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amistre64
Group Title
Show that if a,b,c are integers; c>0; and
a [congruent] b(mod m), then gcd(a,c) = gcd(b,c)
 one year ago
 one year ago
amistre64 Group Title
Show that if a,b,c are integers; c>0; and a [congruent] b(mod m), then gcd(a,c) = gcd(b,c)
 one year ago
 one year ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.0
Heres what I was thinking ...
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
and its a = b (mod c) .. typoed it
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
since a = b (c); then a = b + ck, for some integer k a  b = ck ax1 + bx2 = ck ax1 + bx2 = c(y1+y2) ax1 + bx2 = cy1 + cy2 ax1 + cy1 = bx2 + cy2 ax1 + cy1 = g1, for some integer g g1 = bx2 + cy2 gcd(a,c) = gcd(b,c)
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
If a = b mod c then b = a + ck then gcd(b,c) = gcd(a+ck,c) = gcd(a,c) (I think)
 one year ago

DanielxAK Group TitleBest ResponseYou've already chosen the best response.0
That's true. You might want to give more reasoning though: Let d = gcd(b,c). Then, gcd(b,c) => bd and cd. => a+ckd and cd. Since cd, ckd. And, since ckd, a+ckd => ad. And, then you're done. Easier to see this way, I think.
 one year ago

DanielxAK Group TitleBest ResponseYou've already chosen the best response.0
Oh, those arrows go both ways. It's an equality. Woops.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
(b,c) = d therefore db and dc there is some thrm that states that the gcd of 2 numbers can be written as a linear combination of those numbers d = bx + cy
 one year ago

DanielxAK Group TitleBest ResponseYou've already chosen the best response.0
Oh, I had it backwards. It works the same way, though.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
I was relying on a theorem (or an extension of one, at least I think I was).
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
In fact this is an extension of gcd(m,n) = gcd(n,m%n)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
id have to see a more detailed version of this: gcd(b,c) = gcd(a+ck,c)
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
k, ab = ck like u have...
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
your just subing in b with a+ck ....
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
takes me a few times to read it right :)
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Yes, relying on the theorem above....(so I guess what you want is a proof of that, in effect...
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
ab = ck like u have Let gcd(a,c) = n > n divdes a and c so a = nq and c = nr (some q and r)
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Hope that's right (lots of letters to keep track of).
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
So ab = ck > nq  b = nrk > b = n(qrk) > n divides b Do same for other side and it should drop out (I hope)
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Found a prettier version of the whole thing.....
 one year ago
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