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## Homework Statement

Prove by contradiction: If a and b are rational numbers and b != 0, and r is an irrational number, then a+br is irrational.

In addition, I am to use only properties of integers, the definitions of rational and irrational numbers, and algebra.

You guys should also know that I am new to proofs, so if I'm breaking convention in any blatant way, please lemme know.

## The Attempt at a Solution

First, I figure I need to find the negation of this statement:

Negation: There exists rational numbers a and b, b!=0, and irrational number r, such that a+br is rational.

I'm pretty sure this negation is correct, but I've been wrong before

Anyway, my proof would start like this:

Proof: To prove that for any rational numbers a,b,b!=0, and irrational number r, a+br is irrational, let's suppose not. Suppose there exists rational numbers a,b,b!=0 and irrational number r such that a+br is rational.

Let c,d,e,f be integers, by definition of rational, a+br can be rewritten as:

[tex]\frac{c}{d}[/tex] + [tex]\frac{e}{f}[/tex]*r is rational.

[tex]\frac{e}{f}[/tex]*r = -[tex]\frac{c}{d}[/tex]

r = -[tex]\frac{cf}{de}[/tex]

[tex]\frac{cf}{de}[/tex] can be written as a quotient of integers, so it is rational, therefore, our negation is false, therefore, our theorem is true?

I think I'm on the right track, but I know its messy and I might have made some logic mistakes...Also, my professor said that if we didn't have to explicitly say why b!=0, we probably weren't doing it right, and I never really did, so I have no idea...any help would be greatly appreciated.

edit-I guess if b was 0, that would make the quantity a+

**(br)**0, which leaves you with a, which is just a rational number