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Contrapositive proof of irrational relations 
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#1
Feb712, 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 


#2
Feb712, 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?



#3
Feb712, 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



#4
Feb712, 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. 


#5
Feb912, 07:52 AM

Sci Advisor
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#6
Feb912, 10:36 AM

Sci Advisor
P: 906

well if a and b are integers, then surely k = 7a  10b is also an integer. thus k^{3} and 125b^{3} are also integers, and moreover, 125b^{3} ≠ 0. thus x = k^{3}/(125b^{3}) is rational. but x was assumed irrational, contradiction. 


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