Proof about gcd and remainders

Question

Possible Duplicate:
Why is $\gcd(a,b)=\gcd(b,r)$ when $a = qb + r$?

Prove: $\gcd(a, b) = \gcd(a+bq, b)$

Proof:

By the division algorithm we know:

$a = b\cdot q + r \iff r = a -b\cdot q$

Thus, we see it is equivalent to prove $\gcd(a, b) = \gcd(b, r)$

To me the claim seems straight forward, since the term $a+bq$ in the $\gcd(a+bq, b)$ is "dependent" on the $a$. That is, $\gcd(bq, b) = b$, so adding an $a$ really determines the $\gcd(a+bq, b)$.

How can I formalize this and use more mathematical terms?

Answer 1

I would suggest proving the result like this:

(i) Show that if $x$ divides $a$ and $b$, then $x$ divides $a+bq$ and $b$.

(ii) Show that if $x$ divides $a+bq$ and $b$, then $x$ divides $a$ and $b$.

Once you have done this, you will have shown that the set of common divisors of $a$ and $b$ is the same as the set of common divisors of $a+bq$ and $b$.

From this the result should follow easily. The final details depend on your official definition of $\gcd$. If, as is likely in your course, it is defined as the largest common divisor, then the result follows immediately. For since the sets of common divisors are the same, their largest element must be the same.

Answer 2

Hint $\ $ If $\rm\:c\:|\:b\:$ then $\rm\: c\:|\:a\iff c\:|\:a\!+\!b\,q.\:$ Thus $\rm\:a,\,b\:$ and $\rm\:a\!+\!b\,q,\,b\:$ have the same set C of common divisors $\rm\:c,\:$ hence they have the same greatest common divisor (= max C). $\ $ QED

Or: $\rm\: mod\ c\!:\ if\ \ b\equiv 0\ \ then \ \ a\equiv 0\iff a+b\,q\equiv 0,\ $ using congruence language.

Answer 3

One strategy to show that two positive integers are equal is to show that they divide each other. This is common in number theory.

Let $a, b, q\in\mathbb{N}$. Let $d_1=gcd(a+bq, b)$ and $d_2=gcd(a,b)$. We show that $d_1\mid d_2$ and $d_2\mid d_1$; thus showing they are equal.

Since $d_1=gcd(a+bq, b)$ we have that $d_1\mid b$ and $d_1\mid a+bq$ but since $d_1\mid b$ we have $d_1\mid bq$ thus $d_1\mid a$ (You can show this by writing down the congruence). Since $d_1\mid a$ and $d_1\mid b$ we have $d_1\mid d_2$.

Since $d_2=gcd(a,b)$ we have $d_2\mid b$ and $d_2\mid a$. Since $d_2\mid b$ we have $d_2\mid bq$ and hence $d_2\mid a+bq$ and $d_2\mid b$ hence $d_2\mid d_1$.