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Math Help - Proof by Contradiction help, no clue, lost

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    Proof by Contradiction help, no clue, lost

    Proof by contradictionTheorem: The reciprocal of any irrational number is irrational.
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    Re: Proof by Contradiction help, no clue, lost

    Quote Originally Posted by zhengcl86 View Post
    Proof by contradictionTheorem: The reciprocal of any irrational number is irrational.
    Suppose that \gamma is an irrational number. You know that \gamma\ne 0. WHY?

    Now suppose that \frac{1}{\gamma} is rational.

    What can you do with that?
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    Re: Proof by Contradiction help, no clue, lost

    Lemma: If  \alpha \ne 0 is rational, then  \frac{1}{\alpha} is rational.

    Proof: Let  \alpha \ne 0 be rational. Then  \alpha = \frac{p}{q} where  p, q are integers with  gcd(p, q) = 1 and  q \ne 0 . Then  \frac{1}{\alpha} = \frac{1}{\frac{p}{q}} = \frac{q}{p} . Thus,  gcd(p, q) = gcd(q, p) = 1 and since  \alpha \ne 0 we see that  p \ne 0 . Thus, \frac{1}{\alpha} is rational.

    Now let  \beta be irrational and assume by contradiction that  \frac{1}{\beta} is rational. Then by the Lemma,  \frac{1}{\frac{1}{\beta}} is rational. But,  \beta = \frac{1}{\frac{1}{\beta}} which means a rational number equals an irrational number, a contradiction. Thus, \frac{1}{\beta} is irrational.
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