Slight deviation of proof, would it be correct for integers? Review for exam

[mr_coffee]

Hello everyone.

He told us he could of course change the parameters which he will of the proofs we have been working on so I'm testing out some cases but I want to make sure I'm doing it right.

Here is an example of a proof the boook had:
http://suprfile.com/src/1/3qaagxh/Untitled-1%20copy.jpg


Now where he has the statement: "If a and b are rational numbers..."
I'm changing that to:
"If a and b are integers..."
and now here is my proof, i think its correct but I have to make sure.

If a and b are integers, b != 0, and r is an irrational number, then a+ br is irrational.

Proof by Contradiction:

Suppose not. Suppose that a and b are integers, b != 0, and r is an irrational number such that a+br is rational. We must obtain a contradiction.

Since a, b are integers and a + br are rational, a+br = m/n for some integers m, n with n != 0.

Then

a + br = m/n

br = m/n - a
r = (m-an)/bn

where (m-an) and (bn) are integers since m, a, n, and b are integers, and bn is nonzero since b is nonzero. Therefore, r is rational, contradicting that r is irrational.

Thanks!

 

[courtrigrad]

looks good to me. are you using the book my solow? btw, you you go to UPark?

 

[mr_coffee]

I'm using Discrete Mathematics with appplications 3rd by Susanna S. Epp, and yeah i go to Penn State UPark. CSE260 is the class, hah how did you know.

 

[courtrigrad]

just a guess/facebook/google(thinkquest contest). many of my friend go there.

 

[mr_coffee]

yeah this place is pretty massive, well thanks for the help! I seem to post all questions but don't answer any hah. I did do thinkquest along time ago... I'm supprised that site is still up! well ttyl

 

[HallsofIvy]

Since all integers are rational numbers, once you had proven it for a, b rational, exactly the same proof must hold for a, b integer. Actually, I wouldn't have written the rational numbers as fractions, i/j, etc. It is sufficient to use the fact that the rational numbers are closed under addition and multiplication:
Suppose a+ br, a, b rational b non-zero, is rational: a+ br= c where c is rational. Then br= c- a so r= (c-a)(1/b). Since b is a non-zero rational number 1/b is rational and so r= (c-a)(1/b) is rational- contradiction.

 

[mr_coffee]

That does seem a lot nicer, thanks!