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Math Help - Proof of irrational number

  1. #1
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    Proof of irrational number

    Show that sqrt(3/2) is irrational. What I have so far is (sqrt(3/2))2=a2/b2, so than 3/2=a2/b2. So how does this prove irrational
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  2. #2
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    Re: Proof of irrational number

    $\dfrac{3}{2} = \dfrac{a^2}{b^2}$ is the same as saying:

    $2a^2 = 3b^2$, for integers $a,b$.

    Since $2$ divides $2a^2$, it must divide $3b^2$, and since $2$ does not divide $3$, it must be that $2$ divides $b^2$, and thus $2$ divides $b$.

    So suppose $k > 0$ is the largest power of $2$ that divides $b$, so that $b = 2^km$ so that $b^2 = 2^{2k}m^2$.

    Thus $2a^2 = 3(2^{2k}m^2)$, so that $a^2 = 2^{2k-1}(3m^2)$. Now $k > 0$, so $2k - 1 \geq 1 > 0$. Thus $2$ divides $a^2$, and thus $a$.

    If $a = 2^ts$, where $t > 0$ is the largest power of $2$ that divides $a$, then:

    $a^2 = 2^{2t}s^2 = 2^{2k-1}(3m^2)$.

    Now $s^2$ is an integer, and $2$ does not divide $s^2$, so we conclude that:

    $s^2 = 2^{2k - 2t - 1}(3m^2) = 3m^2$, that is: $2k - 2t - 1 = 0$. But this is the same as: $2(k - t) = 1$, and $1$ is not divisible by $2$.
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  3. #3
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    Re: Proof of irrational number

    Thanks I get it now.
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