Lcm of rationals. Urgent help

[Saumyojit]

Out of desperation I had to create a new thread

https://math.stackexchange.com/questions/44836/rational-numbers-lcm-and-hcf
From first answer

See the writer didn't prove that the particular fraction is a LCM of the fractions so my objective in this post is to show that
lcm of (numerator of fractions) / gcd of (denom of fractions) is a common multiple and then least common multiple of two fractions.(a/b, c/d)

Lets give lcm of numerator of fraction / gcd denom of fraction a name 'k' .
Now I need to prove that this k has to be exactly divisible by both a/b , c/d Which will give me the first part of the proof.

Then I need to prove that 'k' divides any other Common multiple of (a/b, c/d)
which finally tells me that 'k' is the LCM of those two fractions.

But how to approach it.

Now from this part lcm of numerator of fraction / gcd denom of fraction

I have some information which I don't know of what help this will do in the proof.
1: I know that numerator of k is the least common multiple of a&c and it's get divided by a&c.

2: Denominator of k is the highest Common factor of (b,d) , so b & d gets divided by GCD (b,d)

Please help.

 

[Saumyojit]

@Dr.Peterson

 

[Dr.Peterson]

See the writer didn't prove that the particular fraction is a LCM of the fractions

Actually, he did prove that; he just did so in a way that requires some thought. The notation may not be clear to you, and he is trusting the reader to fill in the reasoning in several places. So a different approach may work better for you, and that's what you're trying to do -- you aren't really asking about this proof at all, but about an alternative proof that doesn't rely on the prime factorizations.

Lets give lcm of numerator of fraction / gcd denom of fraction a name 'k' .
Now I need to prove that this k has to be exactly divisible by both a/b , c/d

I don't think that will be helpful in writing a detailed proof. You've hidden the fact that "k" is a fraction.

Instead, keep what is happening visible!

We want to show that the lcm of a/b and c/d is lcm(a,b)/gcd(c,d). Let's say lcm(a,b) = m, and gcd(c,d) = n. You want to show that m/n is a multiple of a/b and of c/d, and that it is the least. Right?

See what you can do with that notation. First show that m/n is an (integer) multiple of a/b. What does that mean?

By the way, you are aware, I hope, that the statement to be proved is false if a/b and c/d are not in lowest terms. That was my first observation when I saw the question, and is stated in the last answer. It will be important in the proof.

Also by the way, starting a new thread for a new question is not a matter of desperation; it is a matter of proper use of a forum, and respect for people who find bloated threads discouraging.

One more thing: I glanced at the other thread, and saw that you were asked,

Did you try working through a concrete example?

Try 6/35 and 22/39.

You never did so. This is excellent advice, and in not following it, you have shown considerable disrespect to those who are trying to help. If you have any trouble doing what I said above, FIRST do this. An example is the best way to start planning a proof, because it can give you a sense of how things work.

 

[Saumyojit]

If you have any trouble doing what I said above, FIRST do this.

Lcm will come 66.

You want to show that m/n is a multiple of a/b and of c/d, and that it is the least

I tried yesterday night but couldn't do that.
Yes a/b and c/d are in lowest terms and a and b are co prime and c and d are co prime.
M and n are co prime.
Now please show me the proof

 

[Dr.Peterson]

Lcm will come 66.

But your goal is a proof, not just a number. Talk me through how the formula works, and think about why.

Since the lcm here is a whole number, maybe you should try a more typical example, maybe 6/35 and 9/20. Again, use the details to think through what the proof you want might do.

I tried yesterday night but couldn't do that.
Yes a/b and c/d are in lowest terms and a and b are co prime and c and d are co prime.
M and n are co prime.
Now please show me the proof

I don't have a proof. You know how the site works: We want you to learn to do things yourself. So TRY. Really try. That's how you learn.

 

[Saumyojit]

Let 2/3, 5/6 and 7/9 be three fractions. Let A/B be their LCM. Now all the fractions must divide A/B, that is, A/B ÷ 2/3, A/B ÷ 5/6 and A/B ÷ 7/9 all must be natural numbers. Or A3/B2, A6/B5 and A9/B7 all must be natural numbers. As a and b are co prime,This requires that A must be divisible by 2,5 and 7 each and B must be a factor of 3,6 and 9 each. In other words A must be a multiple of 2,5 and 7.
How to prove that it is the least ?

 

[Dr.Peterson]

I'll just change a couple details so this looks more normal:

Let 2/3, 5/6 and 7/9 be three fractions. Let A/B be their LCM. Now all the fractions must divide A/B, that is, A/B ÷ 2/3, A/B ÷ 5/6 and A/B ÷ 7/9 all must be natural numbers. Or (3A)/(2B), (6A)/(5B) and (9A)/(7B) all must be natural numbers. As A and B are co prime,This requires that A must be divisible by 2, 5 and 7 each and B must be a factor of 3, 6 and 9 each. In other words A must be a multiple of 2,5 and 7.
How to prove that it is the least ?

Good work. You picked a different example and expanded to three fractions, but showed it clearly; and this reasoning can be easily extended into a proof by using tons of variables.

You've shown that if A/B is the LCM of the fractions, then A must be a multiple of LCM(2, 5, 7), and B must be a divisor of GCD(3, 6, 9). Your work also shows that if A = LCM(2, 5, 7), and B = GCD(3, 6, 9), then A/B is a multiple of the LCM of the fractions. What would it take to find a smaller common multiple?

 

[Saumyojit]

I don't want know any more .
Please ? show it

 

[Dr.Peterson]

I don't want know any more .
Please ? show it

I thought you were trying to learn how to understand a proof. That requires thinking.

Please try answering my question. We have a common multiple, namely 70/3; we know that any common multiple must have at least the factors 2, 5, and 7 in its numerator, and must have a denominator divisible by 3, 6, and 9. So to make a smaller common multiple, would have to remove a factor from the numerator, or multiply by another factor in the denominator. Would it still be a common multiple?

 

[JeffM]

You learn math by doing it.

2/3 and 5/6

Least common multiple of 2 and 5 is 10.

Greatest common divisor of 3 and 6 is 3.

Therefore the theorem says that the least common multiple of 2/3 and 5/6 is 10/3.

Obviously 10/3 = 5 * 2/3 and 10/3 = 4 * 5/6.

So 10/3 is a common multiple.

How might you show that there is no common multiple less than 10/3?

EDIT: I have not used three fractions because the theorem does not. I suggest you prove it for two fractions and then use induction for higher numbers.