Seems to work like a charm!!  Did 2 and 3 numbers filled in, primes and related multiple cousins etc.

Thank You Beth!

Hi;

It is working fine. No problems.

lcm is defined for positive integers only, so why should anyone input anything else, besides their curiosity?

Hi MIF! 

You could allow the minus sign, since according to Wikipedia the lcm of two integers is the smallest
POSITIVE integer that is divisible by both.  In essence the minus signs are just ignored.  Wiki also
says that if either a or b is zero then the lcm is zero.

I've never seen lcm applied to decimals although your program seems to work accepting them as input but treats them as if they had no decimals and then outputs the correct integer accordingly but puts a decimal in the answer so that both inputs divide into it an INTEGRAL number of times.

If I input 2.43 and 8.1 into your program it outputs 24.3 which both 2.43 and 8.1 divide into an
INTEGRAL number of times.  So maybe you've come up with a way to define lcms of terminating
decimals!   Inputting .166 and .333 gives 55.278 which both .166 and .333 divide into an INTEGRAL
number of times.  Perhaps this could be extended to fractions if we write them in a base that makes BOTH of them TERMINATING decimals???

Hmmmmmmmm... 

Hi again!

Take the equation

   x        3x
----- + -----  =  10  and multiply both sides by the lcm 24.3 that your program gives.
2.43     8.1

Then we get  10x + 9x = 243  so 19x=243 so x=12.789.  So maybe there are some decent
applications for the concept.

Of course we could have multiplied the equation through by both 8 .1 and 2.43 and solved, but
then we would have had decimal coefficients for the variable.
    8.1x + 3(2.43)x = 10(8.1*2.43)
        8.1x + 7.29x = 196.83
                 15.39x = 196.83
                         x = 196.83/15.39
                         x = 12.789

Does make the arithmetic a bit easier!

But perhaps we can get lcms for fractions a/b and c/d in the sense that we are looking for the
smallest fraction that both a/b and c/d divide into giving integers.

Example:  15/77 and 25/49  and see if 75/7 does the trick. 
        (75/7)/(15/77)=(75*77)/(7*15)=5*11=55   and  (75/7)/(25/49)=(75*49)/(25*7)=3*7=21
which yields integral values for each division.
                 x           x
So given  ------ + ------- =  7  and multiplying both sides by 75/7 we obtain
              15/77    25/49

                55x  +  21x   =  7*(75/7)  =  75
                               x   =  75/76
                               
It looks like lcm(a/b,c/d) = lcm(a,c)/hcf(b,d)

Example:  lcm(1/4,1/6) = lcm(1,1)/hcf(4,6) = 1/2.    (1/2)/(1/4)=2  and (1/2)/(1/6) = 3.
Example:  lcm(4,6) = lcm(4/1,6/1) = lcm(4,6)/hcf(1,1) = 12/1 = 12.  (Works for integers)
Example:  lcm(2/3,6/15) = lcm(2,6)/hcf(3,15) = 6/3 = 2
                2/(2/3) = 3  and  2/(6/15) = 5

Hmmmmmm.   This might at times be an easier approach to solving equations involving fractions.

But of course for just two fractions we could replace lcm(a,c) with ac/hcf(a,c) so the formula
would become  lcm(a/b,c,d) = ac/(hcf(a,c)*hcf(b,d))

And would lcm(a/b,c/d,e/f) = lcm(a,c,e)/hcf(b,d,f)  etc. for more than two fractions?

Your program is generating some interesting questions!

Yes I have: Greatest Common Factor Calculator

The LCM calculator is a Flash App that uses my "full precision" library

But the GCF calculator is a fairly simple javascript program ... I could re-make it in Flash.

Hi MIF;

This is a very good job. Congratulations!

Hi again!

An interesting note:

Assume a,b,c,d are integers in the following and that the fractions are in reduced form.
(It still seems to work OK even if the fractions are not reduced.)

Using hcf(a/b,c/d)=hcf(a,c)/hcf(b,d)   and   lcm(a/b,c/d)=lcm(a,c)/hcf(b,d)   works for
whole numbers like 10 and 15 written as 10/1 and 15/1.

Example1:
hcf(10,15) = hcf(10/1, 15/1) = hcf(10,15)/hcf(1,1) = 5/1 = 5.
lcm(10,15) =lcm(10/1, 15/1) = lcm(10,15)/hcf(1,1) = 30/1 = 30 and hcf*lcm=5*30=150=10*15

So whole numbers (and so also integers) also work under the definition of hcf for fractions
with the usual M*N=lcm(M,N)*hcf(M,N) formula intact.

BUT for other kinds of fractions this product of the original two numbers equals the product
of the hcf and lcm does NOT necessarily work.

Example2:  hcf(1/10, 1/15) = hcf(1,1)/hcf(10,15) = 1/5.
                 lcm(1/10,1/15) = lcm(1,1)/hcf(10,15) = 1/5
so hcf*lcm = (1/5)(1/5)=1/25  whereas (1/10)(1/15) = 1/150.  The hcf*lcm is missing the other
factor of each of the 10 and 15.  So we only get the 5 and 5 but not the other factors 2 and 3.

Example3: hcf(15/8, 25/6) = hcf(15,25)/hcf(8,6) = 5/2.
                lcm(15/8, 25/6)  = lcm(15,25)/hcf(8,6) = 75/2.
so hcf*lcm = 375/4  whereas (15/8)(25/6)=375/48.  So again we are missing the other factors
in the denominator.  The numerators are always the same since they are a product of the lcm
and hcf of INTEGERS.

So it looks like in the case of integers, we get hcf(M,N)*lcm(M,N)=M*N as a SPECIAL CASE of the
more general definition of lcm and hcf because the denominators are both 1's. 

Example4:  hcf(10/7, 15/7)=hcf(10,15)/hcf(7,7) = 5/7.
                 lcm(10/7, 15/7)=lcm(10,15)/hcf(7,7) = 30/7.
So hcf*lcm = (5/7)(30/7) = 150/49  and  (10/7)(15/7) = 150/49.
So if BOTH denominators are the SAME then the product of the original numbers = lcm*hcf holds.

Example5:  hcf(10/3, 15/7)=hcf(10,15)/hcf(3,7) = 5/1 = 5.
                 lcm(10/3, 15/7)=lcm(10,15)/hcf(3,7) = 30/1 = 30.
so hcf*lcm = 5*30=150  whereas (10/3)(15/7)=150/21  Again these are not equal.

CONCLUSION1:  THE lcm*hcf BEING EQUAL TO THE PRODUCT OF THE original two numbers
                        ONLY WORKS WHEN THE TWO NUMBERS HAVE THE same DENOMINATOR.
                        Of course, integers are written over 1 to make them fractions for the formula.

CONCLUSION2:  The old trick of calculating the lcm by dividing the product of the original numbers
                         by the hcf cannot be used when dealing with fractions unless their denominators
                         are the same.

CONCLUSION3:  Given fractions a/b and c/d with a,b,c,d integral the equality
                          lcm(a/b, c/d) = a*c/(hcf(a,c)*hcf(b,d)) is I believe true because we can
                          substitute lcm(a,c) = a*c/hcf(a,c) since a and c are integers.

CONCLUSION4:  Given a/b and c/d if gcd(b,d)=1 then the lcm and hcf of the two fractions are
                         integers.  See example 5.  Furthermore they are the lcm and hcf of just the
                         numerators.

So MIF, can I blame my lack of sleep on you?  You really got my mind a buzzin' with your lcm
calculator!        

P.S.  The decimals seem to still work for the lcm and gcd calculators and seem to give the same
answer when changed into fractions.