for a, b, c, x, y Є Z and a, b != 0
How do I approach this question?
It is very similar to the proof that shows for all d>0 where d is a common divisor of $\displaystyle a,b\epsilon Z$ that $\displaystyle d=gcd(a,b)$ if and only if $\displaystyle d=x*b*ya$ for some $\displaystyle x,y \epsilon Z$
Prove that c is negative if $\displaystyle c=xa+yb<gcd(a,b)$.
Proof:
Suppose $\displaystyle c=xb+ya$ for some $\displaystyle x,y \epsilon Z$
Any common divisor d of a,b divides $\displaystyle xb+ya \implies d|c$
Since d|c we know that |d| is less than or equal to |c|. This means that gcd(a,b)=|c|<gcd(a,b). This means that c must be -gcd(a,b), and since gcd(a,b) is always positive, then c must be negative.
Where gcd = greatest common divisor
|c|= magnitude of c
I am not guaranteeing its correctness but this is the basic idea... you may have to be more descriptive in areas.