Let $p$ be a prime, and $a$ and $b$ two integers, such that $p\not|$ $a$ and $p\not|$ $b$.
It can be possible that $p$ $|$ $ab$ ?
Asked
Active
Viewed 968 times
1
-
What are your thoughts? Have you tried some examples? Notice any patterns? – John Hughes Dec 14 '16 at 16:13
-
I think it is false :D! – penguina Dec 14 '16 at 16:13
-
Great! Do you have an example that makes you think this? – John Hughes Dec 14 '16 at 16:15
-
@JohnHughes: An example is not quite going to suffice in this case... – barak manos Dec 14 '16 at 16:18
-
1@penguina: Let $[a_1,\dots,a_m]$ be the prime factors of $a$ and $[b_1,\dots,b_n]$ be the prime factors of $b$. What can you say about the prime factors of $ab$? – barak manos Dec 14 '16 at 16:21
-
It will be $[a_1,..,a_m,b_1,..,b_n]$, and I see it now. It will be false, p will not be any of these factors. Thank you! – penguina Dec 14 '16 at 16:26
-
@penguina: Keep in mind that some of these factors may be common to both $a$ and $b$, so the correct way of phrasing it would be using set notation (e.g., let $A={x\mid x\text{ is a prime factor of }a}$ and $B={y\mid y\text{ is a prime factor of }b}$, then $A\cup B$ is the set of prime factors of $ab$)... – barak manos Dec 14 '16 at 16:31
-
Yes, you are right! :D – penguina Dec 14 '16 at 16:32
4 Answers
1
Hint: The contrapositive of
$(p|ab) \Rightarrow (p|a$ or $p|b)$
answers your question.
Rodrigo Dias
- 4,310
- 1
- 12
- 29
-
Right... It's like proving $x \implies y$ by assuming $\neg y \implies \neg x$... But the latter is not given in the question, so you still need to prove it. With your method, we could just as well prove that every integer is a multiple of $5$, by assuming that if a number is not a multiple of $5$ then it is not an integer. – barak manos Dec 14 '16 at 16:24
-
It's just the definition of prime element. Given a ring $R$ and an element $x\in R$, we say that $x$ is an prime element iff $x|ab \Rightarrow x|a$ or $x|b$ for $a,b\in R$. – Rodrigo Dias Dec 14 '16 at 16:29
1
Hint. Recall that a prime $p$ divides an integer $a$ iff $p$ is in the integer factorization (which is unique) of $a$. If you know the integer factorizations of $a$ and $b$, what is the integer factorization of their product? What may we conclude?
Robert Z
- 145,942
0
No, it is not possible. If $p\not|a$ and $p\not|b$ then $p$ is not in prime factorisation of $a$ and $b$. Hence, $p$ is not in prime factorisation of $ab$
Vidyanshu Mishra
- 10,143
0
In $\mathbb{N}$ we have a unique factorization, it means $a=p_1^{n_1}...p_k^{n_k}$. So if $p$ doesn't appear in factorization of a or b, it means it doesn't appear in $ab$.
-
Side note: for the specific question at hand, the multiplicity of each prime factor (i.e., $n_1,\dots,n_k$) is not important. – barak manos Dec 14 '16 at 16:33