# Discrete Math - Prove that if x is irrational, then 1/x is irrational?

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• Jan 28th 2009, 03:37 PM
Grillakis
Discrete 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 PM
Mush
Quote:

Originally Posted by Grillakis
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?

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 PM
Grillakis
Quote:

Originally Posted by Mush
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?

it is not valid b/c our assumption of 1/x being rational is false. Is this correct?
• Jan 28th 2009, 05:25 PM
Mush
Quote:

Originally Posted by Grillakis
it is not valid b/c our assumption of 1/x being rational is false. Is this correct?

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 PM
Grillakis
Quote:

Originally Posted by Mush
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.

I see now...thanks Mush
• Jan 28th 2009, 08:13 PM
Constatine11
Quote:

Originally Posted by Mush
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?

The next step should be to observe that your assumption that $\displaystyle 1/x$ is rational implies that there exist integers $\displaystyle p$ and $\displaystyle q$ such that:

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

.
• Jan 29th 2009, 01:20 AM
Mush
Quote:

Originally Posted by Constatine11
The next step should be to observe that your assumption that $\displaystyle 1/x$ is rational implies that there exist integers $\displaystyle p$ and $\displaystyle q$ such that:

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

.

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 AM
Constatine11
Quote:

Originally Posted by Mush
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.

If q=px then:

x=q/p

which is a direct contradiction to the assumed irrationality of x since q/p is by definition rational.

.