# GCD with remainder theorem proof

I was trying to understand the proof to the GCD theorem. It seems like I'm missing sommething though. I was reading it in this wikiproof article and the first proof uses the GCD with remainder theorem. I didn't really understand it, so I tried to read the proof

gcd{a,b}∖a∧gcd{a,b}∖b            GCD is a common divisor
gcd{a,b}∖(a−qb)          Common Divisor Divides Integer Combination
gcd{a,b}∖r           as r=a−qb
gcd{a,b}≤gcd{b,r}            Definition of gcd{b,r} as the greatest common divisor of b and r

The argument works the other way around:

gcd{b,r}∖b∧gcd{b,r}∖r            GCD is a common divisor
gcd{b,r}∖(qb+r)          Common Divisor Divides Integer Combination
gcd{b,r}∖a           as a=qb+r
gcd{b,r}≤gcd{a,b}            Definition of gcd{a,b} as the greatest common divisor of a and b

Thus  gcd{a,b}=gcd{b,r}


Before we get to how the "argument works the other way around", I didn't understand the last step

gcd{a,b}≤gcd{b,r}

, how do we know the gcd of b,r is greater than that of a,b. Thanks.

First, let's imagine that we have some quantity $q$ such that both of these statements hold: \begin{align} q \backslash r \\ q \backslash b \end{align} That is, $q$ divides $r$, and also $q$ divides $b$. This means that $q$ is a common divisor of $b$ and $r$. Then, because the $\gcd$ means the greatest common divisor, we have: $$q \leq \gcd\left\{b,r\right\}$$
Now the statement you asked about has this statement directly above it: $$\gcd \left\{a,b\right\}\backslash r$$ That says that $\gcd \left\{a,b\right\}$ is a divisor of $r$. But since $\gcd \left\{a,b\right\}$ is also by definition of $\gcd$ a divisor of both $a$ and of $b$, what we have is the situation where $\gcd \left\{a,b\right\}$ is a common divisor of $b$ and $r$. Therefore $$\gcd\left\{a,b\right\} \leq \gcd\left\{b,r\right\}$$
Because, if words have a meaning, the g.c.d. of $a$ and $r$ is the greatest common divisor of $a$ and $r$, and precisely, you've established that $\gcd(a,b)$ is a common divisor of $a$ and $r$.