1. ## Proof by Contradiction help, no clue, lost

Proof by contradictionTheorem: The reciprocal of any irrational number is irrational.

2. ## Re: Proof by Contradiction help, no clue, lost

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

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

What can you do with that?

3. ## Re: Proof by Contradiction help, no clue, lost

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

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

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