Pythagoras's Theorem

Proves that the Diagonal $d$ of a Square with sides of integral length $s$ cannot be Rational. Assume $d/s$ is rational and equal to $p/q$ where $p$ and $q$ are Integers with no common factors. Then

$$d^2=s^2+s^2=2s^2,$$

so

$$\left({d\over s}\right)^2=\left({p\over q}\right)^2=2,$$

and $p^2=2q^2$, so $p^2$ is even. But if $p^2$ is Even, then $p$ is Even. Since $p/q$ is defined to be expressed in lowest terms, $q$ must be Odd; otherwise $p$ and $q$ would have the common factor 2. Since $p$ is Even, we can let $p\equiv 2r$, then $4r^2=2q^2$. Therefore, $q^2=2r^2$, and $q^2$, so $q$ must be Even. But $q$ cannot be both Even and Odd, so there are no $d$ and $s$ such that $d/s$ is Rational, and $d/s$ must be Irrational.

In particular, Pythagoras's Constant $\sqrt{2}$ is Irrational. Conway and Guy (1996) give a proof of this fact using paper folding, as well as similar proofs for $\phi$ (the Golden Ratio) and $\sqrt{3}$ using a Pentagon and Hexagon.

See also Irrational Number, Pythagoras's Constant, Pythagorean Theorem

References

Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, pp. 183-186, 1996.

Pappas, T. ``Irrational Numbers & the Pythagoras Theorem.'' The Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, pp. 98-99, 1989.