1. ## (a,b,c) = ((a,b),c)

Hey! the question is:
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).

This is what i have done:

let d1 = ((a,b),c)

where (a,b) = gcd(a,b) and is defined

hence let e = (a,b)

thus, d1 = (e,c)
since this is the gcd of two numbers, then the gcd(e,c)is also defined.

hence, d1 = ((a,b), c) is defined

now let d = (a,b,c)

d = ax0 +by0 +cz0
let a = d1f, b = d1g, c = d1h
then,

d = d1fx0 + d1gy0 + d1hz0
= d1(fx0 + gy0 + hz0)
therefore, d | d1

i know i need to show that d=(a,b,c) = ((a,b),c)=d1, and dat for this to occur d|d1 and d1|d, but im not sure if what i have done is right especially this part:
d = ax0 +by0 +cz0
let a = d1f, b = d1g, c = d1h
then,

d = d1fx0 + d1gy0 + d1hz0
= d1(fx0 + gy0 + hz0)
therefore, d | d1

Plus i also dont know how to continue -_-'.

2. You got the right idea and not much work is really needed. All you need is the fact that any common divisor of two numbers divides the GCD of those two numbers.

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

You want to show $d_1 \mid d_2$ and $d_2 \mid d_1$ which will imply $d_1 = d_2$.

By definition, we have: $d_1 \mid a, \ d_1 \mid b, \ d_1 \mid c$. But this means $d_1 \mid \gcd (a,b)$. But this also means $d_1$ is a common divisor of both $\gcd(a,b)$ and $c$ which means $d_1 \mid d_2$.

Can you prove the other case?

3. Originally Posted by o_O
You got the right idea and not much work is really needed. All you need is the fact that any common divisor of two numbers divides the GCD of those two numbers.

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

You want to show $d_1 \mid d_2$ and $d_2 \mid d_1$ which will imply $d_1 = d_2$.

By definition, we have: $d_1 \mid a, \ d_1 \mid b, \ d_1 \mid c$. But this means $d_1 \mid \gcd (a,b)$. But this also means $d_1$ is a common divisor of both $\gcd(a,b)$ and $c$ which means $d_1 \mid d_2$.

Can you prove the other case?

... hang on don't you need to prove/show that gcd(a,b,c) exists/is defined? or can u just assume?

really sorry im starting to get confused