How do you prove this question?

Q: Prove that if x is irrational, then 1/x is irrational?

I am thinking Proof by contradiction but I am not sure. Even where to begin?

- Jan 28th 2009, 03:37 PMGrillakisDiscrete Math - Prove that if x is irrational, then 1/x is irrational?
How do you prove this question?

Q: Prove that if x is irrational, then 1/x is irrational?

I am thinking Proof by contradiction but I am not sure. Even where to begin?

- Jan 28th 2009, 03:39 PMMush
Contradiction works.

Hint: If we were to assume that 1/x WAS rational, then it could be written as a ratio of integers p/q.

$\displaystyle \frac{1}{x} = \frac{p}{q} $

$\displaystyle q = xp $

If x is irrational, and 1/x is rational, then we are saying that the rational number q can be written as a product of an irrational number x, and a rationa number. Is this valid? - Jan 28th 2009, 03:59 PMGrillakis
- Jan 28th 2009, 05:25 PMMush
Not quite.

Proof by contradiction works when we assume the opposite of what we want to proove is true. We then derive an erroneous result... a contradiction.

Our problem is... if x is irrational, prove that 1/x is irrational. The first thing we do is assume the opposite. Assume that x is irrational, but conversely, 1/x is rational. From this we will try to derive an impossibility.

A property of a rational number is that it can be written as a ratio of integers. ANY rational number can be written in the form $\displaystyle \frac{p}{q} $, where p and q are both integers.

So if we assume that 1/x is rational then:

$\displaystyle \frac{1}{x} = \frac{p}{q} $

Now multiply both sides by xq

$\displaystyle q = px $

What this statement says is that the integer q can be written as a product of the integer p, and the irrational number q. THIS is invalid, because multiplying an integer by an irrational number gives an irrational number. Hence q can't be an integer, and our initial assumption MUST be wrong.

Hence 1/x is not rational. And any number that is not rational is irrational. - Jan 28th 2009, 06:32 PMGrillakis
- Jan 28th 2009, 08:13 PMConstatine11
- Jan 29th 2009, 01:20 AMMush
But it does not imply that. If we were to assume that x was rational, then you would indeed be correct. But our method maintains that x is irrational, but assumes that 1/x is irrational.

So the next step is to observe that 1/x = p/q, such that p and q are integers. And this leads the contradiction q = px, which cannot be. - Jan 29th 2009, 03:26 AMConstatine11