Exponents of Primes in Prime Decomposition are Less iff Divisor
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Theorem
Let $a, b \in \Z_{>0}$.
Then $a \divides b$ if and only if:
- $(1): \quad$ every prime $p_i$ in the prime decomposition of $a$ appears in the prime decomposition of $b$
and:
Proof
Let $a, b \in \Z_{>0}$.
Let their prime decomposition be:
- $a = p_1^{k_1} p_2^{k_2} \dotsm p_n^{k_n}$
- $b = q_1^{l_1} q_2^{l_2} \dotsm q_n^{l_n}$
Necessary Condition
Let:
- $(1): \quad$ prime in the prime decomposition of $a$ appear in the prime decomposition of $b$
and:
Then:
\(\ds a\) | \(=\) | \(\ds p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}\) | ||||||||||||
\(\ds b\) | \(=\) | \(\ds p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r} \dotsm p_s^{l_s}\) |
where:
- $k_1 \le l_1, k_2 \le l_2, \dotsc, k_r \le l_r, r \le s$
Thus:
- $d = p_1^{l_1 - k_1} p_2^{l_2 - k_2} \dotsm p_r^{l_r - k_r} \in \Z$
and so:
- $b = a d$
So $a \divides b$.
$\Box$
Sufficient Condition
Let $a \divides b$.
Let $a = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$ be the prime decomposition of $a$.
Then:
- $\forall i \in \N_r: p_i^{k_i} \divides a$
Hence by Divisor Relation on Positive Integers is Partial Ordering each $p_i^{k_i}$ also divides $b$.
Thus:
- $\exists c \in \Z: b = p_i^{k_i} c$
The prime decomposition of $b$ is therefore:
- $b = p_i^{k_i} \times \paren {\text {prime decomposition of $c$} }$
which may need to be rearranged.
So $p_i$ must occur in the prime decomposition of $b$ with an exponent at least as big as $k_i$.
The result follows.
$\blacksquare$