- Thread starter
- #1

#### MI5

##### New member

- Sep 8, 2013

- 8

Proof: Consider (ab, ac) = a(b, c) = a. We have c|ab and clearly c|ac so c|a.

It's not so clear to me why c|ac. Perhaps I'm missing something really obvious.

- Thread starter MI5
- Start date

- Thread starter
- #1

- Sep 8, 2013

- 8

Proof: Consider (ab, ac) = a(b, c) = a. We have c|ab and clearly c|ac so c|a.

It's not so clear to me why c|ac. Perhaps I'm missing something really obvious.

- Jan 30, 2012

- 2,492

This is because ac = a * c, i.e., by the definition of the | (divides) relation.It's not so clear to me why c|ac.

- Thread starter
- #3

- Sep 8, 2013

- 8

I still don't understand I'm afraid. Could you say bit more please?This is because ac = a * c, i.e., by the definition of the | (divides) relation.

- Jan 30, 2012

- 2,492

I need to be sure you know the definition of the relation denoted by |. Could you write this definition?I still don't understand I'm afraid. Could you say bit more please?

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- #5

- Sep 8, 2013

- 8

WOW! All I can say is thanks.