Difference between revisions of "Euclid's Lemma"

Revision as of 00:05, 15 November 2007

In Number Theory, the result that

A positive integer $p$ is a prime number if and only iff $p|ab \Longrightarrow p|a$ or $p|b$

is attributed to Euclid

Proof of Euclid's lemma

There are two proofs of Euclid's lemma.

First Proof By assumption $\gcd(a,b)=1$ , thus we can use Bezout's lemma to find integers $x,y$ such that $ax+by=1$ . Hence $c\cdot(ax+by)=c$ and $acx+bcy=c$ . Since $a\mid a$ and $a \mid bc$ (by hypothesis), we conclude that $a \mid acx + bcy =c$ as claimed

Second Proof We have $a\vert bc$ , so $bc=na$ , with $n$ an integer. Dividing both sides by $a$ , we have $\frac{bc}{a}=n$ But $\gcd(a,b)=1$ implies $b/a$ is only an integer if $a=1$ . So $\frac{bc}{a} = b \frac{c}{a} = n$ which means $a$ must divide $c$ .

@@ Line 5: / Line 5: @@
 is attributed to [[Euclid]]
-{{stub}}
+==Proof of Euclid's lemma==
+There are two proofs of Euclid's lemma.
+First Proof
+By assumption <math> \gcd(a,b)=1</math>, thus we can use Bezout's lemma to find integers <math> x,y</math> such that <math> ax+by=1</math>. Hence <math> c\cdot(ax+by)=c</math> and <math> acx+bcy=c</math>. Since <math> a\mid a</math> and <math> a \mid bc </math> (by hypothesis), we conclude that <math> a \mid acx + bcy =c </math> as claimed
+Second Proof
+We have <math> a\vert bc</math>, so <math> bc=na</math>, with <math> n</math> an integer. Dividing both sides by <math> a</math>, we have
+<math>\frac{bc}{a}=n</math>But <math> \gcd(a,b)=1</math> implies <math> b/a</math> is only an integer if <math> a=1</math>. So
+<math>\frac{bc}{a} = b \frac{c}{a} = n </math>
+which means <math> a</math> must divide <math> c</math>.

Page

Toolbox

Search

Difference between revisions of "Euclid's Lemma"

Revision as of 00:05, 15 November 2007

Proof of Euclid's lemma