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 (the square root of 2). We shall show Sqrt is irrational.
Suppose, to the contrary, that Sqrt were rational. Then Sqrt=m/n for some integers m, n in lowest terms, i.e., m and n have no common factors. Then 2=m2/n2, which implies that m2=2n2. Hence m2 is even, which implies that m is even. Then m=2k for some integer k.
So 2=(2k)2/n2, but then 2n2 = 4k2, or n2 = 2k2. So n2 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 is rational) is false, so the Sqrt must be irrational.
This is a classic proof by contradiction.
The Math Behind the Fact:
You may wish to try to prove that Sqrt is irrational using a similar technique. It is also instructive to see why this proof fails for Sqrt (which is clearly rational). The above proof fails for Sqrt because at the point in the proof where we deduce that m2 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>.
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