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.
<http://www.math.hmc.edu/funfacts>.
