Exponents of Primes in Prime Decomposition are Less iff Divisor

Theorem
Let $$a, b \in \mathbb{Z}_+$$.

Then $$a \backslash b$$ iff every prime in the decomposition of $$a$$ appears in the decomposition of $$b$$ and its exponent in $$a$$ is less than or equal to its exponent in $$b$$.

Proof
Let $$a, b \in \mathbb{Z}_+$$. Let their prime decompositions be:


 * $$a = p_1^{k_1} p_2^{k_2} \ldots p_n^{k_n}$$
 * $$b = q_1^{l_1} q_2^{l_2} \ldots q_n^{l_n}$$


 * Suppose every prime in the decomposition of $$a$$ appears in the decomposition of $$b$$ and its exponent in $$a$$ is less than or equal to its exponent in $$b$$.

Then we have:

where $$k_1 \le l_1, k_2 \le l_2, \ldots, k_r \le l_r, r \le s$$.

Thus $$d = p_1^{l_1-k_1} p_2^{l_2-k_2} \ldots p_r^{l_r-k_r} \in \mathbb{Z}$$ and $$b = a d$$.

So $$a \backslash b$$.


 * Now suppose $$a \backslash b$$.

Let $$a = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$$ be the prime decomposition of $$a$$.

Then $$\forall i \in \mathbb{N}_r: p_i^{k_i} \backslash a$$. Hence by Divides is Ordering on Positive Integers it also divides $$b$$.

Thus $$\exists c \in \mathbb{Z}: b = p_i^{k_i} c$$.

The prime decomposition of $$b$$ is therefore:

$$b = p_i^{k_i} ($$ 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.