Difference between revisions of "Euclid's Lemma"

(deleted fallacious proof; mentioned converse in proof; edited a little)
m (Proof of Euclid's Lemma)
Line 8: Line 8:
 
Without loss of generality, suppose <math> \gcd(p,a)=1</math> (otherwise we are done).  By [[Bezout's Lemma]], there exist integers such that <math>x,y</math> such that <math>px+ay=1</math>. Hence <math>b(px+ay)=b</math> and <math>pbx+aby=b</math>. Since <math> p\mid p</math> and <math> p \mid ab </math> (by hypothesis), <math>p \mid pbx + aby =b</math>, as desired.
 
Without loss of generality, suppose <math> \gcd(p,a)=1</math> (otherwise we are done).  By [[Bezout's Lemma]], there exist integers such that <math>x,y</math> such that <math>px+ay=1</math>. Hence <math>b(px+ay)=b</math> and <math>pbx+aby=b</math>. Since <math> p\mid p</math> and <math> p \mid ab </math> (by hypothesis), <math>p \mid pbx + aby =b</math>, as desired.
  
On the other hand, if <math>p>1</math> is not prime, then it must be composite, i.e., <math>p=ab</math>, for integers <math>a,b</math> both greater than 1.  Then <math>p\nmid a</math> and <math>p\nmid b</math>. Thus the lemma's converse holds as well. <math>\blacksquare</math>
+
On the other hand, if <math>p>1</math> is not prime, then it must be composite, i.e., <math>p=ab</math>, for integers <math>a,b</math> both greater than 1.  Then, <math>p\nmid a</math> and <math>p\nmid b</math>. This completes the proof.
  
 
==See Also==
 
==See Also==

Revision as of 04:18, 20 November 2023

Euclid's Lemma is a result in number theory attributed to Euclid. It states that:

A positive integer $p>1$ is a prime number if and only if $p \mid ab$ implies that $p \mid a$ or $p\mid b$, for all integers $a$ and $b$.


Proof of Euclid's Lemma

Without loss of generality, suppose $\gcd(p,a)=1$ (otherwise we are done). By Bezout's Lemma, there exist integers such that $x,y$ such that $px+ay=1$. Hence $b(px+ay)=b$ and $pbx+aby=b$. Since $p\mid p$ and $p \mid ab$ (by hypothesis), $p \mid pbx + aby =b$, as desired.

On the other hand, if $p>1$ is not prime, then it must be composite, i.e., $p=ab$, for integers $a,b$ both greater than 1. Then, $p\nmid a$ and $p\nmid b$. This completes the proof.

See Also