An *irrational* number is a number that cannot be expressed as a fraction. But are there any irrational numbers?

It was known to the ancient Greeks that there were lengths that could not be expressed as a fraction. For instance, they could show that a right triangle whose side lengths (adjacent to the right angle) are both 1 has a hypotenuse whose length is not a fraction. By the Pythagorean theorem this length is Sqrt[2] (the square root of 2). We shall show Sqrt[2] is irrational.

Suppose, to the contrary, that Sqrt[2] were rational. Then Sqrt[2]=m/n for some integers m, n in *lowest terms*, i.e., m and n have no common factors. Then 2=m^{2}/n^{2}, which implies that m^{2}=2n^{2}. Hence m^{2} is even, which implies that m is even. Then m=2k for some integer k.

So 2=(2k)^{2}/n^{2}, but then 2n^{2} = 4k^{2}, or n^{2} = 2k^{2}. So n^{2} is even. But this means that n must be even, because the square of an odd number cannot be even.

We have just showed that both m and n are even, which contradicts the fact that m, n are in lowest terms. Thus our original assumption (that Sqrt[2] is rational) is false, so the Sqrt[2] must be irrational.

**Presentation Suggestions:**

This is a classic proof by contradiction.

**The Math Behind the Fact:**

You may wish to try to prove that Sqrt[3] is irrational using a similar technique. It is also instructive to see why this proof fails for Sqrt[4] (which is clearly rational). The above proof fails for Sqrt[2] because at the point in the proof where we deduce that m^{2} is divisible by 4, we cannot conclude that m is divisible by 4.

**How to Cite this Page:**

Su, Francis E., et al. “Square Root of Two is Irrational.” *Math Fun Facts*. <https://www.math.hmc.edu/funfacts>.

**Fun Fact suggested by:**

Francis Su