LCM AND HCF

Definitions are not truths. Different definitions are useful in different contexts. Different theorems result from using different definitions. You cannot reconcile the irreconcilable. When definitions differ, anything else may differ.

Nor can you learn math by jumping between multiple sources on different topics at different levels of sophistication.
 
Dr.Peterson said:
I can't. As I said, this is related to more advanced math that you are not ready to understand.
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Ok can i just know which part of advanced math? Please

Another thing what i discovered that is for any co prime nos (5,4) gcd(a,b)=1 . Now we dont require the second test condition for that as c will be =1 and c cannot be "less than" as there is no other divisor of 4,5

Saumyojit said:
Now in the line "Put c= gcd(a,b) for (1) "
is he telling to substitute gcd (a,b) in the second condt written using biconditional form.
c | a,b <=> c | gcd(a,b) ; after substituting in place of c i get gcd(a,b) | a,b <=> gcd(a,b) | gcd(a,b)
Am i correct? And why they are telling this .
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I think i am right as u said below

he is defining gcd(a, b) as the (unique) number that (1) is a divisor of both a and b [that is, a common divisor], and (2) is divisible by every other common divisor (or rather, by every common divisor, including itself)
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which is matching my opinion that after substituion of gcd(a,b) in place of c in second equation it looks like this
gcd(a,b) | a,b <=> gcd(a,b) | gcd(a,b)
Right?
 
Saumyojit said:
Ok can i just know which part of advanced math? Please
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Abstract algebra.

Saumyojit said:
Another thing what i discovered that is for any co prime nos (5,4) gcd(a,b)=1 . Now we dont require the second test condition for that as c will be =1 and c cannot be "less than" as there is no other divisor of 4,5
Click to expand...
Yes, if there is only one common divisor, then there is no need to check which is greatest.
Saumyojit said:
which is matching my opinion that after substituion of gcd(a,b) in place of c in second equation it looks like this
gcd(a,b) | a,b <=> gcd(a,b) | gcd(a,b)
Right?
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Yes.
 

Some people think that a factor of b means b divided by any integer "which goes exactly",
for example, the factors of 10 are 10/1, 10/(−2), 10/5etc. This would give the factors of 0 as being 0/1, 0/2 etc, in other words, 0 only, which is not correct.
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He is saying this thats because i cannot divide 0/1 there is nothing to divide right?
 
@Saumyojit

Many of your recent questions have been about abstract algebra. Abstract algebra involves generalizations about different kinds of algebra. A plus sign does not necessarily mean addition in abstract algebra. An asterisk does not necessarily mean multiplication. Symbols and definitions are not consistent between elementary algebra, which is what almost everyone learns in secondary school and has been studied since the seventh century C.E., and abstract algebra, which is studied mostly by mathematicians at the college level and beyond and has been studied for little more than a hundred years.

Trying to grapple with abstract algebra provides no help whatsoever in learning elementary algebra; it is merely confusing. Abstract algebra is hard, and the motivation for studying it only arises when you have learned at least two different kinds of algebra well. The way you are proceeding is frustrating and will not teach you what you need to know to understand regular algebra, calculus, statistics, etc. Abstract algebra is a highly specialized field.
 
Saumyojit said:

Some people think that a factor of b means b divided by any integer "which goes exactly", for example, the factors of 10 are 10/1, 10/(−2), 10/5 etc. This would give the factors of 0 as being 0/1, 0/2 etc, in other words, 0 only, which is not correct.​

He is saying this thats because i cannot divide 0/1 there is nothing to divide right?
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Please observe first that the context amounts to a discussion of different definitions of "factor". This is often the basis of useless arguments, when people think their definition is "the right one". Mathematicians state the definitions they are using before trying to prove anything.

Now, he's defined a factor of b as a number a such that b = ka for some integer k. By that definition, any integer is a factor of 0.

Then he says that if you chose to define a factor of b as the result of dividing b by some integer, when that result is an integer: a = b/k. But his point is that if you divide 0 by anything, you always get 0: 0/k = 0 for any integer k. So by this definition, it would not be true that any integer is a factor of 0. [The difference is that although b = ka is equivalent to a = b/k when k is non-zero, when b = 0 and a is not, it would be necessary for k to be zero -- and you can't divide by zero.] The conclusion is that this is not a useful definition.

You are wrong in saying that one "cannot divide 0/1 [because] there is nothing to divide." The result of 0/1 is 0. What he's saying is not that you can't do the division, but that 0/k is always 0, so it can't be a non-zero number a.

If you don't know that 0/1 = 0, then you really, really should not be reading this! Master the basics before you try to move into higher-level stuff. As you've been told already, what you are doing can only harm you.
 
Saumyojit said:
I think i am right as u said
which is matching my view after susbtiton of gcd(a,b) in place of c in second equation
gcd(a,b) | a,b <=> gcd(a,b) | gcd(a,b)



i got it


Which definition : this one -> c | a,b => c | gcd(a,b).
I dont understand "we replace a literal "greatest" (that any other common divisor must be less) with divisibility, so that we can apply the concept of gcd to entities that have no "less than"
can u explain it to me with a eg.
What entities ?
Please use simple english .


Which one? can u highlight it please.


I need to understand this . I spent from morning 9 am to 5pm . But still these doubts are elft
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You have "reported" some posts as irrelevant - and no-body else has complained. If you have problems with some posts - I suggest either you ignore those posts or pitch tent in some other ground.

Stop complaining.....
 
divides every common divisor
In the link : comment of andre below the accepted answer
"taking the school definition, biggest common divisor, though we can adjust the proof if we use the "divides every common divisor" definition "
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What he is trying to imply?
When talking about the "biggest common divisor" is he implying that biggest common divisor can be only represented by a single number only ; that why only "d" fits .

But if we changed the def to "divides every common divisor" ; then d*a_0 can be included as now the def tells us gcd can be combination of nos in multiplicative form where each term (i.e d and a_0) is dividing every common divisor of a and b. Suppose a=8 b=16 Now the common divisors are { 1,2,4,8} so suppose a_0 =2 and d=4 then a_0 divides every common divisor except 1 and d divides common divisor of a and b except 1 and 2
' 1' is not divided . So where am i wrong
 
In reading that paragraph, I find that the context is hard to follow, and that sentence in particular is incomplete, so if it doesn't quite make sense to me, I just skip it, figuring that either he didn't write what he meant, or I'm misinterpreting what he meant (since he is just making a quick reference to a definition he doesn't fully state), or I'm just missing something and that's okay. I think it's a misstatement on his part, and since he didn't actually state his definition, there's no value in trying to figure it out.

So I would not obsess over it and try to understand every detail, assuming everything I read is correct. I would move on.
 
Andre is replying to Don larynx doubt. It's just there before the paragraph of what I have quoted.
I don't know what hard to follow.
Don larynx asked that "d * a_0" can be a GCD instead of only "d" if a_0 and b_0 are not co prime .
Then Andre is giving reply .
Of which I don't understand " divides every common Divisor"
 
Saumyojit said:
@pka @Jomo can u help me with this part as dr p is offline
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I will be delighted to help you once you state a complete statement of the question to be solved.
Please no referring back to replies in this thread.
 
pka said:
I will be delighted to help you once you state a complete statement of the question to be solved.
Please no referring back to replies in this thread.
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see post 49 . Its already stated clearly @pka @Jomo

after that come to post 54
there are two doubts post 49 and 54
 
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This is a reply to post 27 which i have understood
Dr.Peterson said:
Why do you say they must be?
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m= ha=kb
a=gcd(a,b) * a1
b= gcd(a,b) *b1
substituting in place of a and b in " m=ha=kb" => h*gcd(a,b) * a1 = k* gcd(a,b) *b1
Taking out gcd(a,b) from both the sides i get ( this step they did not show thats why the confusion)
h*a1=k*b1

UNderstood


Dr.Peterson said:
That's part of the proof! They've divided out the GCD in order to get a pair of coprime numbers, a1 and b1
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Okay
 
Saumyojit said:
see post 49 . Its already stated clearly @pka @Jomo

after that come to post 54
there are two doubts post 49 and 54
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This is how you respond to a request not to refer back to previous replies in the thread???

I, too, have no desire to read back through three pages of posts to figure out what you are asking. I'm not going to reply any more to this thread.
Saumyojit said:
View attachment 23790


In that same link ( https://math.stackexchange.com/ques...a-b-times-operatornamelcm-a-b-a/349953#349953) of the second answer given by rob john why did he showed product of two nos / gcd like this
taking out a from b/ gcd and b from a /gcd .
what is he implying
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It should be obvious: this shows that ab/gcd(a,b) is an integer multiple of a (namely, by b/gcd(a,b)) and also of b (namely, by a/gcd(a,b)).
 
Dr.Peterson said:
integer multiple of a
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it can just be a common multiple not a "integer multiple"



In the second answer of the given link ( https://math.stackexchange.com/ques...a-b-times-operatornamelcm-a-b-a/349953#349953)
Dr.Peterson said:
By the minimality of the lcm,
ab/gcd(a,b) ≥ lcm(a,b)
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What is the minimality they are saying ?
They are implying that LCM is the minimum no . something like that?
And ab/GCD gives a no which should be at least equal or greater than LCM.
They cannot directly say it can be equal to LCM as they are trying to prove that only.



Dr.Peterson said:
Because ab and lcm(a,b) are common multiples of a and b, so is r. By the minimality of the lcm, r=0. Therefore, lcm(a,b) divides ab.
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The way they have presented their flow of logic is actually quite the opposite I was thinking.
I was thinking this way
As I am writing this-> ab= q*LCM(a,b)+r
0<_ r < LCM (a,b)

I know that as LCM(a,b) and product of a and b are both Common multiples of a and b so I should get r =0 upon dividing ab/LCM as a*b is also a multiple of lcm(a,b) .
And as r=0 (0 is a mutliple of every integr)so I know that r is a common multiple of a and b.
But here one can say that then LCM must be 0 as it is the least. So to avoid it I can say that "Any common Multiple of a and b € N"
Natural no.

I am not sure!
If u compare my flow of thinking with them it's actually the opposite.
Their flow of thinking is
R is a common multiple of a and b then r=0 then LCM divides ab

My thinking is LCM divides ab ; then r=0 then R is a common multiple of a and b



What am I missing.
 
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