I posted this in puzzles but think it belongs here... my bad.
Can anyone give advice on how to go about proving route 5 is irrational?
only thought that comes to mind is that route 5 minus 1 over 2 is golden ratio, and route 5 plus 1 over 2 is the other golden ratio.
igloo myrtilles fourmis
The way I tackled this so far was to say √5 = a/b
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.
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.
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The square root of 5 is an irrational number because the square root is ? x ? = _. So ? x ? = 5? Also, everything besides the square root of 2, 3, 5, 7, 8, 10, Pi, and lots of other numbers can be expressed as a fraction and a decimal. However, irrational numbers are never ending. For example, the square root of 5 goes on for 2.2360679774997896964091736687313... and it goes on forever.
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