# Contrapositive proof of irrational relations

1. Feb 7, 2012

### FelixHelix

I'm confused with a question and wondered if anyone could help explain where I need to go...

let x ε R. Prove that x is irrational thenI'm confused with a question and wondered if anyone could help explain where I need to go...

let x ε R. Prove that x is irrational then ((5*x^(1/3))-2)/7) is irrational.

I've learnt that this is a proof by contrapositive so set ((5*x^(1/3))-2)/7) to be rational and hence a/b with a,b ε Z and b ≠ 0, and there for x is rational.

But I'm not sure where to go from here. Do I rearrange to find for x? x = ((7a - 10b)/5b)^3

But this is still raational isn't it? Where is the contradiction?

Thanks

2. Feb 7, 2012

### alexfloo

Call that big expression y, and solve for x. If y is rational, what does that tell you about x?

3. Feb 7, 2012

### FelixHelix

I thought I did solve for x. If y is rational then x is rational but where is the contradiction? I don't see

4. Feb 7, 2012

### alexfloo

Ah, I missed that step in your initial post.

Remember that contradiction is different from contraposition. For contraposition, you're actually already done.

Contraposition takes advantage of the fact that a->b is equivalent to (not b) -> (not a).

You showed that y is rational implies that x is rational. Therefore, you've shown that x is irrational implies y is irrational.

5. Feb 9, 2012

### lavinia

the rational numbers are closed under multiplication,division, and addition.

6. Feb 9, 2012

### Deveno

by assumption a, and b are INTEGERS, with b non-zero.

well if a and b are integers, then surely k = 7a - 10b is also an integer.

thus k3 and 125b3 are also integers, and moreover, 125b3 ≠ 0.

thus x = k3/(125b3) is rational.

but x was assumed irrational, contradiction.