Question

Let p be a prime natural number. Prove that \( (\mathbb{Z_p} -\{0\}, \cdot) \)is an associative algebraic system with identity 1
Use Eulicid's Lemma

Answer 1

i have no clue what Eulicid's lemma is

Answer 2

Merely that if \(p\) is prime, and \(p|ab\), then \(p|a\) or \(p|b\).

Answer 3

So we just need to show that it's associative here correct?

Answer 4

yup

Answer 5

I just realized I wasn't doing it a good way. Let me think about this a little bit.

Answer 6

okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

Answer 7

Maybe this will work better.

Let \(s\) be prime. Suppose \(s|a(bc)\). Then we have two cases. \(s|a\) or \(s|bc\). We can split up the second case into two additional cases; \(s|b\) or \(s|c\).

Case 1: \(s|a\). Then \(s|ab\), so \(s|(ab)c\).
Case 2: \(s|b\). Then \(s|ab\), so \(s|(ab)c\).
Case 3: \(s|c\). Then \(s|(ab)c\).

Using a similar argument, we can see that \(s|a(bc)\) iff \(s|(ab)c\). Therefore, these two numbers have the same unique factorization, so they are indeed the same number mod p. This shows associativity.

Answer 8

This seems perfect to meee

Answer 9

Sorry, tab just crashed. The next step is to prove the identity is 1. This follows from associativity and the fact that 1 is the identity within ordinary multiplication in integers.

Answer 10

\[1\cdot(a\cdot(b\cdot c))=(1\cdot a)\cdot(b\cdot c)=a\cdot(b\cdot c)\]

Answer 11

yaaa that is clear

Answer 12

Thanks @KingGeorge

Answer 13

You're welcome.

Answer 14

dw i wont bombard you with questions

Answer 15

I probably won't notice anyways. I've got to go again, and who knows when I'll be back on. I wasn't kidding when I said I'll be busy in the next couple weeks.

Answer 16

lol sounds like it is stressfull, Good Luck

Answer 17

You have no idea :|

I was busy from 8AM-11PM today. Tomorrow it'll start at 9, and hopefully end a little earlier. Anyways, see you later.

Answer 18

See ya around