Let $p$ be a prime number. $ord_p(ab)=ord_p(a)+ord_p(b)$ Let $p$ be a prime number. Prove that $ord_p$ has the following property.
$ord_p(ab)=ord_p(a)+ord_p(b)$. (Thus $ord_p$ resembles the logarithm function, since it converts multiplication into addition!)
In my book, they describe the order as the exponent of the term in the prime factorization. But how do I use this to proof this statement?
 A: Hint $\ \   \overbrace{c\,p^{\large\, j}}^{\large a}\:  \overbrace{d\, p^{\large k}}^{\large b}\, =\, \overbrace{cd\ p^{\large\, j+k}}^{\large ab}.\ $ If $\ p\nmid c,d\ $ then $\ p\nmid cd\ $ (by $\,p\,$ prime)
thus $\ v_p(a)+v_p(b)\, =\, j+k\,=\, v_p(ab)$
Remark $\ $ This can also be deduced directly from the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations). Above is essentially the same but for a single prime $\,p,\,$ i.e. the existence and uniqueness of factorizations of the form $\  n\,p^j\,$ where $\  p\nmid n.$
A: Suppose $\text{ord}_p(a)=m$ and $\text{ord}_p(b)=n$.
Then $a=p^mr$ where $p\not|r$, and $b=p^nq$ where $p\not|q$. So $ab=p^{n+m}$ and $p\not|qr$,
so, $\text{ord}_p(ab)=m+n=\text{ord}_p(a)+\text{ord}_p(b)$.
A: Note that, by the Fundamental Theorem of Arithmetic, there exists unique prime factorizations for $a$ and $b$. If we let $p_{1},p_{2},\ldots$ be all prime numbers, then we may write their factorizations as $$a=p_{1}^{r_{1}}p_{2}^{r_{2}}\cdots$$
and $$b=p_{1}^{s_{1}}p_{2}^{s_{2}}\cdots$$
where all but finitely many of the $r_{i}$ and $s_{i}$ are zero. Then, we have $$ab=\left(p_{1}^{r_{1}}p_{2}^{r_{2}}\cdots\right)\left(p_{1}^{s_{1}}p_{2}^{s_{2}}\cdots\right) = p_{1}^{r_{1}+s_{1}}p_{2}^{r_{2}+s_{2}}\cdots.$$
Finally, if we let $p=p_{i}$ for some $i$, we have that $\text{ord}_{p}\left(a\right)=r_{i}$, $\text{ord}_{p}\left(b\right)=s_{i}$, and $\text{ord}_{p}\left(ab\right)=r_{i}+s_{i}=\text{ord}_{p}\left(a\right)+\text{ord}_{p}\left(b\right)$, as required.
A: Let $a=p^{\alpha} p_1^{\alpha_1}\dots p_n^{\alpha_n}$ and $b=p^{\beta} p_1^{\beta_1}\dots p_n^{\beta_n}$.(some exponents can be zero so the primes coincide)
Then $ab=p^{\alpha+\beta} p_1^{\alpha_1+\beta_1}\dots p_n^{\alpha_n+\beta_n}$.
From here $ord_p(a)+ord_p(b)=\alpha+\beta=ord_p(ab)$
