Euclid's Lemma for Prime Divisors/Proof 2
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Lemma
Let $p$ be a prime number.
Let $a$ and $b$ be integers such that:
- $p \divides a b$
where $\divides$ means is a divisor of.
Then $p \divides a$ or $p \divides b$.
Proof
Let $p \divides a b$.
Suppose $p \nmid a$.
Then from the definition of prime:
- $p \perp a$
where $\perp$ indicates that $p$ and $a$ are coprime.
Thus from Euclid's Lemma it follows that:
- $p \divides b$
Similarly, if $p \nmid b$ it follows that $p \divides a$.
So:
- $p \divides a b \implies p \divides a$ or $p \divides b$
as we needed to show.
$\blacksquare$
Source of Name
This entry was named for Euclid.
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $3$: The Integers: $\S 12$. Primes: Theorem $21 \ \text{(i)}$
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-2}$ Divisibility: Corollary $\text {2-3}$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 23$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 12.5$: Highest common factors and Euclid's algorithm
- 1982: Martin Davis: Computability and Unsolvability (2nd ed.) ... (previous) ... (next): Appendix $1$: Some Results from the Elementary Theory of Numbers: Theorem $8$