# Euclid's Lemma

Jump to navigation
Jump to 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 1

Follows directly from Integers are Euclidean Domain.

$\blacksquare$

## Proof 2

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

We have that $a \perp b$.

That is:

- $\gcd \set {a, b} = 1$

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:

- $c = c \paren {a x + b y} = c a x + c b y$

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

So, since:

- $a \divides a c$

and:

- $a \divides b c$

it follows from Common Divisor Divides Integer Combination that:

- $a \divides \paren {c a x + c b y}$

However:

- $c a x + c b y = c \paren {a x + b y} = c \cdot 1 = c$

Therefore:

- $a \divides c$

$\blacksquare$

## Source of Name

This entry was named for Euclid.

## Also see

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $2$: Some Properties of $\Z$: Exercise $2.5$