How do you prove that the square root of 7 is an irrational number?

Answer 1

The proof is by the method of reductio ad absurdum.

We start by assuming that sqrt(7) is rational.

That means that it can be expressed in the form p/q where p and q are co-prime integers.

Thus sqrt(7) = p/q.

This can be simplified to 7*q^2 = p^2

Now 7 divides the left hand side (LHS) so it must divide the right hand side (RHS).

That is, 7 must divide p^2 and since 7 is a prime, 7 must divide p.

That is p = 7*r for some integer r.

Then substituting for p gives,

7*q^2 = (7*r)^2 = 49*r^2

Dividing both sides by 7 gives q^2 = 7*r^2.

But now 7 divides the RHS so it must divide the LHS.

That is, 7 must divide q^2 and since 7 is a prime, 7 must divide q.

But then we have 7 dividing p as well as q which contradict the requirement that p and q are co-prime.

The contradiction implies that sqrt(7) cannot be rational.