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<c

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<r<c.

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

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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<r

0\le (a-bt)-b<r

if

0<a-b(t+1)<r

this contradicts the choice for   r

then

0\le r<b

Now for uniqueness:

suppose

a=bt+r,\qquad 0\le r<b

and

a=bt'+r',\qquad 0\le r'<b

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

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