Proof that Square Root of Three is Irrational


In The Intuitive Proof of the Infinitude of Primes, I showed you a proof strategy called proof by contradiction.  In this post, we use this strategy to prove that \sqrt{3} is irrational.  In proof by contradiction, if the statement P is true, you have to assume the contrary, and then find a contradiction somewhere. Note that the proof in this post is very similar to the proof that \sqrt{2} is irrational.

square root of three

Theorem:  \sqrt{3} is irrational.


We assume the contrary; that is, we assume that \sqrt{3} is rational. Then

\sqrt{3} = \displaystyle\frac{a}{b}

where a and b are integers with no common factor except 1. This means that \frac{a}{b} is in lowest terms.

Squaring both sides, we have

3 = \displaystyle\frac{a^2}{b^2}.

 which implies that

3b^2 = a^2.

Since 3 is a factor of 3b^2, this means that a^2 is divisible by 3. Now, if a^2 is divisible by 3 then a is also divisible by 3 (we still have to prove this, but you can try) (1).

Now, since a is divisible by 3, 3 is a factor of a. That is, a = 3k, where k is some integer. Now,

3b^2 = a^2 \implies 3b^2 = (3k)^2 \implies 3b^2 = 9k^2

Dividing both sides by 3, we have b^2 = 3k^2 which means that b^2 is divisible by 3. This implies that b is also divisible by 3 (2).

Now, in (1), a is divisible by 3 and in (2) b is also divisible by 3 This contradicts the fact that a and b have no common factors except 1.

Therefore, the assumption is false. So, \sqrt{3} is irrational. \blacksquare

By the way, David Feinberg has written a nice poem about the Square Root of Three.

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