Proving that no cancellation is possible in the fraction (a+c)/(b+d), if ab-cd=1

**mbmstudent** · February 27th 2011, 10:40 AM

Hi,

I have the following problem, but I don't know what to do. I have been told by someone to think about gcd(a+c, b+d), but I still don't know how to prove this. Thanks for your help!

"Prove that for any positive integers a, b, c, d satisfying

$ab-cd=1$

no cancellation is possible in the fraction

$(a+c)/(b+d)$ , where the result is a rational number."

**DrSteve** · February 27th 2011, 10:52 AM

The set of rationals is closed under addition and nonzero division. Thus there is no way that the final expression you gave could be irrational.

**mbmstudent** · February 27th 2011, 11:00 AM

I am sorry, I got confused with another exercise so I typed it wrong... It is supposed to be rational :-/

**DrSteve** · February 27th 2011, 11:47 AM

Perhaps start by solving the equation for a, and substituting to get $\frac{1+c(b+d)}{b(b+d)}$ .

**Bruno J.** · February 27th 2011, 05:21 PM

What you want to show is that $a+c, b+d$ are relatively prime.

It suffices to notice that $b(a+c)-c(b+d)=1$ .

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