Let $a, b, d \in\mathbb{Z}$. Suppose that $d \mid ab$. Prove that there exists $a_1, b_1 \in \mathbb{Z}$ such that $d = a_1 b_1$ and $a_1 \mid a$ and $b_1 \mid b$.
Asked
Active
Viewed 75 times
0
-
1Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – YuiTo Cheng Feb 24 '20 at 11:42
-
It seems very obvious but here is a constructive proof. lets write $d$ as product of its primes and the same for $a$ and $b$. now each prime in $d$ is also in either $a$ or $b$. This implies you can write $d$ as a product of primes either in $a$ or $b$. QED. – Aven Desta Feb 24 '20 at 11:47
1 Answers
0
Consider $$a=p_1^{\alpha _1}...p_n^{\alpha _n}\quad \text{and}\quad b =q_1^{\beta _1}...q_m^{\beta _m},$$ decomposition of $a$ and $b$ in prime numbers. Then $$d\mid ab\iff d=p_1^{\gamma _1}...p_n^{\gamma _n}q_1^{\delta _1}...q_m^{\delta _m},$$ for some $\gamma _i\in \{0,...,\alpha _i\}$ and $\delta _i\in\{0,...,\beta _i\}.$
Take $$a_1=p_1^{\gamma _1}...p_n^{\gamma _n}\quad \text{and}\quad b_1=q_1^{\delta _1}...q_m^{\delta _m}.$$ The claim follows.
J. W. Tanner
- 60,406
Surb
- 55,662