LCM Misconception
For integral choices of x and y , LCM ( x , y ) ≤ x y .
Is the above statement true or false?
Clarification: The LCM is the Lowest Common Multiple of two numbers.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
4 solutions
Moderator note:
Great! The LHS is positive / non-negative, so we just need to force the RHS to be negative.
Exactly! The same is for the fractions lying between 0 and 1. For an example, consider 1/2 and 1/4 and their LCM is 1/2. But 1 / 2 ≤ ( 1 / 2 ) ⋅ ( 1 / 4 ) is false.
Log in to reply
But then one could argue that LCM of 1 and 947 is -1894, since -1894 * -(1/1894) = 1, and -1894 * -(1/2) is 947, so... LCM of any number is ∞-0.000...01
he say for integral
If I go with definition of LCM I remember it can be applied for integers , I am not aware of this, also going by the example say addition or subtraction of fractions we consider lcm, even then lcm (2 and 4) is 4. I may be wrong but with some clear explanations one could clarify my doubt with this example.
I think the LCM must be 1 because it's integral
Log in to reply
I disapprove of what you claim, Arulx. LCM can be in fractions too. And the LCM of 1/2 and 1/4 is definitely 1/2.
Log in to reply
@Sandeep Bhardwaj – Yes, I figured it out using similar examples as yours. :)
@Sandeep Bhardwaj – Oh! Sorry for the comment earlier. I figured out that the definition which I learned was probably wrong. Thanks for correcting me!
Log in to reply
@Arulx Z – Thinking that the LCM is an integer, most likely arose because you were looking at the instance where the 2 values were an integer. In this case, because we are extending the definition, you have to be aware of what else could change.
Log in to reply
@Calvin Lin – What is the significance of LCM outside the context of positive integers?
You could argue any z (integer or otherwise) as a common multiple of non-zero x and y by using the multipliers z/x and z/y.
Log in to reply
@Tim Thielke
–
Yes, that argument is valid if it was defined by allowing for
rational multiples
. However, since the definition is for
integer multiples
, the interpretation is incorrect.
IE We are looking at the smallest positive number that is in the set
{
…
,
−
2
x
,
−
x
,
0
,
x
,
2
x
,
…
}
∩
{
…
,
−
2
y
,
−
y
,
0
,
y
,
2
y
,
…
}
. By convention, if the intersection is
{
0
}
, then we define the lcm to be 0.
Surely it's absurd to 'extend' the LCM to non positive integers? The LCM is a concept only designed for positive integers
Log in to reply
And it also doesn't make sense to extend it to non-integers. What is the LCM of pi and e?
People are citing 1/2 and 1/4, but if you extend LCM to fractions, I can say the LCM of 2 and 3 is 1 because 1/2 2=1 and 1/3 3=1.
LCM is a concept that becomes absurd and useless outside of the context of positive integers. You could argue any z as a common multiple of non-zero x and y by using the multipliers z/x and z/y.
Log in to reply
That is changing the definition, not extending it. We already know that the definition of lcm is LCM(2,3)=6.
There are many uses for extending the definition of LCM, which allows us to talk about the underlying structure of the set. For example, given functions f and g that are periodic with periods x and y , what can we say about the function f + g ? Is it periodic with period L C M ( x , y ) ? If so, does it matter if x , y are not integers, or even rational numbers?
In abstract algebra terms, we can ask for the LCM of 2 elements in a Commutative Ring , which extends our idea of the "lowest common multiple" into wider areas. It is defined as the generator of the intersection of the ideals ( x ) , ( y ) .
In a similar vein, the complex numbers help extend our idea of the real numbers, so that we can have solutions to the equation x 2 + 1 = 0 .
Log in to reply
@Calvin Lin – Let's take note that the problem is under Number Theory which focus on the study of integers and natural numbers ...
Log in to reply
@Remogel Pilapil – I believe one of the points Calvin Lin was making is that, although number theory may focus on the study of integers/natural numbers, this doesn't mean that one cannot use non-integers to accomplish that goal. As a more general example, how could one study light without considering darkness, or have an idea of good without an idea of evil?
Aside from that, the fact that a sub-discipline may focus on the study of a particular set of numbers (or, more generally, properties or ideas) neither implies nor necessitates that numbers (or properties, or ideas) outside that set are somehow off-limits or against the rules. Indeed, the very study of integers/natural numbers requires an understanding of the properties of negative numbers, irrational numbers, complex numbers, etc.
It is quite often (perhaps even always) the case that applying or extending the principles, methods of analysis, properties etc. of a particular discipline/field of study to objects beyond the common or traditionally-used set of circumstances and universes of discourse provides deeper, fuller insight into those principles. I find that it is rarely a good idea to restrict oneself to only a subset of possible applications, as it hinders one's ability to discover the deeper and more fundamental essence or nature of the particular theory in question.
Sorry for the tangential, philosophical exposition on the study of mathematics, but I simply couldn't help myself. :)
@Calvin Lin – Fair enough. I don't follow the commutative ring thing, but my knowledge of mathematics isn't nearly as deep as that of some around here.
The periodic functions explanation makes sense, though.
Log in to reply
@Tim Thielke – Great to hear! Unfortunately I couldn't come up with more simple obvious extensions, because it goes into deeper mathematical theory.
Do think about the periodic functions case, and especially what happens if we consider the fundamental period.
They said that about complex numbers too. Extending the lcm doesn't seem remotely useful, but it certainly makes sense and that's what math is about.
The LCM of real numbers a and b is the smallest positive real number that is an integer multiple of both.
LCM(a,b) = LCM(-a,-b) = LCM(a,-b) = LCM(-a,b)
LCM(xa,xb)=|x| LCM(a,b)
LCM(pi,e) = undefined
Yeah..., Nice solution , @Arulx Z
I though, that since lcm(a,B) x gcd(a,B) = aB, lcm(a,B) should be lesser than aB... :(
Log in to reply
i guess they involved -ve values too. but even then LCM of -8 & -6 => 24 or -24. [-6 = -2 * 3 or -6= 2 * -3] both are less than the product.
But if -8 & 6, LCM = -24 , product is -48. here product is less than LCM. So, not always.
Lowest Common Multiple of 8 and 6, for example is 24, which is less than 48. product of 8 and 6 (xy). So may answer to the question as TRUE seems to be correct!
Log in to reply
i guess they involved -ve values too. but even then LCM of -8 & -6 => 24 or -24. [-6 = -2 * 3 or -6= 2 * -3] both are less than the product.
But if -8 & 6, LCM = -24 , product is -48. here product is less than LCM. So, not always.
For positive integers, LCM * GCD = product. & GCD >= 1 Thus LCM <=product ..
For coprime integers(a,b) doesn't this work? LCM(5,8)<=(5)(8) 40<=40??
Log in to reply
You're making an implicit assumption about the integers.
so, the answer is ‘True’, right?! Why hasn’t Brilliant changed it?!
Log in to reply
The solution explains when the statement is not true:
Consider one positive integer a and one negative integer b . Here, the LCM is LCM ( a , b ) which is positive. However, a b is negative. Hence LCM ( a , b ) > a b .
The problem specified integral choices of x and y. 1/10 is not an integer.
i love you
marry me!!!
LCM(x, y)
if x=2, and y=10, LCM(x, y)=10
xy=2*10=20
10>20
10>xy
LCM(x, y) can be smaller than xy
What!!!!!! 10>20! Seriously??And what do you want to prove?
Will u please delete your answer
this made me lloose my brain
What is your point? The problem is asking if every LCM of two numbers is smaller or equal as the product of those numbers. It just makes nonsense what you are proving because your conclusion doesn't solve the main question.
10<20 ....A round of applause
You are copying the question statement. This means you accept that the answer should be yes but it is not.
Does LCM even have any meaning for irrational numbers?
LCM dont exist for irrational and rational numbers.
Log in to reply
LCM exists for some rational numbers: you only can obtain the LCM of a set of numbers if they are integers (and some complex numbers called Gaussian numbers). Integers are rational numbers.
Log in to reply
I wanted to say LCM(rational,irrational) dont exist
Log in to reply
@Gagan Sharma – Yes, sorry about that, I understand.
It depends on the numbers
One of the things to understand here is that LCM is always positive.
Consider one positive integer a and one negative integer b . Here, the LCM is LCM ( a , b ) which is positive. However, a b is negative. Hence LCM ( a , b ) > a b .
Sometimes, LCM doesn't exist and hence it can't be compared.