Euclid's lemma
Dependencies:
- GCD is the smallest Linear Combination Used in proof
Let $p$ be a prime. If $p \mid ab$, then $p \mid a \vee p \mid b$.
Proof by contradiction
Assume $p \not\mid a \wedge p \not\mid b$.
\begin{align} &\Rightarrow \gcd(p, a) = 1 \\ &\Rightarrow \exists r \exists s \textrm{ such that } pr + as = 1 \\ &\Rightarrow prb + asb = b = p(rb) + s(ab) \\ &\Rightarrow p \mid b \Rightarrow \bot \end{align}
This is a contradiction, which means $p \mid a \vee p \mid b$.
Dependency for:
- ϕ is multiplicative
- ab is coprime to x iff a and b are coprime to x
- Fundamental Theorem of Arithmetic Used in proof
- Zp is an integral domain
- Eisenstein's criterion
- Gauss' Lemma
Info:
- Depth: 2
- Number of transitive dependencies: 2