater than 0".

[Note: the choice of which statement is R and which is ¬R is arbitrary.]

It is common to use this first type of argument with propositions such as the one above, concerning the non-existence of some mathematical object. One assumes that such an object exists, and then proves that this would lead to a contradiction; thus, such an object does not exist. For other examples, see the proof that the square root of 2 is not rational and Cantor's diagonal argument.

On the other hand, it is also common to use arguments of the second type concerning the existence of some mathematical object. One assumes that the object doesn't exist, and then proves that this would lead to a contradiction; thus, such an object must exist. Although it is quite freely used in mathematical proofs, not every school of mathematical thought accepts this kind of argument as universally valid. See further nonconstructive proof.

### In mathematical logic

In mathematical logic, the proof by contradiction is represented as:

If
${\displaystyle S\cup \{P\}\vdash \mathbb {F} }$
then
${\displaystyle S\vdash \neg P.}$

or

If
${\displaystyle S\cup \{\neg P\}\vdash \mathbb {F} }$
then
${\displaystyle S\vdash P.}$

In the above, P is the proposition we wish to disprove respectively prove; and S is a set of statements, which are the premises — these could be, for example, the axioms of the theory we are working in, or earlier theorems we can build upon. We consider P, or the negation of P, in addition to S; if this leads to a logical contradiction F, then we can conclude that the statements in S lead to the negation of P, or P itself, respectively.

Note that the set-theoretic union, in some contexts closely related to logical disjunction (or), is used here for sets of statements in such a way that it is more related to logical conjunction (and).

## Notation

Proofs by contradiction sometimes end with the word "Contradiction!". Isaac Barrow and Baermann used the notation Q.E.A., for "quod est absurdum" ("which is absurd"), along the lines of Q.E.D., but this notation is rarely used today.[1] A graphical symbol sometimes used for contradictions is a downwards zigzag arrow "lightning" symbol (U+21AF: ↯), for example in Davey and Priestley.[2] Others sometimes used include a pair of opposing arrows (as ${\displaystyle \rightarrow \!\leftarrow }$ or ${\displaystyle \Rightarrow \!\Leftarrow }$), struck-out arrows (${\displaystyle \nleftrightarrow }$), a stylized form of hash (such as U+2A33: ⨳), or the "reference mark" (U+203B: ※).[3][4] The "up tack" symbol (U+22A5: ⊥) used by philosophers and logicians also appears, but is often avoided due to its usage for orthogonality.

## Exercises

Exercise 1.1

Prove the following statements using contradiction, that is to say find an example where the contrary is false.

## Quotations

In the words of G. H. Hardy (A Mathematician's Apology), "Reductio ad absurdum, which Euclid loved so much, is one of a mathematician's finest weapons. It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game."

## References

1. Hartshorne on QED and related
2. B. Davey and H.A. Priestley, Introduction to lattices and order, Cambridge University Press, 2002.
3. The Comprehensive LaTeX Symbol List, pg. 20. http://www.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf
4. Gary Hardegree, Introduction to Modal Logic, Chapter 2, pg. II–2. http://people.umass.edu/gmhwww/511/pdf/c02.pdf