For $x < y$ and $x$ and $y$ are coprime, is it true that $y-x$ and $y$ are coprime?

Question

I'm working Problem 72 on Project Euler and had this thought that if $x/y$ is a proper fraction then naturally $(y-x)/y$ is as well such as 3/8 and 5/8 or 4/15 and 11/15. Growing up I've always noticed this symmetry but never have I questioned it from this perspective.

I know the fraction can't be reduced if $GCD(x, y) = 1$ or, in other terms, if $x$ and $y$ are coprime. So, while in the world of fractions I understand that if $x/y$ is a proper fraction it means that $(y-x)/y$ is also proper, I can't make sense of why $x$ and $y$ being coprime means that $y-x$ and $y$ are coprime. I don't know of any properties of coprime numbers that would make this true without thinking of fractions where it just seems magically true.

Is my assumption true? If so, how can it be presented so I can wrap my head around it?

Yes. $\gcd(x,y)=\gcd(x-y,y)=\gcd(y-x,y)$ using Euclidean Algorithm. — user236182, Oct 05 '15 at 17:45

score 1 · Accepted Answer · answered Oct 05 '15 at 17:48

1

If $x$ and $y$ are coprime, but a prime $p$ divides $y-x$ and $y$, then $p$ also divides $y-(y-x)=x$; contradiction.

answered Oct 05 '15 at 17:48

ajotatxe

65,084

For $x < y$ and $x$ and $y$ are coprime, is it true that $y-x$ and $y$ are coprime?

1 Answers1