Prove that there is no rational number whose square is 8?

Answer 1

The proof is based on proving that sqt(2) is irrational, by the method of reductio ad absurdum.

We start by assuming that sqrt(2) is rational.

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

Thus sqrt(2) = p/q.

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

Now 2 divides the left hand side (LHS) so it must divide the right hand side (RHS).

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

That is p = 2*r for some integer r.

Then substituting for p gives,

2*q^2 = (2*r)^2 = 4*r^2

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

But now 2 divides the RHS so it must divide the LHS.

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

But then we have 2 dividing p as well as q which contradicts the requirement that p and q are co-prime.

The contradiction implies that sqrt(2) cannot be rational.

Finally, sqrt(8) = 2*sqrt(2) and the product of a rational and an irrational is irrational. Therefore sqrt(8) is irrational.

Alternatively, you could go through 2 more cycles of the above process of 2 divides LHS/RHS and so must divide the other side, etc.