Math Help - b=aq+r then gcd(b,a)=gcd(a,r)

1. b=aq+r then gcd(b,a)=gcd(a,r)

If $b=aq+r$, then the gcd(a,b)=gcd(a,r)

Not to sure with what to do here.

2. \begin{aligned} \gcd(a, b) & = ax+by \\& = ax+(aq+r)y \\& = ax+aqy+ry \\& = a(x+qy)+ry \\& = \gcd(a, r). \end{aligned}

3. This was not quite clear to me, so I found a way that might be easier

* c|b notation reads c divides b
Let (b,a) = c and (a,r) = d

c|b and c|a, then c|b-aq
r = b - aq, so c|r
c (less than or equal to) d

d|a and d|r, then d|aq + r
b = aq + r, so d|b
d (less than or equal to) c

Therefore d=c