Proof: Suppose $a,b \in \mathbb{N}$. Then a = gcd(a,b) iff $a|b$.

Question

So I've constructed a proof for this answer and I'm having trouble completeing it. To prove iff statements we need to prove the statment $P \rightarrow Q$ then $Q \rightarrow P$. Convert the statement into a conditional statement then prove the conditional statement and prove its converse.

If $a = \textrm{gcd}(a,b)$ then $a|b$. Assume $a = \textrm{gcd}(a,b)$ by definition of $\textrm{gcd}$ $a$ is a divisor of both $a$ and $b$. So by definition of divides $a|b$ must be true.

Prove conversely, Assume $a|b$ by definition of divides $a(n)=b$ and from our definition of $\textrm{gcd}$ then b(k) = a for some integer $n$ and $k$. Consider, $a(n)(k) = a = n(k) = 1$ by the identity property we know that $(n)(k)$ are both 1. Considering that $n=1$ and $k =1$ and $a$ divides $b$ then...

Is the above statement correct? What does $n = 1$ and $k = 1$ tell me about about $a$?

How do you reach $b(k) = a$ from the definition of gcd? Try $a = 2, b = 4$ — Manifoldski, Sep 20 '23 at 17:11
No, it seems like you are confusing $lcm(a,b)$ (least common multiple, the smallest $c$ such that $b\mid c$ and $a\mid c$) with $gcd(a,b)$(the largest $d$ such that $d \mid a$ and $d\mid b$) — Manifoldski, Sep 20 '23 at 17:33
@ThomasAndrews the natural numbers. I can see it in black board bold, latex. — Ziggy, Sep 20 '23 at 17:47
@Ziggy Usually, it should be capital $\mathbb N.$ My browser did not make it clear it was not just $n.$ — Thomas Andrews, Sep 20 '23 at 18:23
As here (case $r=0),,$ if $,a\mid b,$ then $,a,b,$ and $,a,$ have the same set $S$ of common divisors so they have the same greatest common divisor $(= \max S = \gcd(a) = a)\ \ $ — Bill Dubuque, Sep 20 '23 at 19:05
The above actually proves both directions since the converse holds too, i.e. with $\rm cd = $ common divisors, $\gcd(a,b) = a,\overset{!\rm\color{#c00}U}\Rightarrow, {\rm cd}(a,b) = {\rm cd}(a),\Rightarrow, a\mid b,$ (by $,a\in {\rm cd}(a)= {\rm cd}(a,b)\Rightarrow a\mid b),\ $ where we used that $,d\mid \gcd(a,b)!\iff! d\mid a,b,,$ i.e. $\rm\color{#c00}U=$ gcd universal property. But w/o knowing this property, for this easy direction it's simpler to just use $,a=\gcd(a,b)\mid b,$ as you did. — Bill Dubuque, Sep 20 '23 at 19:58

Proof: Suppose $a,b \in \mathbb{N}$. Then a = gcd(a,b) iff $a|b$.

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