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?

**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\
mid c,d\ $ then $\ p\
mid 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 $\ p\
mid n.$