Euclid's Lemma/Proof 2

From ProofWiki
Jump to: navigation, search

Theorem

Let $a, b, c \in \Z$.

Let $a \divides b c$, where $\divides$ denotes divisibility.


Let $a \perp b$, where $\perp$ denotes relative primeness.

Then $a \divides c$.


Proof

Let $a, b, c \in \Z$.

We have that $a \perp b$.

That is:

$\gcd \left\{{a, b}\right\} = 1$

where $\gcd$ denotes greatest common divisor.

From Bézout's Lemma, we may write:

$a x + b y = 1$

for some $x, y \in \Z$.

Upon multiplication by $c$, we see that:

$c = c \left({a x + b y}\right) = c a x + c b y$


Now note that $c a x + c b y$ is an integer combination of $a c$ and $b c$.

So, since:

$a \mathrel \backslash a c$

and:

$a \mathrel \backslash b c$

it follows from Common Divisor Divides Integer Combination that:

$a \mathrel \backslash \left({c a x + c b y}\right)$


However:

$c a x + c b y = c \left({a x + b y}\right) = c \cdot 1 = c$

Therefore:

$a \mathrel \backslash c$

$\blacksquare$


Source of Name

This entry was named for Euclid.


Also see


Sources