Please could someone run me through the process of proving why the square root of 3 is irrational. I get the sqrt3 = a/b but when I started further reading I saw odd and even come into the proof and something being a multiple and I basically have no idea. Thanks
Square root of 3 as irrational watch
- Thread Starter
- 18-01-2010 20:03
- 19-01-2010 14:27
You are assuming that it's rational. If this is the case, it can be written as a/b (we can assume that a and b have no common factors (except 1,-1), because if they do we can cancel this factor in the expression above).
Square both sides to get 3b^2 = a^2. This tells us that a^2 is divisible by 3.
Now, every number a is either a) a multiple of 3, b) 1 more than a multiple of 3 or c) 1 less than a multiple of 3.
In case a), this means we can write a=3k for some integer k
In case b), this means we can write a=3k+1 for some integer k
In case c), this means we can write a=3k-1 for some integer k
We have worked out that a^2 is a multiple of 3. So, since
(3k+1)^2 = 9k^2 +6k +1 = 3(3k^2 + 2k) +1
and (3k-1)^2 = 3(3k^2 - 2k) +1
we must have a=3k for some integer k.
But then 3b^2 = 9k^2, so b^2 = 3k^2.
Then, with the same argument as before, we can show that b=3j for some integer j.
But then we have shown that 3 is a factor of both a and b, which contradicts our initial assumption that they had no non trivial common factors.
So root 3 must then be irrational.