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Math Help - proof by contradiction

  1. #1
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    proof by contradiction

    Hi guys, having problems solving this problem. Question is prove by contradiction that \sqrt{3} is irrational. Any help would be highly appreciated.
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  2. #2
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    Ok here is my attempt.
    Suppose that \sqrt{3} is rational. Therefore \sqrt{3} = \frac{m}{n} for some integers m, n with n  \neq 0.
    We may suppose without loss of generality (wlog) that m, n have no common factor (i.e. are coprime). Now m^2 = 3n^2 and so m is even. Put m = 2k. Then 4k^2 = 2n^2 and so n^2 = 2k^2
    which implies that n is even. This contradicts the assumption that m, n are coprime.
    Hence \sqrt{3} must be irrational.

    Is this correct?
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  3. #3
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    Hello, jvignacio!

    You've mixed in 2's, 4's, and even numbers.


    Prove by contradiction that \sqrt{3} is irrational.
    Assume that \sqrt{3} is rational.

    Then: . \sqrt{3} \:=\:\frac{a}{b} for positive integers a and b, where a and b are coprime. .[1]


    Square the equation: . \left(\sqrt{3}\right)^2 \:=\:\left(\frac{a}{b}\right)^2 \quad\Rightarrow\quad 3 \:=\:\frac{a^2}{b^2} \quad\Rightarrow\quad 3b^2 \:=\:a^2

    Since the left side is divisible by 3, the right side is also divisible by 3.
    . . Hence: . a \,=\,3c .[2]


    The equation becomes: . 3b^2 \:=\:(3c)^2 \quad\Rightarrow\quad 3b^2 \:=\:9c^2 \quad\Rightarrow\quad b^2 \:=\:3c^2

    Since the right side is divisible by 3, the left side is also divisible by 3.
    . . Hence: . b \,=\,3d .[3]


    But [2] and [3] say that \text{GCD}\,(a,b) \,=\,3 . . . which contradicts [1].

    . . Q.E.D.

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  4. #4
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    Quote Originally Posted by Soroban View Post

    Since the left side is divisible by 3, the right side is also divisible by 3.
    . . Hence: . a \,=\,3c .[2]
    dont understand how you got to the a = 3c part?

    thanks
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  5. #5
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    Square the equation: .

    Since the left side is divisible by 3, the right side is also divisible by 3.
    . . Hence: . .
    dont understand how you got to the a = 3c part?
    That's because, if a^2 is divisible by 3, then so is a. In order for 3 to divide a\cdot a, one could try breaking 3 into two factors: 3=pq so that p\mid a and q\mid a, but pq\not\;\mid a ( x\mid y means " x divides y"). However, 3 is prime, and this is impossible.

    Since 3\mid a, we have 9\mid a^2, and we can divide both sides of 3b^2=a^2 by 3.
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