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Contrapositive proof of irrational relations

by FelixHelix
Tags: contrapositive, irrational, proof, relations
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FelixHelix
#1
Feb7-12, 01:45 PM
P: 28
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
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alexfloo
#2
Feb7-12, 01:55 PM
P: 192
Call that big expression y, and solve for x. If y is rational, what does that tell you about x?
FelixHelix
#3
Feb7-12, 02:09 PM
P: 28
I thought I did solve for x. If y is rational then x is rational but where is the contradiction? I don't see

alexfloo
#4
Feb7-12, 02:35 PM
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.
lavinia
#5
Feb9-12, 07:52 AM
Sci Advisor
P: 1,716
Quote Quote by FelixHelix View Post
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.
Deveno
#6
Feb9-12, 10:36 AM
Sci Advisor
P: 906
Quote Quote by FelixHelix View Post
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.


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