Chapter 2.4, Problem 78E

Solution

To find: $\sqrt{2}$ is rational number.

The explanation is given.

Given information

In this case, we start by supposing that v2 is a rational number. Thus, there will exist integer p and q (where q is non-zero) such that $\frac{p}{q}=\sqrt{2}$ . We also make the assumption that p and q have no common factors. As even if they have common facors we would cancel them to write it in the simplest form. So, let us assume that p and q are co-prime number here having no common factor other than 1.

Now, squaring both sides, it gives $\frac{p^2}{q^2}=2$ which can be rewritten as,

  $p^2=2q^2\mathrm{...}(1)$

Note that the right-hand side of the equation is multiplied by 2, which means that the left-hand side is a multiple of 2. So, it can be said that p² is a multiple of 2. This further means that p itself must be a multiple of 2, as when a prime number is a factor of a number, let's say, m², it is also a factor of m. Thus, we can assume that,

p=2m, m∈Z [Set of Integers]

⇒ (2m)2= 2q² [From (1)]

⇒ 4m²=2q²

⇒ q²=2m²

Now, the right-hand side is a multiple of 2 again, which means that the left-hand side is a multiple of 2, which further means that q is a multiple of 2, i.e., q = 2n, where n ∈ Z. We have thus shown that both p and q are multiples of 2. But is that possible? This can only mean one thing: our original assumption of assuming v2 as $\frac{p}{q}$ (where p and q are co-prime integers) is wrong:

  $\sqrt{2}\ne \frac{p}{q}$

Thus, v2 does not have a rational representation v2 is irrational.