Question

Is the lcm/gcd always an integer?

Answer 1

\[\Large \frac{lcm(a,b)}{\gcd(a,b)} \in \mathbb{Z}\]

Answer 2

I checked on internet , ( I am sure you must have did it too haha)
If it is or it is not how will you genralise it for all numbers

Answer 3

The GOOGLE is silent on this issue

Answer 4

are a,b integers ?

Answer 5

Yes, I suppose I didn't even know you could find the LCM or GCD of nonintegers, but I guess you could haha.

Answer 6

I would imagine that if you could find the GCD of nonintegers then it would always just be the smallest of the two numbers, since you can find something that multiplies always, so it doesn't seem very interesting.

Answer 7

lcm of two integers `a`, `b` is the positive integer `m` satisfying below properties :
1) a|m and b|m
2) If a|c and b|c, then m<= c

Answer 8

@No.name But if there was a "noninteger" gcd it would just be this: \[\gcd ( 5/2,7)=5/2\] since 7 is divisible by 5/14.

Answer 9

Also using this fact: \[\gcd(a,b) lcm (a,b)=ab\] we can see that \[\frac{lcm(a,b)}{\gcd(a,b)}=\frac{lcm(a,b)}{a}*\frac{lcm(a,b)}{b}\] which is an integer because lcm is divisible by a and b.

Answer 10

What ganeshie is saying, but yea you have to break it up, so the gcd of a,b divides both a and b

Answer 11

the lcm, even if is only in terms of one will be able to be divided

Answer 12

right, next you want to prove gcd(a,b) * lcm(a,b) = a*b ?

Answer 13

No, but I guess I would like to prove in general for \[\frac{lcm(a,b,...,z)}{\gcd(a,b,...,z)} \in \mathbb{Z}\] if it is true, which I'm not sure it is.

Answer 14

could do prime factorization then show it, but I believe conceptually it must be true

Answer 15

you just need to extrapolate the previous argument

Answer 16

use lcm and gcd definition
they dont bother about number of operands

Answer 17

lcm(a,b,c) = lcm(lcm(a,b), c)
gcd(a,b,c) = gcd(gcd(a,b),c)

Answer 18

Yeah, it seems like it should be true. I believe that this might be related to the determinant.

Is this true @ganeshie8 \[\gcd(a,b,...,z)*lcm(a,b,...,z)=a*b*...*z\]

I guess I don't understand the proof of this in the 2 argument case I guess, it just kind of makes sense to me, but not completely.

Answer 19

suppose all a,b,...,z are prime
then certainly that relation holds

Answer 20

proof is trivial, i am just trying to convince myself it holds in general by considering examples

Answer 21

it wont hold when the gcd of numbers is not 1
we need to use the previous split up when we have more than 2 arguments

Answer 22

Ahhh I like that, if they're all prime it is kind of obvious that it will be true.

Answer 23

lcm(a,b,c) * gcd(a,b,c) = lcm(lcm(a,b), c) * gcd(gcd(a,b), c)

it ends here.

Answer 24

lcm(2,2,2) * gcd(2,2,2) = lcm(lcm(2,2), 2) * gcd(gcd(2,2), 2) = lcm(2,2) * gcd(2,2) = 2*2 = 4 \(\ne\) 8 = 2*2*2

Answer 25

yeah simply extrapolating could be dangerous sometimes >.<

Answer 26

Ahhh yeah ok so it doesn't work in general. Oh well.

Answer 27

it works when all numbers are coprime so that no common factors get dropped and you get a full prime factorization of the product a*b*...*z on left hand side

Answer 28

I feel like it should work in general because by def. the lcm is the lowest common multiple of the numbers ie a|m and b|m and then the gcd is the biggest number s.t g|a and g|b

Answer 29

so then you have at the smallest a/g or b/g which divides evenly

Answer 30

yeah it works always when there are only two numbers

Answer 31

should work for more too by definition

Answer 32

counter example

lcm(2,2,2) * gcd(2,2,2) = 2*2 = 4

Answer 33

? that's not a counterexample...?

Answer 34

2/2 would equal 1 which is indeed an integer

Answer 35

we want it equal to 2*2*2 = 8 right ?

Answer 36

oh are you trying to prove your earlier assertion?

Answer 37

Ohkay i got you :) ratio of lcm and gcd of any number of integers is always an integer

Answer 38

that holds by definition yeah

Answer 39

yea, I was just proving that part

Answer 40

yours, I agree doesn't work by your counterex then

Answer 41

gcd(a,b,...,z) | a
gcd(a,b,...,z) | b
....
gcd(a,b,...,z) | z


so it divides their product also :
gcd(a,b,...,z) | a*b*...*z

Answer 42

So is there some way to create extra function(s) that have meaning that take what's leftover? \[lcm(a,b,...,z)*\gcd(a,b,...,z)*???(a,b,...,z) =a*b*...*z\]

Answer 43

prime factorization would be a good start in pursuit of that function

Answer 44

\[a = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}\]
\[b = p_1^{b_1}p_2^{b_2}\cdots p_r^{b_r}\]
\[\cdots\]
\[z = p_1^{z_1}p_2^{z_2}\cdots p_r^{z_r}\]


\[\gcd(a,b,\cdots,z) = p_1^{G_1}p_2^{G_2}\cdots p_r^{G_r},~~~G_i = \min\{a_i,b_i,\cdots,z_i\}\]
\[\mathrm{lcm}(a,b,\cdots,z) = p_1^{L_1}p_2^{L_2}\cdots p_r^{L_r},~~~L_i = \max\{a_i,b_i,\cdots,z_i\}\]

Answer 45

just defined gcd and lcm in terms of prime factorization so that we have some concrete stuff to play with

Answer 46

now its easy to define your desired function in terms of \(G_i,L_i \) and \(a_i,b_i,\cdots, z_i\)

Answer 47

Well since gcd and lcm and all the numbers multiplied together are already defined, it's just algebra to define it, however this might help in trying to understand what the function represents.

Answer 48

right, its just algebra

Answer 49

So in my kinda hacky way of doing it that conveys the idea: \[\Large f(\bar x)=\frac{\prod( \bar x)}{\gcd(\bar x) lcm(\bar x)}\]

Answer 50

Okay what good is this function ?

Answer 51

what we can do with this ? like what properties it has ?

Answer 52

I don't know, all I know is that when f has two arguments it always returns 1.

Answer 53

also you could say it returns 1 when gcd(a,b,...z) = 1

Answer 54

all other cases are complicated i think

Answer 55

So does it mean anything to say that f=gcd when the arguments are mutually prime?

Answer 56

your function is a number theoretic function as it works on integers and spits out an integer

Answer 57

is this true :
gcd(a,b,...z) * lcm(a,b,...,z) <= a*b*...*z

?

Answer 58

If all the arguments are x and there are n number of arguments then f(x)=x^(n-2)

Answer 59

Yes, that is true @ganeshie8

Answer 60

yes then your function qualifies as a number theoretic function

Answer 61

Another function we might want to consider that I just made up might or might not help us is: \[plum(p^aq^br^c)=a+b+c\]

so this is the prime logarithm sum that takes the exponents on the prime factorization and adds them. I guess we could make a product version and call it the ploduct. Sorry for the silly names haha.

Answer 62

there is a number theoretic function similar to this already
let me pull up my notes

Answer 63

Ok, cool, I would like to learn more. You might be interested in the arithmetic derivative as well.

Answer 64

Ahh they are so juicy.. idk abc of them
they are part of algebraic number theory... i only took elementary number theory

Answer 65

\(\omega(n)\) gives number of distinct prime factors of \(n\)

Answer 66

Have you heard of the primorial function? It's like the factorial function: \[10 \# = 2*3*5*7\]

Answer 67

there is Lioville \(\lambda \) function
it is defined based on your plum function :
\[\lambda(n) = (-1)^{\mathrm{plum}(n)}\]

Answer 68

Hmm what's the purpose of this function? Seems interesting. Just sort of figuring stuff out, like plum(x)=1 iff x is prime.

Answer 69

Oh factorial of all the primes less than or equal to a number ?
never used it much Kai

Answer 70

it is used in many places, used for testing perfect squares i guess
let me take a screenshot and attach

Answer 71

I found a use for primorial for a couple things. One is the proof that there are infinitely many prime numbers and the other is with the arithmetic derivative: \[\LARGE \frac{a}{a\#}\le \gcd(a,a') \le a\]

Answer 72

Although now that I think about it, I think that might be wrong.

Answer 73

Oh can i see the proof of infinite primes using primirial function ?

Answer 74

It's sort of a superficial use of it, it's just that if you think there is a finite number of primes, then if you multiply them all together and add 1 to it, you have now created a number that isn't divisible by any of the primes in that set so you have to add another one. So you could use the primorial here to say that this was all the primes in your set multiplied together.

Answer 75

okay looks a lot like euclid proof

Answer 76

Yeah, that's all it is, Euclid's proof except slightly less general since you could say your finite number of primes multiplied together is 2*5*7 and leave out 3.

Answer 77

hmm il need to think about that statement. Euclid's contradiction proof olny requires us to consider all the finitely many primes if they exist. so missing/not missing few is never a problem..

Answer 78

here is a cool property of your function

\[\sum\limits_{d|n}(-1)^{\mathrm{plum}(d)} = \left\{\begin{array}{} 1 &:&\text{if n is a
perfect square} \\0 &:&\text{otherwise}\end{array}\right.\]

Answer 79

thats trivial to see/prove i think
but certainly i feel we can apply that function in many other areas, it looks so general...

Answer 80

is plum multiplicative ?

Answer 81

plum(m*n) \(\ne\) plum(m)*plum(n)

but this holds
plum(m*n) = plum(m) + plum(n)

Answer 82

here is another property

\[\sum\limits_{i=1}^n (-1)^{\mathrm{plum}(i)}\left\lfloor \frac{n}{i}\right\rfloor =
\left\lfloor \sqrt{n}\right\rfloor\]

Answer 83

Hmm interesting, but what does d|n mean for this:

\[\sum\limits_{d|n}(-1)^{\mathrm{plum}(d)} = \left\{\begin{array}{} 1 &:&\text{if n is a
perfect square} \\0 &:&\text{otherwise}\end{array}\right.\]

Answer 84

I really like the plum(a*b)=plum(a)+plum(b) property, that is pretty cool. Definitely feels like a logarithm.

Answer 85

d is a divisor of n

Answer 86

Oh I see this is basically saying that squares have an odd number of divisors.

Answer 87

\(\sum \limits_{d|n}f(d)\) is the sum of f(d) s as d ranges over all the divisors of n

Answer 88

\[\sum\limits_{d|6} f(d) = f(1) + f(2) + f(3) + f(6)\]

Answer 89

Exactly Kai !!

Answer 90

Not sure if you guys wanted this still...

\(gcd(a,b)lcm(a,b)=ab\implies \frac{lcm(a,b)}{gcd(a,b)}=\frac{a}{gcd(a,b)}\frac{b}{gcd(a,b)}=k_0*k_1\in \mathbb{Z}\)

Answer 91

Makes sense, since usually factors come in pairs, but squares are the one case where it has two factors that are identical. Interesting.

Answer 92

@zzr0ck3r Yeah we basically figured it out but the other way around, where we solved for gcd and plugged that in, and noting that lcm is divisible by a and b.

Answer 93

\(\color{blue}{\text{Originally Posted by}}\) @Kainui
Also using this fact: \[\gcd(a,b) lcm (a,b)=ab\] we can see that \[\frac{lcm(a,b)}{\gcd(a,b)}=\frac{lcm(a,b)}{a}*\frac{lcm(a,b)}{b}\] which is an integer because lcm is divisible by a and b.
\(\color{blue}{\text{End of Quote}}\)

Answer 94

perfect squares are just one case, yes

Answer 95

word, sorry I got lost in the convo:) gn

Answer 96

no prob, take it easy unless you read further and get interested haha.

Answer 97

@ganeshie8 the reason I am interested in this is because I sort of imagine treating exponents on primes as spanning a vector space.