Math Is Fun Forum

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.

The gcd calculator seems to work fine for positive integer inputs.  I input negative integers, decimals and fractions and it just returned 1 in every case although it allows inputting these (fractions in form a/b) forms.  I have written many programs in BASIC and have had to try to "idiot-proof" them.  It's difficult to anticipate what other people might do.

Somewhere down the line when you have a good chunk of time (if ever) it would be nice to have a
calculator that allows the input of integers, decimals and fractions as well as positive integers and
then have the program calculate both the lcm and gcd  for the input set of numbers.  You could
have the first site on the internet that does these calculations for all these kinds of inputs.  It might
generate a good bit of curiosity and cause a bit more membership and traffic on the site.

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!

Hi MIF!

Did you mean for it to work with decimals too?  The few I tested worked.
Did you mean for it to work with negative numbers?  That didn't work.

Works nice for positive integers.

You might consider limiting the inputs to just positive integers.
Zero poses a problem, since any number times zero equals zero.  multiplies of zero are 0,0,0,...
The multiples of 2 are 0,2,4,6,...  So accordingly lcm(0,2) would be zero.

People are going to try to input all kinds of numbers if there are no restrictions. 
It may be better to simply not accept their inputs rather than accept the inputs and then
not get an answer or get an answer they can't understand. 

Yes indeedy! 

I agree.  After all math is a language.   We don't have to speak the language if we don't want to.  Or we can use only the parts of the language that we choose.  It all depends on our individual axioms of choice.

Hi!

The general approach to absolute value equalities is to consider where the quantities inside the
absolute values are positive, zero, or negative.  Try the following:

Where both of 2x+1 and 2x-1 are positive replace the absolute values with these and solve.
Where both of 2x+1 and 2x-1 are negative replace the absolute values with the opposite of these
    and solve.
Where 2x+1 is positive and 2x-1 is negative replace |2x+1| with 2x+1 and |2x-1| with -(2x-1)
    and solve.
You will also find solutions where each of these quantities individually are zero.

You will find the solutions in [-1/2, 1/2].

Hi!  I like E best except for the baby.  It is hard to tell which way it is facing compared to that in V.
     Also it might be better to use something like 300 liters instead of 30000 liters.  Big numbers
     are scary to some folks.

     To me the music in S is distracting and S seems to move a bit fast since it seems to do more
     showing of answers on the screen without verbalization.

     Multiplication of signed numbers is difficult to explain.  I like the walking on the number line
     and money examples much better than the tank example. 

     I don't know whether it would be worth mentioning that the calculation of the size of the answer
     is independent of finding the sign of the answer.  The videos certainly follow that pattern.  But
     in addition and subtraction the size of the two numbers can affect the sign of the result.  So
     addition and subtraction are typically harder for people to learn.  Perhaps the subject of
     another video?

     I congratulate you on producing these videos.  Seeing the operations in real time and being
     explained in real time certainly beats plane jane black and white pages of explanations.
     Your videos are more "colorful and inviting" than Khanacademy's but probably take longer to
     produce and are probably are more expensive to produce.

     These videos "beg" for more on addition, subtraction and division.

     Keep up the good work!!!

Hi!

Here's an interesting approach to averaging.  Averages work similar to game scores.  Let the Home
team be on top and the Visitors on the bottom, supposing that we are dealing with football.  By
quarters suppose we have:
                    1st Quarter  2nd Quarter  3rd Quarter  4th Quarter   Totals
Home Team         7                 3                  14                6             30
       Visitors        3                 6                    7                7             23

We can write this more concisely as "scores" as follows:
                            7 + 3 +14 + 6     30
                            ~   ~   ~~   ~ = ~~   so the Home team won by 7 points.
                            3 + 6 + 7  + 7     23
Note that we are interested in the difference not the quotient of the two numbers, hence we use
the "~" to distinguish these from fractions which use the "-" or "/".

Now averages work similarly.  To average test scores of 80, 70 and 90 we obtain
                                    80  + 70 +  90     240                                        80
                                    ~~    ~~    ~~ = ~~~ which reduced by 3 gives ~~.
                                     1   +  1  +   1       3                                           1

The reducing is like asking what same score should be made on all three tests to get 240 points.
Obviously it is 80, 80 and 80 (which is the 240 divided by 3).

Now for a more difficult question we could ask what must  one average on the next two tests
to raise the overall 5 test average to 85?

                               80 + 70 + 90 + x +  x     240 + 2x       85
                               ~~   ~~   ~~    ~    ~  = ~~~~~~  =  ~~
                                 1 +  1  +  1 +  1 + 1           5              1

We set the sum equal to the desired five-test average.
Cross multiplication of the last two expressions yields 240+2x = 425
                                                                                     2x = 185
                                                                                       x = 92.5
Of course it is not likely that one would make 92.5 or each of the two tests.  More likely it would
occur as 92 and 93,  or 91 and 94,  or 90 and  95, etc.

The idea and notation of "scores" also works well with some other types of algebra word
problems, for example mixture problems and some rate/time/distance problems.
(Scores are to signed numbers as fractions are to rational numbers.  Fractions can be viewed
as "pre-rational numbers."  So scores can be viewed as "pre-signed numbers."
                                                                                                     0  1   2  3
   .5 corresponds to {1/2, 2/4, 3/6, ...}  whereas  -2 corresponds to { ~, ~, ~, ~, ...}
                                                                                                     2  3   4  5
                                                                                                     2  3   4  5
   2. corresponds to {2/1, 4/2, 6/3, ...}  whereas +2 corresponds to { ~, ~, ~, ~, ...}
                                                                                                     0  1   2  3
-2 is like the Home team being 2 behind and +2 is like the Home team being 2 ahead.

Example of a rate/time/distance problem:
   A man made a trip in two legs.  The first leg he went two miles in 3 hours and the second leg
he went 3 miles in 4 hours (pushing his car perhaps!).  What was his average speed on the trip?
                                                                 2mi       3mi
If we treat these ratios as fractions we have  ---- and ---- .  If we add, subtract, multiply or
                                                                 3hr        4hr
divide these fractions, none of the results are what we need to solve the problem.  On the other
hand if we treat them as scores we get
       2 + 3     5mi                                                    5/7 mi 
       ~    ~ = ~~~  and reducing this by 7 we obtain  ~~~~ ; that is, 5/7 mph.
       3 + 4     7hr                                                       1 hr

So this problem which is difficult for many people to figure out is only TWO steps.
1)  Add the scores and 2) reduce by the bottom number to get 1hr in the bottom.

An additional question: How long should a 3rd leg of 5mi take to raise the 3leg average to 1mph?
      2 + 3 + 5     1                   10      1
      ~   ~    ~ =  ~  becomes  ~~~ = ~  on cross multiplying becomes 7+x=10 so x = 3hrs.
      3 + 4 + x     1                  7+x     1

The language of fractions is the wrong language to work this kind of problem as well as averages.
We probably compare two numbers by differences as much as or more than we do as quotients.
The language of fractions is available for the quotient comparisons.
The language of scores for difference comparisons is analogous.

Here I am getting a bit "long winded" but one more observation might be helpful.

The idea of WEIGHTED averages works well with scores also.  For example if we have
three 90's and two 80's and a 70 to average we can write out all six scores individually and
add them or we can write
        x 90   +    x 80   +  70       270  +   160  +  70      500                           500/6       83.3
      3   ~~      2   ~~      ~~  = ~~~      ~~~     ~~  = ~~~ which reduces to ~~~~  = ~~~~
        x  1    +    x  1    +  1         3     +    2    +   1         6                              5/5           1
Then we might ask how low an average can we get on the next 4 tests to keep at least an 80
average on the ten tests altogether.
                   x  z              4z                                      80                    500+4z     80
Just add in  4   ~; that is, ~~ and set the total equal to ~~ and solve:   ~~~~~ = ~~
                   x  1              4                                        1                        10           1

500+4z = 800 so 4z = 300 so z = 300/4 = 75.

Grade point averages used so much at colleges are basically weighted averages also.

Scores are a "whole different ball game" so to speak.   

Hi stefy!

Perhaps we were taught differently.  I was taught that using the set W={0,1,2,3,...} of whole numbers, the set of fractions derived thereby is F = {a/b | a and b are whole numbers, b not 0}.
Then the rational number .75 would be the class {3/4, 6/8, 9/12,...} of equivalent fractions.

Having differing viewpoints often leads to interesting conversations, and occasionally to new ideas
in mathematics.  You apparently have a quite varied and extensive background in mathematics. 
More power to you!

Interesting side note:  Most people say you can't have zero in the denominator of a fraction
because division by zero is not defined.  I see the reason for not having zero in the denominator
as being the definition precludes it from the very beginning.  Just writing something like 2/0 is
violating the definition.

Hi!

You might write the remainder as -2/3.  The quotient is 3x^2 + 13/3. 

Hi Y'all!

Indeed a nice way to get the n numbers in between! However the following question arises.

Are we talking about n fractions between two fractions or n rational numbers between two rational
numbers? (or fractions between two rational numbers, or perhaps rational numbers between two
fractions?)  It seems that the question is for fractions.  Given two rational numbers 2.5 and 3.0 we
can get any number of rational numbers between these by just starting with 2.5 and tack on any
digits we wish beyond the 5 in 2.5.  For example 2.53, 2.56, 2.568, 2.5809 are four rational
numbers between 2.5 and 3.0.  And there are many other options starting with 2.6, 2.7, 2.8 or 2.9.

For four fractions between 5/2 and 3/1 we can use  11/4, 22/8, 33/12, and 44/16.  These are
different fractions between 5/2 and 3/1, but all correspond to only one rational number 2.75
between 2.5 and 3.

We tend to use for each rational number the reduced fraction that corresponds to it.  .5 and 1/2
are used more or less interchangeably although they are not technically speaking the same
thing.  But for .285714... (repeating infinitely this 6 digit block) we tend to prefer the fraction 2/7.

.75 represents the total amount represented by each fraction in {3/4, 6/8, 9/12, 12/16...} but
each of these fractions indicate how that amount is divided up (For example, 3 fourths  pie,
6 eights pie, 9 twelfths pie, 12 sixteenths pie, ) We especially pick on 75/100 when reading .75 but
have you ever seen a pie cut into 100 pieces? A nice pecan pie with whole pecans  would become
a pecan pudding with diced pecans!

Picky, picky, picky, eh? 

Just had to get my 2/100 dollar in (or is it 4/200 or 6/300 etc. dollar?).

Hi again!

You might check out "infinitesimal calculus" via Google.  Lots of interesting sites.  Robinson introduced
his views in 1960's.  Kiesler has a text about it.  If I recall correctly infinitesimal was introduced as an
infinite sequence that converges to zero.