2

This is the way I prove $(b, m) = (b + mx, m)$:

Suppose $(b, m) = t$.

Then $b = pt$, $m = qt$, where $(p, q) = 1$.

$b + mx = pt + qtx$.

So $(b + mx, m) = (pt + qtx, qt) = t = (b, m)$

But if I am given $(b,m)$, I can't come up with $(b, m) = (b + mx, m)$ right away.

I need some better explanation about the relation between $(b, m)$ and $(b + mx, m)$.

davidlowryduda
  • 91,687
yashirq
  • 505

2 Answers2

1

If $x$ is an integer, then $(b,m) = (b+mx,m)$.

If $c = (b,m)$ you can easily show that $c\mid (b+mx)$ and $c\mid m$. So $c$ is a common divisor of both $b+mx$ and $m$. Thus $(b,m) \leq (b+mx,m)$ (by definition of $\gcd$).

On the other hand if $d = (b+mx,m)$, then first note $d\mid m$. Then $d\mid b$ as well, since $d\mid (b+mx)$ and $d\mid mx$. So $(b+mx,m) \leq (b,m)$.

Combining both inequalities yields $(b,m) = (b+mx,m)$.

Camille
  • 678
  • I know this is another way to prove it. But my problem is if I am given $(b, m)$ alone, the other part $(b + mx, m)$ won't even come up to my mind. You know if we are give $a > b$, we will think $b < a$ very intuitively. – yashirq Nov 01 '16 at 10:59
  • Intuitively, anything that divides $b$ and $m$ simultaneously must divide $b+mx$ and $m$ simultaneously, and vice-versa. So the GCD remains unchanged under addition of any multiple (including negatives) of either number, be it $b$ or $m$, to any of the numbers. Does this help? – Camille Nov 01 '16 at 14:59
  • Ya, I get the point. – yashirq Nov 01 '16 at 15:30
  • @yashirq I have explained a geometric viewpoint in my answer. – Bill Dubuque Nov 01 '16 at 15:31
1

Key idea: we can replace a gcd argument with one that yields the same set of common divisors.

Note if $\,d\mid m\,$ then $\,d\mid b\iff d\mid b\!+\!mx.\,$ Thus $\,m,b\,$ and $\,m,\,b\!+\!mx\,$ have the same set $S$ of common divisors $\,d,\,$ so they have the same greatest common divisor $(= \max S)$.

Using modular alrithmetic we can express this equationally, enabling a geometrical view:

${\rm mod\ d}\!:\ $ if $\, m\equiv 0\,$ then $\ b\equiv 0\iff b+mx\equiv 0,\,$ i.e. the systems $\,m\equiv 0,b\equiv 0\,$ and $\,m\equiv 0,\, b+mx\equiv 0\,$ are equivalent.

Geometrically, if we rename to customary variables, then the equivalent systems of equations is $\ x\equiv 0, y\equiv 0\iff x\equiv 0,\, y+mx\equiv 0.\,$ This says the the origin $\, (x,y) \equiv (0,0)\,$ can be specified either as the intersection of the $y$ axis $(x\equiv 0)$ with the $x$ axis $(y\equiv 0)$ or, equivalently, as the intersection of the $y$ axis $(x\equiv 0)$ with the line $\, y\equiv -mx.\,$

If you know linear algebra you will recognize this equivalence as the effect of an elementary linear transformation. This connection between linear algebra and number theory will be made more explicit when one studies module theory in abstract algebra.

Bill Dubuque
  • 272,048