-1

So I am supposed to be proving that if $p_1$ and $p_2$ are distinct primes and $p_1\mid a$ and $p_2\mid a$ then $p_1p_2\mid a$, and I need to use Euclid's Lemma except as far as I understand Euclid's lemma is the converse of this statement and I have tried for the last few hours to work with Euclid's and GCDs to figure this one out and I just don't know where to start since I can't wrap my head around this one. Can anyone help me out?

Bill Dubuque
  • 272,048
Jade Rose
  • 11
  • 2
    Please use MathJax to typeset mathematics. Here’s a tutorial – Arturo Magidin Apr 17 '21 at 23:02
  • 6
    Say $p_1|a$. Write $a=p_1b$. Then $p_2|p_1b$ and $\gcd(p_2,p_1)=1$. Therefore... – Arturo Magidin Apr 17 '21 at 23:02
  • Please search first for answers to avoid posting duplicate question. Please delete this question since we alteady have too many on this topic. – Bill Dubuque Apr 18 '21 at 00:45
  • 1
    @BillDubuque I'm sorry but I checked out other questions and still couldn't work out the answers so I feel as though my question isn't a duplicate and getting these answers helped me. – Jade Rose Apr 18 '21 at 02:01
  • But the proof in the answer you accepted is already given in the linked dupes. It is best for the site if you delete the duplicate question (you will need to unacccept first). – Bill Dubuque Apr 18 '21 at 02:04
  • 1
    Just because it has the same answer does not mean the questions were the same :(, if I myself could not find the answer then maybe someone else also could not find the answer. I understand limiting questions that are duplicates are needed but I struggled to figure this out until I got an answer in my own question to help me understand using the language I am used to using. I am new to this type of mathematics so my understanding is very limited. – Jade Rose Apr 18 '21 at 02:08

1 Answers1

1

Euclid's lemma: If $p$ is a prime number and $p\mid bc$, then $p\mid b$ or $p\mid c$.

Given that $p_1\mid a$ we have $a=p_1b$ for some integer $b$. Given that $p_2\mid a$ we have $p_2\mid p_1b$, and so by Euclid's lemma $p_2\mid p_1$ or $p_2\mid b$. But $p_1$ and $p_2$ are distinct primes, so $p_2\mid p_1$ is impossible. Hence $p_2\mid b$ and so $b=p_2c$ for some integer $c$. It follows that $a=p_1b=p_1p_2c$, and so $p_1p_2\mid a$.

Servaes
  • 63,261
  • 7
  • 75
  • 163
  • Thank you for answering my question it helped me out so much, I could not figure the question out from other posts because they were all different questions, so thank you so much. – Jade Rose Apr 18 '21 at 02:03