The

square root of 2, denoted

sqrt(2), is an

irrational number.

Proof:

Suppose sqrt(2) is a

rational number. Then sqrt(2)=a/b for some

integers a and b. Assume that a/b is in

lowest terms; that is, assume a and b have no

common factors other than 1 or -1.

sqrt(2)=a/b => b*sqrt(2)=a => 2b

^{2}=a

^{2}
So a is

even because 2 is a factor and by

Lemma below. Therefore a=2c for some integer c. So a

^{2}=(2c)

^{2}=4c

^{2} and since a

^{2}=2b

^{2}, we have 4c

^{2}=2b

^{2}. It follows that 2c

^{2}=b

^{2}. Thus b

^{2} is even. By

earlier result, since b is an integer and b

^{2} is even, it follows that b is even. Thus b=2d for some integer d. But the fact that 2 is a factor of both a and b is a

contradiction. Therefore the

assumption that sqrt(2) is rational is

incorrect. Thus sqrt(2) is not rational.

Q.E.D.
Lemma: Let a be an integer and let a

^{2} be even. Then a is even.