Euclid's lemma: Difference between revisions
imported>Michael Underwood No edit summary |
imported>Michael Hardy |
||
Line 6: | Line 6: | ||
Let ''a'', ''b'', ''p'' <math>\in\mathbb{Z}</math> with ''p'' prime, and | Let ''a'', ''b'', ''p'' <math>\in\mathbb{Z}</math> with ''p'' prime, and | ||
suppose that ''p'' is a divisor of ''ab'', ''p''|''ab''. | suppose that ''p'' is a divisor of ''ab'', ''p''|''ab''. | ||
Then there exists an [[integer]] ''n'' such that | Then there exists an [[integer]] ''n'' such that ''ab = pn''. But this means that the fraction | ||
''ab=pn''. But this means that the fraction | |||
:<math>\frac{ab}{p}=n</math> | :<math>\frac{ab}{p}=n</math> | ||
must be an integer as well. | |||
Now, since ''p'' is prime the [[greatest common divisor]] of ''p'' and ''a'' is either 1 or ''p''. | must be an integer as well. Now, since ''p'' is prime the [[greatest common divisor]] of ''p'' and ''a'' is either 1 or ''p''. If it is ''p'' then the proof is complete since ''p''|''a'', so suppose that ''p'' does not divide ''a''. In this case we write | ||
If it is ''p'' then the proof is complete since ''p''|''a'', so suppose that ''p'' does not divide ''a''. | |||
In this case we write | |||
:<math>a\frac{b}{p}=n\ ,</math> | :<math>a\frac{b}{p}=n\ ,</math> | ||
and since gcd(''a'',''p'')=1 and ''n'' is an integer, ''b/p'' must also be an integer, say ''m''. | |||
But then ''b=mp'' so ''p'' divides ''b''. | and since gcd(''a'', ''p'') = 1 and ''n'' is an integer, ''b/p'' must also be an integer, say ''m''. | ||
But then ''b = mp'' so ''p'' divides ''b''. | |||
Therefore in either case we have ''p'' dividing (at least) one of ''a'' and ''b'', and the proof is complete. | Therefore in either case we have ''p'' dividing (at least) one of ''a'' and ''b'', and the proof is complete. | ||
[[category:Mathematics Workgroup]] | [[category:Mathematics Workgroup]] |
Revision as of 19:26, 3 August 2007
In number theory, Euclid's lemma, named after the ancient Greek geometer and number theorist Euclid of Alexandria, states that if a prime number p is a divisor of the product of two integers, ab, then either p is a divisor of a or p is a divisor of b (or both).
Euclid's lemma is used in the proof of the unique factorization theorem, which states that a number cannot have more than one prime factorization.
Proof of the lemma
Let a, b, p with p prime, and suppose that p is a divisor of ab, p|ab. Then there exists an integer n such that ab = pn. But this means that the fraction
must be an integer as well. Now, since p is prime the greatest common divisor of p and a is either 1 or p. If it is p then the proof is complete since p|a, so suppose that p does not divide a. In this case we write
and since gcd(a, p) = 1 and n is an integer, b/p must also be an integer, say m. But then b = mp so p divides b.
Therefore in either case we have p dividing (at least) one of a and b, and the proof is complete.