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Math Help - Contradiction proof

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    Contradiction proof

    I have to prove the square root of 3 is irrational using proof by contradiction.
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    Quote Originally Posted by natewalker205 View Post
    I have to prove the square root of 3 is irrational using proof by contradiction.
    Say p/q = \sqrt{3} where p,q are reduced. Then p^2 = 3q^2 since RHS is divisible by 3 it means LHS is divisible by 3 but then p^2 makes LHS divisible by 3 and so q needs to be divisible by 3. Thus, it is impossible, by infinite descent.
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    Quote Originally Posted by natewalker205 View Post
    I have to prove the square root of 3 is irrational using proof by contradiction.
    Say that \sqrt{3} is not irrational. Then it can be put into the form:
    \sqrt{3} = \frac{p}{q}
    where p and q are integers ( q \neq 0). For the sake of the argument, let this fraction be in lowest terms.

    The we have that
    p^2 = 3q^2

    Now, this implies the 3 divides p^2 evenly. But we assumed that the fraction \frac{p}{q} is already in lowest terms, so 3 does not divide q. So the prime factorization of p^2 only contains a single 3, which is impossible.

    Thus etc.

    -Dan
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