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## Homework Statement

if (a,c)=1 and (b,c)=1 prove that (ab,c)=1

## Homework Equations

I know that (a,c)=1 says that au+cv=1 and bs+ct=1 prove abq+cr=1

## The Attempt at a Solution

I set au+cv=bs+ct now I don't know what to do

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- Thread starter j9mom
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- #1

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if (a,c)=1 and (b,c)=1 prove that (ab,c)=1

I know that (a,c)=1 says that au+cv=1 and bs+ct=1 prove abq+cr=1

I set au+cv=bs+ct now I don't know what to do

- #2

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do you know the theorem that if d|ab, then d|a or d|b?

- #3

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Yes I did learn that, but how do I use that?

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- #5

Dick

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do you know the theorem that if d|ab, then d|a or d|b?

That's false. 6|(3*4) but 6 doesn't divide 3 or 4. You had better say the word 'prime' at some point.

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- #7

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Wait... x is a prime number greater than 1 that divides c

- #8

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That's false. 6|(3*4) but 6 doesn't divide 3 or 4. You had better say the word 'prime' at some point.

You're right. Thanks for the correction

- #9

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Thanks

- #10

Dick

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Wait... x is a prime number greater than 1 that divides c

Slow down. Can you state your reasoning in complete sentences? (a,c)=1 means a and c have no common prime divisors. Now continue.

- #11

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Wait... x is a prime number greater than 1 that divides c

you don't need to specify 'greater than 1' because 1 isn't prime.

so if d|(ab, c), then there exists some prime p such that p|ab.

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- #13

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- #14

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- #15

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A philospher: my x equals a prime number that divides c.

- #16

Dick

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Why are you rushing this? Think about it. You can't assume BOTH x|c AND (ab,c)=x without justifying it. Look, start from this: If (ab,c) is not 1, then let x be a prime that divides (ab,c). Go from there. SLOWLY.

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- #18

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Ok let me start again.

- #19

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You can't assume BOTH x|c AND (ab,c)=x without justifying it.

Yes you can. By definition. What can't be assumed is that (ab, c)=x and that x is prime.

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- #21

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that's the idea. looks good.

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- #23

Dick

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Yes you can. By definition. What can't be assumed is that (ab, c)=x and that x is prime.

On that, you are right. Oh, I see you've both already finished. Sorry to be so slow.

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- #25

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You originally used the theorem that (m,n)=d implies mx+ny=d for some integers x, y. Even though the converse is not true in general, it is true that for the special case d=1, this is an "if and only if."

That is, (m,n)=1 if and only if mx+ny=1 for some integers x, y. If you have this theorem, then you can give a nice solution to your problem.

So, if (a,c)=1 and (b,c)=1 prove that (ab,c)=1.

You know au+cv=1 and bs+ct=1 for some integers u, v, s, and t.

Now multiply those two equations and put into the form ab(some integer) + c(big mess)=1, where big mess is an integer.

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