Why is the square root of 27 irrational?

Answer 1

There is a general form of such proofs based on the method of reduction ad absurdum. We assume the opposite of what we wish to prove and, using logical steps, show that it leads to a contradiction. It then follows that the supposition must be false.
The following proof is somewhat abbreviated.

So, suppose the square root of 27 is rational. Therefore it can be written as a ratio of two integers in its lowest form.

So suppose sqrt(27) = p/q where p and q have no factors in common, and q > 0.

Then 27 = p2/q2 or 27q2 = p2.Now 3 divides the left hand side so 3 must divide the right hand side. Since 3 is a prime, 3 must divide p.So let p = 3r where r is an integer. And since p and q are coprime, q and r are also coprime.

Substituting p = 3r gives

27q2 = (3r)2 = 9r2

that is 3q2 = r2

Repeat the same argument to show that 3 must divide r and so r = 3s.

So 3q2 = (3s)2 = 9s2

So q2 = 3s2

Now 3 divides the right hand side and therefore must divide the left hand side. And then, because 3 is a prime, it must divide q.

But that means p and q have a common factor: 3, which is a contradiction. Therefore the supposition, that sqrt(27) is rational, must be false.