Proof by contrapositive

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September 14th 2010, 03:19 PM
robasc

Proof by contrapositive

I am having a terrible time understanding this one:

Prove that for all x E r, and y E R, if x is rational and y is irrational, then x + y is irrational. use proof by contradiction known as contrapositive.

OK I understand that contrapositive (~p --> ~q) so

Proof: if x is not rational then y is not irrational

Now I have no idea what to do next?

Please help, I have a test coming up in two days and I need to grasp this stuff.(Headbang)
September 14th 2010, 03:24 PM
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Quote:

Originally Posted by robasc

I am having a terrible time understanding this one:

Prove that for all x E r, and y E R, if x is rational and y is irrational, then x + y is irrational. use proof by contradiction known as contrapositive.

OK I understand that contrapositive (~p --> ~q) so

Proof: if x is not rational then y is not irrational

Now I have no idea what to do next?

Please help, I have a test coming up in two days and I need to grasp this stuff.(Headbang)

So assume the negation of the conclusion.

Original conclusion: x + y is irrational
Negation of it: x + y is rational

So under this assumption you can write x + y = r, where r is in Q (rationals). Now what happens when you solve for y?
September 14th 2010, 03:42 PM
robasc

I have no idea, I am assuming that the answer will be false because a rational number and a irrational number will conclude to be an irrational number but how do I prove that?
September 14th 2010, 03:45 PM
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Quote:

Originally Posted by robasc

I have no idea, I am assuming that the answer will be false because a rational number and a irrational number will conclude to be an irrational number but how do I prove that?

What does "answer will be false" mean?

x + y = r with r in $\mathbb{Q}$ , by assumption
$\implies y = r - x$

recall that $\mathbb{Q}$ is closed under + and -. Thus y is rational. Contradiction.