# Contrapositive proof of irrational relations

by FelixHelix
Tags: contrapositive, irrational, proof, relations
 P: 22 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
 P: 192 Call that big expression y, and solve for x. If y is rational, what does that tell you about x?
 P: 22 I thought I did solve for x. If y is rational then x is rational but where is the contradiction? I don't see
P: 192

## Contrapositive proof of irrational relations

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.
P: 1,667
 Quote by 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
the rational numbers are closed under multiplication,division, and addition.
P: 905
 Quote by 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
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.

 Related Discussions General Math 2 Calculus 2 Calculus 1 Precalculus Mathematics Homework 7 Calculus & Beyond Homework 4