Corollary on Divisors of prime

Question

Good day! I'm currently working on the divisors of a prime along with its properties. Eventually, I encountered this corollary: If p is a prime, and $p^{a}|p^{b}$ if and only if $0\leq a \leq b$ it seems obvious at first but as i tried to prove, i was hang into something unthinkable and still don't know if I'm still doing OK...any help? Thanks a lot in advance!

Answer 1

The fact that $p$ is a prime is irrelevant. Suppose $p$ is any integer at all. It is also required for $a$ and $b$ to both be non-negative integers for this problem to make sense and be correct.

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Suppose that $a$ and $b$ are non-negative integers satisfying $0\leq a\leq b$

Then $b-a$ is a nonnegative integer implying that $p^{b-a}$ is an integer.

Then $p^b = p^a\cdot p^{b-a}$ implying by definition of divisibility that $p^a\mid p^b$

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Suppose instead that $a$ and $b$ are non-negative integers not satisfying $0\leq a\leq b$. That implies that they instead satisfy $0\leq b<a$

Then $b-a$ is an integer satisfying $b-a<0$ and therefore $p^{b-a}$ is not an integer.

Since the only possibility for $k$ in $p^b=p^a\cdot k$ is $k=p^{b-a}$ is not an integer, this implies that there is no valid integer choice for $k$ to make the equality. This implies then that $p^a\not\mid p^b$, implying by contrapositive that $p^a\mid p^b\implies 0\leq a\leq b$

Still to prove or correctly cite: why is $\frac{1}{n}$ not an integer for $n$ an integer other than $\pm 1$. Proving this will imply the step that $p^{b-a}$ is not an integer in the case that $b-a<0$. Also needed is knowledge that for $n$ an integer and $m$ a non-negative integer that $n^m$ is also an integer. This follows from the closure of integers under multiplication and induction.