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

Printable View

- May 6th 2013, 09:14 AMzhengcl86Proof by Contradiction help, no clue, lost
Proof by contradictionTheorem: The reciprocal of any irrational number is irrational.

- May 6th 2013, 09:35 AMPlatoRe: Proof by Contradiction help, no clue, lost
- May 6th 2013, 09:44 AMmathguy25Re: 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.