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=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[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 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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Francis Su