Proving LCM Formula and GCD for Non-Integers

[rukawakaede]

Hi,

I encountered the following formula while reading, can anyone prove this:

lcm(a,b)=gcd(a^{-1},b^{-1})^{-1}

Also, how could one do the gcd for non-integer?

for example we know that lcm(1/3,2/5)=2. then if we use the formula above then lcm(1/3,2/5)=1/gcd(3,5/2). then gcd(3,5/2) should be 1/2 but it does not make sense (to me at least).
Can anyone explain? Is this depends on which integral domain and its field of fractions that we work on?

 

[tiny-tim]

hi rukawakaede!

… gcd(3,5/2) should be 1/2 but it does not make sense (to me at least).

1/2 divides both 3 (6 times) and 5/2 (5 times)

and nothing higher than 1/2 divides both

so gcd(3,5/2) is 1/2, isn't it? ​

 

[agentredlum]

hi rukawakaede! 1/2 divides both 3 (6 times) and 5/2 (5 times)

and nothing higher than 1/2 divides both

so gcd(3,5/2) is 1/2, isn't it? ​

I have a more simple question, how is lcm(1/3, 2/5) = 2 ?

never mind i figured it out, 2/(1/3) is whole number, 2/(2/5) is whole numbed and 2 is smallest positive integer that works.

 

[rukawakaede]

hi rukawakaede!


1/2 divides both 3 (6 times) and 5/2 (5 times)

and nothing higher than 1/2 divides both

so gcd(3,5/2) is 1/2, isn't it? ​

Yes, so what is the general approach for this?

 

[rukawakaede]

I have a more simple question, how is lcm(1/3, 2/5) = 2 ?

never mind i figured it out, 2/(1/3) is whole number, 2/(2/5) is whole numbed and 2 is smallest positive integer that works.

well, 1/3, 2/3, 1, 4/3, 5/3, 2, ... then 2/5, 4/5, 6/5, 8/5, 2, ...
the lowest common multiple of 1/3 and 2/5 is 2.

 

[tiny-tim]

Yes, so what is the general approach for this?

just use common-sense, really …

find a number that's obviously a divisor of both, and then see if you can make it any larger ​

 

[agentredlum]

Yes, so what is the general approach for this?

I have derived a way, i think it works for all fractions but i am not sure. Here is the procedure.

To find gcd(a/b, c/d) first get (ad)/(bc) then reduce this to lowest terms call it s/t

the gcd(a/b,c/d) is then a/(bs) or b/(ct) they both give the same answer

1) your example: gcd(3, 5/2) get (ad)/(bc) = (3*2)/(5*1) = 6/5 already in lowest terms so

gcd(3, 5/2) = 3/(1*6) or 5/(2*5) both give 1/2

Example2: gcd(4, 12/3) get (ad)/(bc) = (4*3)/(12*1) = 12/12 reduce to 1/1 so

gcd(4, 12/3) = 4/(1*1) or 12/(3*1) both give 4

Example3: gcd(10/12, 15/6) get (ad)/(bc) = (10*6)/(15*12) reduce to (1/3)

gcd(10/12, 15/6) = 10/(12*1) or 15/(6*3) both give 5/6

I will assume you got the hang of it now so I'll just post results

gcd(2/3, 5/7) (2*7)/(3*5)= 14/15 so

gcd(2/3, 5/7) = 2/(3*14) or 5/(7*15) both give 1/21

gcd(3/5, 11/7) 21/55

3/(5*21) or 11/(7*55) = 1/35

gcd(5/6, 17/16) get (5*16)/(6*17)and reduce 40/51

5/(6*40) or 17/(16*51) both give 1/48

Can someone confirm these gcd using computer program?

Or can someone provide a counter example where this procedure fails?

 

[saim_]

GCD for non-integers??! Say we define GCD on rationals; we know all rationals divide all rationals. Infinity would be the GCD for any pair 'a' and 'b' ! :D

 

[rukawakaede]

GCD for non-integers??! Say we define GCD on rationals; we know all rationals divide all rationals. Infinity would be the GCD for any pair 'a' and 'b' ! :D

so do you mean that it is nonsensical to do the gcd for non-integers, in particular rationals?

 

[agentredlum]

GCD for non-integers??! Say we define GCD on rationals; we know all rationals divide all rationals. Infinity would be the GCD for any pair 'a' and 'b' ! :D

Can you find a rational larger than 1/2 that divides 3 AND 5/2 without remainder?

I never heard of this before today so thanks for the post rukawakaede, please don't be too eager to give up on your own question.

 

[micromass]

Hi,

I encountered the following formula while reading, can anyone prove this:

lcm(a,b)=gcd(a^{-1},b^{-1})^{-1}

Also, how could one do the gcd for non-integer?

for example we know that lcm(1/3,2/5)=2. then if we use the formula above then lcm(1/3,2/5)=1/gcd(3,5/2). then gcd(3,5/2) should be 1/2 but it does not make sense (to me at least).
Can anyone explain? Is this depends on which integral domain and its field of fractions that we work on?

Can you tell me where you encountered the formula? Taking the gcd in that fashion makes little sense to me I would like to see the reference to see what exactly they're talking about.

 

[micromass]

Can you find a rational larger than 1/2 that divides 3 AND 5/2 without remainder?

Yes: 1, 2, 3/5, etc. all divide 3 and 5/2.

 

[micromass]

Ah, perhaps they mean that a rational q divides p if there exists an integer n such that qn=p. That would actually make sense. It's something the book should have mentioned.

 

[saim_]

Can you find a rational larger than 1/2 that divides 3 AND 5/2 without remainder?

I never heard of this before today so thanks for the post rukawakaede, please don't be too eager to give up on your own question.

Yes. Take 10000 for example. It divides both 3 and 5/2. You could define division as micromass suggests in the post above, with the multiplying number 'n' being integer even though all other numbers are rationals, but, that seems like a pretty pointless exercise.

 

[agentredlum]

Yes. Take 10000 for example. It divides both 3 and 5/2. You could define division as micromass suggests in the post above, with the multiplying number 'n' being integer even though all other numbers are rationals, but, that seems like a pretty pointless exercise.

10000 does not divide either of them, try again, with this hint, if you are looking for rationals larger than 5/4 that divide both 3 and 5/2 WITHOUT REMAINDER, you are wasting your time.

When you define 'n' to be integer and all others rational you get gcd of fractions which is something you, micromass, me, and probably many others never heard of before today. IMHO this is not a pointless excercise.

 

[agentredlum]

Yes: 1, 2, 3/5, etc. all divide 3 and 5/2.

No, they don't, none of them do

 

[micromass]

10000 does not divide either of them, try again, with this hint, if you are looking for rationals larger than 5/4 that divide both 3 and 5/2 WITHOUT REMAINDER, you are wasting your time.

Sigh, you again

You are looking at rationals numbers, and in rational number 100000 divides both 3 and 5/2. The result of the division is 3/100000 and 5/200000.
The definition of division in the rationals (or any commutative ring) is classicaly that "a divides b if there exists a c such that ac=b". So according to this definition (which really is the standard definition), you will get that every nonzero number will divide any other number.

It is only when demanding c to be an integer that you will get the division that seems to be used here.

The notion of divisibility is very precisely defined in mathematics. And you MUST follow the definitions ALWAYS.

 

[agentredlum]

Sigh, you again

You are looking at rationals numbers, and in rational number 100000 divides both 3 and 5/2. The result of the division is 3/100000 and 5/200000.
The definition of division in the rationals (or any commutative ring) is classicaly that "a divides b if there exists a c such that ac=b". So according to this definition (which really is the standard definition), you will get that every nonzero number will divide any other number.

It is only when demanding c to be an integer that you will get the division that seems to be used here.

You are missing the point, the point is nobody knows everything there is to know about divisibility.

 

[micromass]

You are missing the point, the point is nobody knows everything there is to know about divisibility.

Can you do me a favor? Pick up an abstract algebra book and actually read through it. If you still disagree with me after that, feel free to discuss it with me.

Or if you are unwilling to do so, look at http://en.wikipedia.org/wiki/Greatest_common_divisor and show me one sentence that supports your point.

You can't just go claiming anything, you know.

 

[agentredlum]

Can you do me a favor? Pick up an abstract algebra book and actually read through it. If you still disagree with me after that, feel free to discuss it with me.

Or if you are unwilling to do so, look at http://en.wikipedia.org/wiki/Greatest_common_divisor and show me one sentence that supports your point.

You can't just go claiming anything, you know.

Do me a favor, when you come across a new idea that seems to make sense, don't close your mind to it because you haven't read about it in a book.

I have read many books on abstract algebra.

 

[micromass]

Do me a favor, when you come across a new idea that seems to make sense, don't close your mind to it because you haven't read about it in a book.

You know that this forum only supports discussions that are published somewhere, you know? So if what you are discussing is not in a book, then it is not allowed here.

I have read many books on abstract algebra.

OK, which ones? Pick any abstract algebra book, and show me a paragraph that supports your point. Shouldn't be too difficult will it?

Also, pick the abstract algebra book that you have read. Look for the section on divisibility in integral domains, and quote the definition of divisibility. Can you do that for me?

 

[agentredlum]

You know that this forum only supports discussions that are published somewhere, you know? So if what you are discussing is not in a book, then it is not allowed here.
OK, which ones? Pick any abstract algebra book, and show me a paragraph that supports your point. Shouldn't be too difficult will it?

Also, pick the abstract algebra book that you have read. Look for the section on divisibility in integral domains, and quote the definition of divisibility. Can you do that for me?

Your idea that something is not correct if it is not in a book is unsatisfactory. rukawakaede will post the title of the book.

You misunderstand me, i have read many abstract algebra books but i never heard of this before today, so obviously it wasn't in any of the books i read. That doesn't make it wrong.

 

[micromass]

Your idea that something is not correct if it is not in a book is unsatisfactory.

It's the way math works. Something is correct only if it is published (or it about to be published). Like it or not.

rukawakaede will post the title of the book.

Nonono, you said something. So you must back it up with references.

You said you already read abstract algebra books, which one?? I probably have that book, so I'll put the quote here.

 

[agentredlum]

It's the way math works. Something is correct only if it is published (or it about to be published). Like it or not.
Nonono, you said something. So you must back it up with references.

You said you already read abstract algebra books, which one?? I probably have that book, so I'll put the quote here.

Look, the definition of divisibility in abstract algebra is constrained to non zero integers but if you loosen the constraint a little bit you get something very interesting. That's all I'm saying. It appears to me you want to dismiss it prematurely.

Two books i read thoroughly were by Mcoy and Gallian. I have read at least 10 others that i don't remember titles now, also read Niven and Zuckerman, Anderson, Posemintere, and many other number theory books, including Recreations in the theory of numbers The Queen of mathematics entertains by Bieler

I don't remember reading about this approach in any of the books i read, like i said before, that is one of the reasons i find it fascinating.

 

[agentredlum]

I think I am wasting my time posting anything. From now on I will just read other peoples posts and figure it out myself.

agentredlum at rest.

 

[micromass]

Look, the definition of divisibility in abstract algebra is constrained to non zero integers but if you loosen the constraint a little bit you get something very interesting. That's all I'm saying.

Yes, you do get something extremely interesting by losening the constraints. But you should always say with which definitions you are working. You can't simultaniously say that I am wrong and refuse to give your definitions. You could at least have said that you were losening the constraints...

It appears to me you want to dismiss it prematurely.

What happened is that I told you what definition I was working with, and you told me that I was wrong. I dismissed nothing.

Two books i read thoroughly were by Mcoy and Gallian. I have read at least 10 others that i don't remember titles now, also read Niven and Zuckerman, Anderson, Posemintere, and many other number theory books, including Recreations in the theory of numbers The Queen of mathematics entertains by Bieler

I don't remember reading about this approach in any of the books i read, like i said before, that is one of the reasons i find it fascinating.

You mean "contemporary abstract algebra" by Gallian, right?? Let me quote that:

The following terminology and notation is convenient. If a and b belong to
a commutative ring R and a is nonzero, we say a divides b (or a is di factor of
b) and write a | b, if there exists an element c in R such that b = ac.[

This is the standard terminology. So with this terminology, 10000 will divide 2/5. In fact, every nonzero number will divide 2/5.

If you're losing constraints and making new definitions, then please say so! If you don't, then this will lead to pointless discussions. I hope you understand that.

 

[saim_]

10000 does not divide either of them, try again, with this hint, if you are looking for rationals larger than 5/4 that divide both 3 and 5/2 WITHOUT REMAINDER...

Remainders don't come up in rational multiplication and division. I understand you are trying to modify it but then you should have defined what you are trying to do instead of having people guess at it. I can't believe you have actually read abstract algebra books and never learned how division is defined on rationals?! That is strange. Pointless debate.

 

[agentredlum]

Remainders don't come up in rational multiplication and division. I understand you are trying to modify it but then you should have defined what you are trying to do instead of having people guess at it. I can't believe you have actually read abstract algebra books and never learned how division is defined on rationals?! That is strange. Pointless debate.

I can't believe you went to grade school and never learned what division is. A child would get the answer before any of you got it. So far in this thread tiny-tim is the only one who understands division.

Of course fractional division can have remainders. Just because you don't know how doesn't mean they don't exist.

 

[agentredlum]

Yes: 1, 2, 3/5, etc. all divide 3 and 5/2.

1 divides 3 but leaves remainder .5 for 5/2

2 leaves remainder 1 for 3 and leaves remainder .5 for 5/2

3/5 divides 3 remainder 0 but leaves remainder .1 for 5/2

Don't just dismiss this as nonsense, PICK UP PENCIL AND PAPER AND DO THE CALCULATIONS!

 

[agentredlum]

What happened is that I told you what definition I was working with, and you told me that I was wrong. I dismissed nothing.

NOOOOOOOO...nooooo...no. You posted solutions first, i told you they were wrong, then you told me what definition you were working with. Look at the posts you cannot rewrite history.

 

[saim_]

I can't believe you went to grade school and never learned what division is. A child would get the answer before any of you got it. So far in this thread tiny-tim is the only one who understands division.

Of course fractional division can have remainders. Just because you don't know how doesn't mean they don't exist.

They don't even teach that in grade school. Remainders are defined for integer division only. Quote one grade school level book that does this :D. For more information see:

Wikipedia: Division
http://en.wikipedia.org/wiki/Division_(mathematics)

You can see remainders are there only in division for integers, not rationals . How you are defining division is fine and understandable but once you understand that this is not the standard way of defining division go back and review the objections that were raised.

 

[agentredlum]

They don't even teach that in grade school. Remainders are defined for integer division only. Quote one grade school level book that does this :D. For more information see:

Wikipedia: Division
http://en.wikipedia.org/wiki/Division_(mathematics)

You can see remainders are there only in division for integers, not rationals . How you are defining division is fine and understandable but once you understand that this is not the standard way of defining division go back and review the objections that were raised.

I never said it was standard, i never said fractional remainders are taught in grade school. Distorting my words is not helping your case. You say i don't know the definition of division, that is incorrect. I say you don't know how to divide. Whether or not you have memorized definitions means little to me. Your insistance that considering gcd between fractions is pointless confirms my point that you do not know how to divide. I also suspect you do not know what a remainder is or could be.

 

[saim_]

I never said it was standard

Then you should define what you mean!

At least you are happy :p

 

[tiny-tim]

Come on guys , it's obvious from the original post that division by an integer is intended …

Hi,

I encountered the following formula while reading, can anyone prove this:

lcm(a,b)=gcd(a^{-1},b^{-1})^{-1}

Also, how could one do the gcd for non-integer?

for example we know that lcm(1/3,2/5)=2. then if we use the formula above then lcm(1/3,2/5)=1/gcd(3,5/2). then gcd(3,5/2) should be 1/2 but it does not make sense (to me at least).
Can anyone explain? Is this depends on which integral domain and its field of fractions that we work on?

hi rukawakaede!


1/2 divides both 3 (6 times) and 5/2 (5 times)

and nothing higher than 1/2 divides both

so gcd(3,5/2) is 1/2, isn't it? ​

 

[agentredlum]

Then you should define what you mean!

At least you are happy :p

I am not responsible for peoples misunderstandings, i try to make things work instead of throwing around definitions why things shouldn't work.

You would be happy too if you realized, like i have, that your horizons have been expanded. If you realize all those math books you read, not one of them mentioned something so simple and obvious. Instead they tell you what can or can't be done. Gallians book was o-k but Mccoy was a NIGHTMARE, now you know why i am suspicious of abstract algebra, not to mention the fact that Godel proved no mathematical system strong enough to give rise to arithmetic can be completely consistent. Do you see where I'm comming from now? I don't trust the definitions (even though i know many, not all) and I am mildly annoyed anytime someone mentions a definition as if it was 'etched in stone'. I ALWAYS look at any definition with suspicion and my motto is question EVERYTHING.

I felt like i owed you an explanation, thanks for being a good sport.

You know i once said the same thing to a math professor friend of mine and he asked 'what should we do? say to hell with rigor!' and i said 'yes, that's what Euler did, that's what Fermat did and look at what they accomplished. I'm not saying get rid of all rigor, just loosen it up a little bit so you can get something done. Rigor is like a stranglehold on creativity'

 

[saim_]

@agentredlum: You don't want to define everything, fine, don't do it. It was friendly advice.

 

[micromass]

I am not responsible for peoples misunderstandings

Yes you are. You are making statements that are obviously false. And only after a few posts you explain that you are working with a different definition. You should have explained immediately under what definition you are working with. That is your responsibility.

I can post here that I know for certain that 1+1=0. People will come in arguing with me, showing me references of why this is not true, explaining why it is not true. And only after many posts I will exclaim that I was working in modulo 2. This is pointless, I should have said that immediately.

You would be happy too if you realized, like i have, that your horizons have been expanded. If you realize all those math books you read, not one of them mentioned something so simple and obvious. Instead they tell you what can or can't be done. Gallians book was o-k but Mccoy was a NIGHTMARE, now you know why i am suspicious of abstract algebra, not to mention the fact that Godel proved no mathematical system strong enough to give rise to arithmetic can be completely consistent.

OK, now you give me the impression that you didn't actually understood the math books you read. And your statement of Godels theorem is also false.

Do you see where I'm comming from now? I don't trust the definitions (even though i know many, not all) and I am mildly annoyed anytime someone mentions a definition as if it was 'etched in stone'. I ALWAYS look at any definition with suspicion and my motto is question EVERYTHING.

What's not to trust with a definition?? A definition is what it is, a definition of a term. There's nothing to trust here. You may argue that there are better definitions or that some modification makes a more efficient definition. But definitions and axioms are etched in stone. From the moment on that you accept to work in a certain axiom-system, then this axiom system is the truth. If you choose not to work in a certain axiom-system, then it becomes false. Either way is fine, but you need to mention which definitions and axioms you are working with!

You know i once said the same thing to a math professor friend of mine and he asked 'what should we do? say to hell with rigor!' and i said 'yes, that's what Euler did, that's what Fermat did and look at what they accomplished. I'm not saying get rid of all rigor, just loosen it up a little bit so you can get something done. Rigor is like a stranglehold on creativity'

Now I think you are not understanding the place of rigor in mathematics. Rigor is very important in mathematics, why? Not because it's fun to be rigorous, but because it is necessary.

When doing mathematical research, there are two important processes that play:
- intuition
- rigor
Intuition is important because it let's you find new results and because it let's you find solutions for problems. However, after you find these solutions, you must always check them with rigor. Why? To see if what you did is not false!
There are a lot of intuitively obvious things that turn out to be false.

 

[agentredlum]

Yes you are. You are making statements that are obviously false. And only after a few posts you explain that you are working with a different definition. You should have explained immediately under what definition you are working with. That is your responsibility.

I can post here that I know for certain that 1+1=0. People will come in arguing with me, showing me references of why this is not true, explaining why it is not true. And only after many posts I will exclaim that I was working in modulo 2. This is pointless, I should have said that immediately.
OK, now you give me the impression that you didn't actually understood the math books you read. And your statement of Godels theorem is also false.
What's not to trust with a definition?? A definition is what it is, a definition of a term. There's nothing to trust here. You may argue that there are better definitions or that some modification makes a more efficient definition. But definitions and axioms are etched in stone. From the moment on that you accept to work in a certain axiom-system, then this axiom system is the truth. If you choose not to work in a certain axiom-system, then it becomes false. Either way is fine, but you need to mention which definitions and axioms you are working with!
Now I think you are not understanding the place of rigor in mathematics. Rigor is very important in mathematics, why? Not because it's fun to be rigorous, but because it is necessary.

When doing mathematical research, there are two important processes that play:
- intuition
- rigor
Intuition is important because it let's you find new results and because it let's you find solutions for problems. However, after you find these solutions, you must always check them with rigor. Why? To see if what you did is not false!
There are a lot of intuitively obvious things that turn out to be false.

Do you see what your rigor is doing to you? YOU ARE STILL ARGUING AGAINST SOMETHING THAT SHOULD HAVE BEEN OBVIOUS TO YOU FROM POST #2

I resign from this nonsense!

http://www.youtube.com/watch?v=ZbNXyfPsPwk&feature=related

Let others decide if i am wrong about Godels theorem, i clearly do not trust your judgement anymore

http://en.wikipedia.org/wiki/Gödel's_incompleteness_theorems

http://mathworld.wolfram.com/GoedelsIncompletenessTheorem.html

 

[saim_]

@agentredlum: So because of Godel's theorem we shouldn't be rigorous? You know that theorem is absolutely rigorously proven; everything is defined and all. If rigor could do Godel good, it can sure do us good.

 

[agentredlum]

This is what you should do. You should apologize for posting incorrect solutions too quickly and without thinking. Show a little humility and admit you can't know everything micromass.

 

[agentredlum]

@agentredlum: So because of Godel's theorem we shouldn't be rigorous? You know that theorem is absolutely rigorously proven; everything is defined and all. If rigor could do Godel good, it can sure do us good.

Well it didn't do you any good here did it? You know for people who place so much value on rigor, you guys didn't seem to read post #1 and post #2 and give it ANY rigorous consideration, yet you have the nerve to demand absolute rigor from me.

You can still be rigorous and you can still question everything, even definitions set in boldface type.

Books don't teach you to question them, you learn that through other ways, example working in a math center solving hundreds of problems every day, 6 days a week, with pencil and paper for many years.

 

[micromass]

This is what you should do. You should apologize for posting incorrect solutions too quickly and without thinking. Show a little humility and admit you can't know everything micromass.

Show me when I was incorrect.

Well it didn't do you any good here did it? You know for people who place so much value on rigor, you guys didn't seem to read post #1 and post #2 and give it ANY rigorous consideration, yet you have the nerve to demand absolute rigor from me.

I did give it rigorous consideration:

Ah, perhaps they mean that a rational q divides p if there exists an integer n such that qn=p. That would actually make sense. It's something the book should have mentioned.

It is you who starts asserting things without defining anything.

What I don't know is whether you are clueless or deliberatly obfusciating things.

 

[agentredlum]

Show me when I was incorrect

Post # 12 you give incorrect solutions to a problem that should have been obvious from post #2. The fact that you believed the solutions were correct based on your misunderstanding is irrelevent, your solutions were wrong.

To your credit, and i admire this about you, 8 minutes later you figure out what is going on in post #13 but for some reason known only to you, you do not retract the incorrect solutions.

Post # 16 i point out the solutions in post #12 are incorrect so here we are 40 posts later still arguing. If i bruised your ego i apologize. Because that's what it looks like to me we are arguing about.

 

[micromass]

Post # 12 you give incorrect solutions to a problem that should have been obvious from post #2. The fact that you believed the solutions were correct based on your misunderstanding is irrelevent, your solutions were wrong.

To your credit, and i admire this about you, 8 minutes later you figure out what is going on in post #13 but for some reason known only to you, you do not retract the incorrect solutions.

Post # 16 i point out the solutions in post #12 are incorrect so here we are 40 posts later still arguing. If i bruised your ego i apologize. Because that's what it looks like to me we are arguing about.

Well, if you think that my post 12 is incorrect, then I'm afraid that you did not understand me very well. I am of course to blame for that, I should have expressed better what I meant.

Anyway, I'm done discussing to you, since (by your own admission) you are unwilling to listen to rigor and reason.

 

[agentredlum]

Well, if you think that my post 12 is incorrect, then I'm afraid that you did not understand me very well. I am of course to blame for that, I should have expressed better what I meant.

Anyway, I'm done discussing to you, since (by your own admission) you are unwilling to listen to rigor and reason.

I never said that about rigor, i said loosen it up a little bit. You misinterperate every statement i make.

We are not going to resolve this issue between us. Let others decide if your solutions in post 12 are right or wrong.

I am also done discussing with you.

 

[rukawakaede]

Can you tell me where you encountered the formula? Taking the gcd in that fashion makes little sense to me I would like to see the reference to see what exactly they're talking about.

Hi micromass,

The name of the book is algebraic theory of numbers by pierre samuel. on page 14.

sorry, i missed this message.

i was wondering if gcd(a,b) is an integer which divides a and b? so if a and b are not integers, how could we do gcd of such numbers? if there is no way or it is meaningless to do gcd of rationals which are not integers, then the formula is not well defined. Could you comment further?

 

[rukawakaede]

Ah, perhaps they mean that a rational q divides p if there exists an integer n such that qn=p. That would actually make sense. It's something the book should have mentioned.

the book does not specific the domains, just a general integral domain and its field of fractions.

 

[micromass]

the book does not specific the domains, just a general integral domain and its field of fractions.

Yes, I looked in the book and I saw that. But in the case of the integers, it reduces to that.

Hi micromass,

The name of the book is algebraic theory of numbers by pierre samuel. on page 14.

sorry, i missed this message.

i was wondering if gcd(a,b) is an integer which divides a and b? so if a and b are not integers, how could we do gcd of such numbers? if there is no way or meaningless to do gcd of rationals which are not integers, then the formula is not well defined. Could you comment further?

The key concept here is divisibility. Let me quote from the book

Let A be an integral domain, K its field effractions, x and y elements of K. We shall
say that x divides y if there exists a e A such that y = ax.

So divisibility is well defined by this. So a fraction q divides a fraction p if p/q is in the integral domain.

Now, how is a gcd defined. Well, a is called a gcd of b and c if

• a|b and a|c
• If d|b and d|c, then d|a

So, for example gcd(1/2,1/3)=1/6, since

• 1/6 divides 1/2, because (1/2)/(1/6)=3 is an integer
• 1/6 divides 1/3, because (1/3)/(1/6)=2 is an integer
• If 1/a divides 1/2 and 1/6, then a/2 and a/3 are integers, so a must be a multiple of 6. This means that 1/a will always divide 1/6

In general, if you are asked to find the gcd of two elements, then you should find all its divisors and see if there is a greatest one. This will be the gcd.

In the case of rational numbers, I think we can do something of the following to find the gcd. I'll do an example:

Take rationals 60/7 and 8/42. Take the prime decomposition of these numbers:

\frac{2^2*3*5}{7}~\text{and}~\frac{2^2}{3*7}

take the gcd of the numerators and the lcm of the denominator, this yields

\frac{2^2}{3*7}=\frac{4}{21}

this is how to find the gcd of 60/7 and 8/42.

So in general I conjecture that

gcd(a/b,c/d)=\frac{gcd(a,c)}{lcm(b,d)}

I would be pleased if somebody would check this formula for me and tell me if it is indeed correct.

 

[agentredlum]

Try gcd(30/7, 8/42) your method gives 2/42 or 1/21

Try gcd(30/7, 4/21) your method gives 2/21

8/42 = 4/21 but 1/21 does not equal 2/21



You can fix this if you reduce all fractions before computing but that does not gaurantee there are no other difficulties with your method.

I invite everyone to test my method in post #7

 

[rukawakaede]

Yes, I looked in the book and I saw that. But in the case of the integers, it reduces to that.



The key concept here is divisibility. Let me quote from the book



So divisibility is well defined by this. So a fraction q divides a fraction p if p/q is in the integral domain.

Now, how is a gcd defined. Well, a is called a gcd of b and c if

• a|b and a|c
• If d|b and d|c, then d|a

So, for example gcd(1/2,1/3)=1/6, since

• 1/6 divides 1/2, because (1/2)/(1/6)=3 is an integer
• 1/6 divides 1/3, because (1/3)/(1/6)=2 is an integer
• If 1/a divides 1/2 and 1/6, then a/2 and a/3 are integers, so a must be a multiple of 6. This means that 1/a will always divide 1/6

In general, if you are asked to find the gcd of two elements, then you should find all its divisors and see if there is a greatest one. This will be the gcd.

In the case of rational numbers, I think we can do something of the following to find the gcd. I'll do an example:

Take rationals 60/7 and 8/42. Take the prime decomposition of these numbers:

\frac{2^2*3*5}{7}~\text{and}~\frac{2^2}{3*7}

take the gcd of the numerators and the lcm of the denominator, this yields

\frac{2^2}{3*7}=\frac{4}{21}

this is how to find the gcd of 60/7 and 8/42.

So in general I conjecture that

gcd(a/b,c/d)=\frac{gcd(a,c)}{lcm(b,d)}

I would be pleased if somebody would check this formula for me and tell me if it is indeed correct.

A counterexample to the formula
gcd(a/b,c/d)=\frac{gcd(a,c)}{lcm(b,d)}
is:
a=2, b=4, c=4, d=8

gcd(a,c)=2
lcm(b,d)=8
gcd(a,c)/lcm(b,d)=1/4

but gcd(a/b,c/d)=gcd(1/2,1/2)=1/2

Moreover, you can see the error if you investigate the formula further:
gcd(a/b,c/d)=\frac{gcd(a,c)}{lcm(b,d)}
gcd(ad,cb)=gcd(a,c)\cdot gcd(b,d)
this is certainly not true: take the same example as above.