1. ## Define gcd(a,b,c)

Question:

Assuming that the gcd(a,b) of any two integers a and b is defined, show that the gcd(a,b,c) of any three integers is also defined, and that (a,b,c)=((a,b),c).

My Answer so far: ( i think its totally wrong)

Let gcd(a,b)=d so d|b, d|a.

If gcd(a,b,c) = e exists then e|a , e|b and e|c ((1))

We want to prove that
gcd(a,b,c)=gcd(gcd(a,b),c)

=gcd(d,c)
=e

Since (a,b)=d and e|a and e|b => e|d

and from ((1)) e|c

As e|d and e|c we have gcd(d,c)=e

But i think i still need to show that (a,b,c) is definded and im not sure that i proved the statement correctly, i think i need to show that it is the greatest common divisor but i only showed that it was a divisor, im not sure. If anyone has a completly different solution that actually makes sense that would be super helpful!

Thanks

2. Let $\displaystyle d_1 = \gcd(a,b,c) \geq 1$ and $\displaystyle d_2 = \gcd ( \gcd(a,b), c) \geq 1$.

The idea is to show that $\displaystyle d_1 \mid d_2$ and $\displaystyle d_2 \mid d_1$ which implies $\displaystyle d_1 = d_2$.

It shouldn't be too troublesome to see why this is the case.