Welcome to the forum.

You are trying to adapt the root 2 proof I think.

At this step

So let a = 5x

you need to justify that a will have 5 as a factor.

You know that (a squared) has 5 as a factor, so you've got to argue that a single 5 cannot occur as a factor, but rather that they must come in pairs. It's OK, they do, but it is due to the properties of prime number decomposition and square numbers having prime factor pairs.

To show what I'm getting at, try using the same approach to prove that root 4 is irrational. Obviously, it isn't. So where in the proof does root 4 differ from root 2 ?

Oh yes, and one more thing. You need to start by declaring that a/b is a fraction in its lowest terms (is fully cancelled down) so that when 5 is shown to be a factor of both a and b you have a contradiction.

Bob

Bob

]]>so 5 = (a/b) (a/b) or 5x (a²/b² )

5b² = a²

So let a = 5x

a² = 25x

25x = 5b²

5x = b²

I have no idea if this is the right approach but it made sense when I initially wrote it, seems a bit iffy looking back as I dont really know where I was going with this. I think I was going to look for even numbers of prime factors but cant recall why.

]]>Can anyone give advice on how to go about proving route 5 is irrational?

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