/sci/ - Science & Math » Thread #14368891

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given two natural numbers a and b

Anonymous Wed 06 Apr 13:16:45 2022 No.14368891 View Reply Original Report

Quoted By: >>14368899 >>14368959 >>14370308

why is it that LCM(a,b) x GCD(a,b) = a x b ?

LCM(a,b) is the Least Common Multiple of a and b
GCD(a,b) is the Greatest Common Divisor of a and b

Anonymous

Anonymous Wed 06 Apr 2022 13:19:00 No.14368899 Report

Quoted By: >>14368922 >>14368950

>>14368891
Because of the second isomorphism theorem for groups applied to Z/nZ, where n is any common multiple of a and b

Anonymous

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Anonymous Wed 06 Apr 2022 13:32:07 No.14368922 Report

Quoted By: >>14368926

>>14368899

Anonymous

Anonymous Wed 06 Apr 2022 13:34:21 No.14368926 Report

Quoted By: >>14368933

>>14368922
I did explain it simply though
if you need me to spell it out, let S be the subgroup of order a and let N be the subgroup of order b
Literally high school shit

Anonymous

Anonymous Wed 06 Apr 2022 13:38:10 No.14368933 Report

Quoted By: >>14368938

>>14368926
>I did explain it simply though
what is even an isomorphism?

Anonymous

Anonymous Wed 06 Apr 2022 13:41:04 No.14368938 Report

Quoted By: >>14368950

>>14368933
a structure-preserving bijection

Anonymous

Anonymous Wed 06 Apr 2022 13:45:00 No.14368943 Report

Quoted By:

Okay, I will bite. Let p be any prime, and let s be the power of p in the factorization of a, and t be
the power of p in the factorization of b. Then the power of p in LCM is max{s,t}, and its power
in GCD is min{s,t}. This makes the power of p in the product LCM x GCD equal to
max{s,t} + min{s,t} = s + t,
which is the same as the power of p in a x b.

Anonymous

Anonymous Wed 06 Apr 2022 13:49:05 No.14368950 Report

Quoted By: >>14368955

>>14368938
>>14368899
what is a structure? what is a group? what is nZ?

Anonymous

Anonymous Wed 06 Apr 2022 13:50:23 No.14368955 Report

Quoted By:

>>14368950
>structure
a set with some functions, relations or distinguished constants
>group
a structure with a single binary function satisfying three axioms which may be found on wikipedia
>nZ
all integers which are multiples of n

Anonymous

Anonymous Wed 06 Apr 2022 13:51:57 No.14368959 Report

Quoted By:

>>14368891
The least common multiple lcm(a, b) is defined as a*b/gcd(a, b), so the claim follow immediately since
lcm(a b)*gcd(a b)=(a*b/gcd(a b))*gcd(a, b).

So the real question is, why is lcm defined as lcm(a b)=a*b/gcd(a b)? Because if k=gcd(a b), then k divides both a and b, i.e.
a=km and b=kn
for some pair of integers m, n. Their product is then given by
ab=(km)(kn)=k^2 mn.
You will note that this product has an extra k term. We can factor that out, and the resuld
ab/k = kmn
Is still divisible my both a and b.

Anonymous

Anonymous Wed 06 Apr 2022 13:52:55 No.14368965 Report

Quoted By: >>14368999 >>14369008

Bunch of pseuds on this board, lol. Op here's the real reason. a and b both have prime factorizations. Let the product of all shared primes be S, and the product of all unshared primes be R. GCF(a,b) = S while LCD(a,b) = R/S (can you see why?)

Anonymous

Anonymous Wed 06 Apr 2022 13:57:51 No.14368977 Report

Quoted By: >>14369002

a=cA
b=cB
Where c is greatest common element
Now the least common multiple must be cAB
So axb=ccAB=gce x lcm

Anonymous

Anonymous Wed 06 Apr 2022 14:06:03 No.14368999 Report

Quoted By:

>>14368965
Fuck typo. LCD(a,b) = RS

Anonymous

Anonymous Wed 06 Apr 2022 14:06:46 No.14369002 Report

Quoted By: >>14371765

>>14368977
>Now the least common multiple must be cAB
why? you didn't specify what is A or B besides that they're the factor that remains when you divide the number a and the number b by their greatest common divisor

Anonymous

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Anonymous Wed 06 Apr 2022 14:08:28 No.14369008 Report

Quoted By: >>14369015 >>14369046

>>14368965
>a and b both have prime factorizations.
prove it
i will give a coutner example
what is the prime factorization of the number in the picture?

Anonymous

Anonymous Wed 06 Apr 2022 14:13:14 No.14369015 Report

Quoted By: >>14369041

>>14369008
$\prod_i=1^n \prod_j=1^m p_j^i$

For unique primes p_j. This is by the fundamental theorem of arithmetic. Now fuck off pseud.

Anonymous

Anonymous Wed 06 Apr 2022 14:21:42 No.14369041 Report

Quoted By: >>14369048

>>14369015
>no prime factorization givenj

Anonymous

Anonymous Wed 06 Apr 2022 14:23:00 No.14369046 Report

Quoted By:

>>14369008
That's actually a perfect power of 2, surprisingly

Anonymous

Anonymous Wed 06 Apr 2022 14:23:36 No.14369048 Report

Quoted By: >>14369057

>>14369041
I don't need to know what that prime factorization is. Fundamental theorem of arithmetic guarantees such a unique factorization exists, which can be written down as a product of primes (ignore my mangled LaTeX.. you try writing that shit on mobile). Are you trying to assert the fundamental theorem of arithmetic is wrong, or invalid, for large integers?

Anonymous

Anonymous Wed 06 Apr 2022 14:27:58 No.14369057 Report

Quoted By: >>14369216

>>14369048
>I don't need to know what that prime factorization is
>I don't need to show the prize factorization to prove it exists

Anonymous

Anonymous Wed 06 Apr 2022 15:37:46 No.14369216 Report

Quoted By: >>14370249

>>14369057
I don't need to prove it exists. It already exists.

Anonymous

Anonymous Wed 06 Apr 2022 22:39:14 No.14370249 Report

Quoted By:

>>14369216
>It already exists
Then write it down for me without using multiplication, division, addition or subtraction, just the digits

Anonymous

Anonymous Wed 06 Apr 2022 23:00:38 No.14370308 Report

Quoted By: >>14371707

>>14368891
Here are some hints:
Let g = GCD(a, b) and l = LCM(a, b). Obviously, there exist positive integers x and y such that a = xg and b = yg.
Hint 1) Write ab in terms of these numbers x, y, and g
Hint 2) The factorization of positive integers is unique
Hint 3) Since the GCD of two numbers divides both of them, the GCD should divide BOTH the LCM as well as their product.
Hint 4) Examine the discrepancy between the product of ab and LCM(a,b) both expressed as a multiple of the GCD.

If you understand the problem, then the hints above are basically a proof sketch. If you don't, think hard about hints 3 and 4. The whole point of this problem is in hint 4.

Anonymous

Anonymous Thu 07 Apr 2022 11:52:31 No.14371707 Report

Quoted By:

>>14370308
ok thanks

Anonymous

Anonymous Thu 07 Apr 2022 12:17:59 No.14371765 Report

Quoted By:

>>14369002
A and B are unique to a and b, respectively. This follows that a number that is divisible to both a and b must have factors c, A and B. The least of them is of course cAB

Anonymous

Anonymous Thu 07 Apr 2022 23:50:33 No.14373507 Report

Quoted By:

bump

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