Question

Show that the value given is irrational
Sqrt5 (H)

Answer 1

To prove: The square root of 5 is irrational. In other words, there is no rational number whose square is 2.

Proof by contradiction: Begin by assuming that the thesis is false; that is, that there does exist a rational number whose square is 2.

By definition of a rational number, that number can be expressed in the form c/d, where c and d are integers, and d is not zero. Moreover, those integers, c and d, have a greatest common divisor, and by dividing each by that GCD, we obtain an equivalent fraction a/b that is in lowest terms: a and b are integers, b is not zero, and a and b are relatively prime (their GCD is 1).

Now we have (a/b)^2 = 5, so a^2 = 5b^2

If b is even then b^2 is even and so 5b^2 is even. If b is odd then b^2 is odd, but 5b^2 is even. So since 5b^2 is even, a^2 is even. Thus a is even.

So we can say a = 2r and thus (2r)^2 = 5b^2
4r^2 = 5b^2
.8r^2 = b^2

Since .8r^2 is even, b^2 must be even, and thus b is even.

But since both a and b are even, a/b is not in its lowest terms which contradicts the assumption that a and b are
relatively prime. Therefore the assumption is incorrect, and there must NOT be a rational number whose square is 5.