Question

Prove that gcd(a,b) | lcm(a,b) @hartnn @ParthKohli @phi @TuringTest @ash2326 @RadEn @robtobey @jhonyy9 @Luigi0210 @Compassionate @KingGeorge @e.mccormick @shamil98 @timo86m @hba @beccaboo333 @SithsAndGiggles @Preetha @jigglypuff314 @ash2326 @ace
@austinL @inkyvoyd @Jamierox4ev3r @tHe_FiZiCx99

Answer 1

Where's abb0t when you need him ._.

Answer 2

uhm….

Answer 3

This is what I did: This would normally occur when gcd(a,b) = d*lcm(a,b) where d is an integer.

I let gcd(a,b) = d, then df = a and dg = b where d,g are integers.
By a theorem, gcd(a,b)*lcm(a,b)=ab, then dlcm(a,b) = dfdg = ab by substitution. Then lcm(a,b) = d*fg, so since gcd(a,b) = d, then lcm(a,b) = gcd(a,b)*fg where fg is the extra constant, proving the identity.

Answer 4

Yup, no problems there!

Answer 5

You may also use prime factorization.

Answer 6

Aye, our number theory book doesn't teach that yet so we aren't allowed to use that yet! xD

Answer 7

Hehe. Then generalise a factorization for a and b.

Answer 8

How much relation is there between number theory and abstract algebra?

Answer 9

There is a lot of overlap between the two.

Answer 10

For example, using some ring theory, I can give you a one line proof of the fundamental theorem of arithmetic, if you're given the fact that the Euclidean Algorithm works.

Answer 11

We are studying EA right now

Answer 12

gcd(a,b)=d
d|a,d|b...... d|a*b
lcm(a,b)|a*b
but lcm(a,b)<=a*b( by diffenition)
a*b=k lcm(a,b)
a*b/k=lcm(a,b)
nw u need to prove that k=d.