# Tag Archives: arithmetic

## the greatest common divisor exists!

for the sake of all of us, let us assert and demonstrate:

$\forall a,b\in{\mathbb{Z}}$ the $GCD(a,b)$ exists and its is unique

Proof:

Consider the auxiliar set $T_{ab}=\{am+bn>0\ :\ m,n\in{\mathbb{Z}}\}$

as a piece of ${\mathbb{N}}$ there exists $c=\min\ T_{ab}$.

But if $c\in T_{ab}$ then $c=a\mu+b\nu$ for some $\mu,\nu\in{\mathbb{Z}}$.

We are going to check that $c$ satisfy the $GCD$ for $a,b$:

Claim: $c|a$

Applying the divisions’ algorithm to this  pair we get

$a=ct+r$ with $0\le r

Since $r=a-ct=a-(a\mu+b\nu)t=a(1-\mu t)+b(-\nu t)\ge0$

If $r>0$ then $r\in T_{ab}$ and $0.

This contradicts the choice for $c$, then $r=0$ and the claim follows.

Similarly $c|b$.

So $c$ is a common divisor of $a,b$.

Uniqueness:

Now suppose that $d$ is a positive integer that $d|a$, $d|b$ and for any other $l$ which divides $a,b$ implies $l|d$.

Then, since $c$ divides both $a,b$ then $c|d$.

Additionally $d|a,\ d|b$ implies $d|a\mu,\ d|b\nu$ and $d|(a\mu+b\nu)$, hence $d|c$

Then $c|d$ and $d|c$. That is $d=\pm c$, but $d=c$ is the only possible.

$\Box$

Filed under cucei math, group theory, numbers

## division’s algorithm

let us write for the sake of young mathematicians this important arithmetic fact:

PROPOSITION:

$\forall a,b\in{\Bbb{Z}}\ \exists t,r\in{\Bbb{Z}}\ : a=bt+r,\quad 0\le r<|b|$

PROOF:

Let $a>b>0$ be the case and we set

$S_{ab}=\{a-bm\ge0\ |\ m\in{\Bbb{Z}}\}$

then there exists

$\exists$ $r=\min S_{ab}$.

If

$r=0$

then

$a=bt$.

For the rest of the argument we consider

$r>0$.

So

$r=a-bt,\qquad \exists t\in{\Bbb{Z}}$

and $a=bt+r$.

We are going to establish

$0\le r< b$.

Suppose

$r\ge b$

$r-b\ge0$

$0\le r-b

$0\le (a-bt)-b

if

$0

this contradicts the choice for   $r$

then

$0\le r

Now for uniqueness:

suppose

$a=bt+r,\qquad 0\le r

and

$a=bt'+r',\qquad 0\le r'

$r\ge r'$

$r-r'\ge 0$

but

$b(t'-t)=r-r'$ and

$b(t-t')=-r'+r>-r>-b$

are  integer multiples of $b$ between

$-b$ and $b$

then

$b(t-t')=0$

and

$t'-t=0$

since $b\neq0$.

So $t=t'$ and $r=r'$

$\Box$