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    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

    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.
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Updated: January 19, 2010
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