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Thread: prove that root 6 is irrational

  1. #1
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    prove that root 6 is irrational

    plzz prove that root 6 is irrational.

    i have to show the whole prove not just say that 6 is not a perfect square so root 6 is irrational. i have to prove it wholly.

    thanks in advance
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    Suppose $\displaystyle \sqrt{2\cdot3} = n/m$. Then $\displaystyle 2\cdot3\cdot m^2 = n^2$. Factor both sides into primes. Then the power of 2 on the RHS is even because it is a square, but the power of 2 on the LHS is odd. This contradicts unique factorization.
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    extremely sryy but i can't understand it properly plzz give me detail.
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  4. #4
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    Quote Originally Posted by saha.subham View Post
    plzz prove that root 6 is irrational.

    i have to show the whole prove not just say that 6 is not a perfect square so root 6 is irrational. i have to prove it wholly.

    thanks in advance
    Or note that $\displaystyle \sqrt{6}$ is a solution to $\displaystyle x^2-6=0$ and the only rational solutions to that are...

    P.S. It follows that $\displaystyle \sqrt{2}+\sqrt{3}\notin\mathbb{Q}$
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    srryy but i can;t understand without details
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  6. #6
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    But I thought you said you had prove it. What do you understand about proving that roots are not rational?
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  7. #7
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    haha! If you square root anything but a perfect square then the answer will be irrational!

    PROVEN!!
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  8. #8
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    Quote Originally Posted by jgv115 View Post
    haha! If you square root anything but a perfect square then the answer will be irrational!

    PROVEN!!
    I have a feeling that if the OP is going to say that then that statement has to be proven in the process.


    Assume that $\displaystyle \sqrt{6}$ is rational.

    Then $\displaystyle \sqrt{6} = \frac{a}{b}$, where $\displaystyle a, b$ are integers, and $\displaystyle \frac{a}{b}$ is in simplest form (since every rational number can be reduced to simplest form).


    Therefore $\displaystyle 6 = \frac{a^2}{b^2}$

    $\displaystyle 6b^2 = a^2$

    $\displaystyle 2\cdot 3b^2 = a^2$.


    Since $\displaystyle a^2$ is even, that means $\displaystyle a$ is even. So write it as $\displaystyle a = 2c$. Then you have

    $\displaystyle 2\cdot 3b^2 = (2c)^2$

    $\displaystyle 2\cdot 3b^2 = 4c^2$

    $\displaystyle 3b^2 = 2c^2$

    $\displaystyle b^2 = 2\left(\frac{c^2}{3}\right)$.


    Remembering that $\displaystyle b$ is an integer, so is $\displaystyle b^2$. Also, since 2 is a factor, $\displaystyle b^2$ is even, and so $\displaystyle b$ is even. Write it as $\displaystyle b = 2d$.


    Remember that $\displaystyle \sqrt{6} = \frac{a}{b}$

    but $\displaystyle a = 2c$ and $\displaystyle b = 2d$

    So $\displaystyle \sqrt{6} = \frac{2c}{2d}$

    $\displaystyle \sqrt{6} = \frac{c}{d}$.


    But we said that $\displaystyle \frac{a}{b}$ was already in simplest form.

    So we have a contradiction.

    This means that $\displaystyle \sqrt{6}$ can not possibly be rational.

    Therefore $\displaystyle \sqrt{6}$ is irrational.
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