Can the GCD and LCM of [a,b] be proven to equal [ac,bc] with c>0?

[1+1=1]

hey check out this ?. with c>0, prove [ac,bc] = to c[a,b]. so far, i am saying that c/a and c/b. so, after that i get stuck. any help?

 

[AKG]

What exactly are you talking about? What is [a,b] and [ac,bc] (I mean are they points, vectors, what?) What do you mean you're "saying that c/a and c/b?" Actually, looking at the title, are you saying you need to prove that the lowest common multiply of ac, and bc is c times the lowest common multiple of a and b?

 

[1+1=1]

a,b,c are just numbers in the pos. int. and the ? is show that [ac,bc]=c[a,b]. i started off the proof saying that c divides a and c divides b, but can i say that? if so, how does that help me with the [ac,bc] part ?

 

[Gokul43201]

What does the notation [x,y] mean ? Is it the lcm of x and y ? Or is it the gcd ? Or what ?

 

[AKG]

a,b,c are just numbers in the pos. int. and the ? is show that [ac,bc]=c[a,b]. i started off the proof saying that c divides a and c divides b, but can i say that? if so, how does that help me with the [ac,bc] part ?

You can't say c divides a or c divides b. Moreover, we don't even know what [ac,bc] means, but I guess it means the lowest common denominator of ac and bc. You really need to be clear when asking questions, it's asking a little much of people here to answer your questions if they have to work just to figure out what the question is.

 

[Gokul43201]

If [x,y] is the lcm(x,y) then use the fact that lcm(x,y)=xy/gcd(x,y).

Then you are left with having to prove that gcd(ab,ac) = a*gcd(b,c) which is straightforward, from the prime factorization of b,c.

 

[AKG]

I think I have a solution assuming I know what the question is. I will use a slightly less ambiguous notation, however, that is:

\mathop{\rm lcm}\nolimits (x,y) \equiv the lowest common multiple of x and y

\mathop{\rm gcd}\nolimits (x,y) \equiv the greatest common denominator of x and y

I don't have a proof yet, but I'm pretty sure:

\mathop{\rm lcm}\nolimits (x,y) = \frac{x \times y}{\mathop{\rm gcd}\nolimits (x,y)}

I'll come back and try and prove it. However, assuming it's true, then we need to prove that:

\frac{ac \times bc}{\mathop{\rm gcd}\nolimits (ac,bc)} = c \frac{a \times b}{\mathop{\rm gcd}\nolimits (a,b)}

Now, if we have the prime factorizations of ac and bc, then we can choose a common denominator of ac and bc by determining the product of some set of prime factors that are common to both. The greatest common denominator is the product of all primes common to both. For example, if we had 28 and 20, we'd have:
2, 2, 7
2, 2, 5
{2,2} is the set of prime factors common to both, and it gives us a greatest common denominator of 2 x 2 = 4. Now, let the set of primes common to a and b be C, and thus the product of all elements in C gives us gcd(a,b). Let A and B represent the set of prime factors of a and b respectively, therefore \mathbf{C} = \mathbf{A} \cap \mathbf{B}. Now, let c represent the set of prime factors of c. The set of prime factors of ac is thus \mathbf{A} \cup \mathbf{c}. Now, the set of prime factors common to ac and bc would be(\mathbf{A} \cup \mathbf{c}) \cap (\mathbf{A} \cup \mathbf{c}) = (\mathbf{A} \cap \mathbf{B}) \cup \mathbf{c}. Note that A, B, and c, would be disjoint in this treatment. So, from the last line, we find that the gcd of ac and bc would be equal to gcd(a,b) * c, and this result gives us what we wanted to prove.

 

[matt grime]

Once you've done it in terms of the prime factorizations, try it without resorting to it. Arguably in an exam you'd have to prove that the integers formed a unique factorization domain before you were allowed to use that argument.

(lcm(x,y))= {ax+by| a,b in Z}
(lcm(cx,cy))={acx+bcy | a,b in Z}

show that the second ideal is "c times" the first ideal. The lcms are the minimal positive elements in each ideal.

 

[1+1=1]

ok [a,b] is notation for the least common multiple. and, what matt was saying, if i can show that "c times " what ever then any multiple of that would be = to those letters. this is rather perplexing i know, but for sure and for further reference [x,y] means least common multiple and (x,y) means greatest common divisor.

AKG- i had no intentions on confusing anyone here. i just assumed that everyone thought that [ ] means the least common multiple. again, sorry for the confusion and frustration. i will try to be more clear from now on.

 

[matt grime]

correction to myself, the gcds are the minimal positive elements in those ideals. once you've done the gcd, you may then figure out the lcm result.