√2 is irrational: proof by poem

This entry was posted by on Tuesday, 7 October, 2008 at

i found this scribbled in chalk one day, on the sidewalk outside the University of Chicago math department:

Double a square is never a square,
and here is the reason why:
If m-squared were equal to two n-squared,
then to their prime factors we’d fly.
But the decomposition that lies on the left
had all of its exponents even,
But the power of two on the right must be odd,
so one of the twos is “bereaven”!

Here’s my translation of this poem. It’s a simple proof by contradiction.

  1. Assume sqrt(2) is rational, so sqrt(2) = (m/n)
  2. By squaring both sides and dividing, we get m^2 = 2*n^2
  3. Reduce m and n into sequences of prime factors:

    (p1 * p2 * …. * pv)^2 = 2 * (q1 * q2 * …. * qw)^2

  4. Apply the square to each element in parenthesis:

    p1^2 * p2^2 * … * pv^2 = 2 * ( q1^2 * q2^2 * …. * qw^2)

  5. Examine the right side, remembering that all p’s and q’s are prime numbers. One of the following things must be true:
    • There exists a q = 2. In which case, we can combine our lone “2” into it, giving us a 2^3 term.
    • There is no q = 2. In which case, our power of 2 is simply 1, i.e. 2^1.

    Either way, we’ve proven the the right side contains an odd power of 2.

  6. But, the right side contains only even powers of primes. Contradiction.

Isn’t math fun?


  1. meneame.net