Math 20 - Lesson 26

A Portland Community College mathematics telecourse. A course in arithmetic review produced at
Portland Community College. This lesson will tackle the problem of adding two fractions
that are not like fractions. What to do. You know, even if this were not a review course
for you, I'd bet that you would probably know what's going to be done. If so, then you are
beginning to think like a scientist. That is, you're asking what have you known before
this that would relate to this problem? You'd probably tell yourself, "I can't add until
these are the same, but through fraction building I can make these be many, many things." So
I ask myself, "What's the one number that both of these will divide into?" That is,
what is the least common multiple of these two? And that of course is 12. Then, you would
say, well I could build 3/4 to 12ths by multiplying top and bottom by 3. And I could build 6 to
12ths by multiplying the top and bottom by 2. So, my two fractions have been changed
to equivalent fractions 9/12 and 2/12. And now the denominators are the same. They are
like fractions, so my denominator stays the same and I add the tops, check to see if it's
reduced and I'm done. So, in principle, adding fractions with unlike denominators are simple.
Simply get them to have common denominators. Then, go ahead and add. Again, to add unlike
fractions, build them to have common denominators, then add. Then, check to see if reduction
is possible. Quite frequently one can simply see what the least common multiple will be.
In this case, what's the least number that these three will divide into? And of course,
we can just see that it's 30 and in this case, we'll build with a factor of 6 to get 30.
This already is 30 and here we'll build by a factor of 5 to make it 30. And my three
fractions now become the three equivalent fractions 12, 7 and 5/30. But, since the denominators
are now the same, we keep it the same and add the numerators. So, this is 12 plus 12
is 24 and this would reduce. 4 goes into. . . 6 goes into 24 4 times. 6 goes into 30
5 times. So, 4/5 is the sum of these two fractions. Most problems the average person runs into
will have denominators so simple you can just see it. Most of them. However, in technologies
and engineering and sometimes business, it isn't quite that simple. But first, let's
answer the question, what would have happened had we not picked the least common denominator
which was 30? We could have picked 60 or 90 in this case. In such a case, and as a matter
of fact any common multiple of the denominators would work, but any but the lowest will always
present you with more reducing on the answer than you would encounter otherwise. In short,
you would simply get more work to do and that's something generally we try to avoid. Now,
here's a beast of a problem that most people would feel they couldn't do. But, it really
reduces to just this. How do you find the lowest common denominator given enough time
to work with it with a calculator or however you wish. Now, with your background so far,
this should be a very routine problem. However, it will take a bit of time. Could you instruct
a friend, step by step, through this problem? What would you tell them to do first in finding
the least common multiple of those three numbers? You'd tell them to prime factor each denominator
and express each prime factorization in exponential form. Let's do each one very quickly as one
last review. Factoring, or by repeated divisions by prime numbers, we would get that 196 is
2 to the second power times 7 to the second power. So, let's write that. 2 to the second
power times 7 to the second power. And by repeated divisions by prime numbers into 616,
you'd find 616 is 2 to the third power, 7 to the first, 11 to the first. So, let's express
that. Then, finally by repeated divisions by prime numbers into 182, we get that it's
2 to the first times 7 to the first times 13 to the first. So, here we have the three
denominators in prime factored form. Now, what would you instruct your friend to do
next? List all the different bases as factors to be. So, let's see. I've got a base of 2.
I've got a base of 7, 11 and 13, putting them in sequential order. Then, you would tell
them give to each of those listed bases the highest exponent used. So, on the base 2,
there's a 2 there, a 3 there, a 1 there. So, the the 3 was the highest one used. On 7,
there's a 2, 1, 1, so 2 is the highest. 11 is 1. 13 is 1. So, this is the prime factorization
of my least common multiple and in this case, my least common denominator. A terribly messy
number if we were to multiply it out. As a matter of fact, it's value is 56,056. That's
the smallest number that exists that all three of these would divide into. There are many,
many, in fact an infinite number greater than this, but this, as big as it is, is the smallest.
However, we didn't really need to know that just yet. As it turns out, this factored form
of the lowest common denominator also makes it easier to build fractions. So, let's just
leave it as it is and watch what happens. Let's just observe and count. I want this
to look like this. This one has three 2's. This one has only two. So, I'm going to have
to multiply it by 1 more and I'll list it down underneath here. Okay. This has two 7's,
but so does this. This has one 11. This has none, so I'm going to need one more 11. This
has one 13. This has none, so I'll need a 13. So, in fact, to make this look like this,
I need to multiply it by 2 times 11 times 13 both top and bottom. If I do that, the
top becomes 286. That is, that times that times that times 1, over the lowest common
denominator which is this large creature here. Now, let's do the same thing for this one.
I want this to be like this. Well, let's compare to this. This has three 2's. So does this.
This has two 7's. This has only one, so I'll have to multiply it by one more 7. This has
11. So does that. This has 13. This one doesn't, so I'll have to multiply it by 13 to make
it become exactly that. Then, I have to multiply the top by 7 times 13, so I'd have 7 times
13, whatever that is, times 5 and this fraction would become 455 over that common denominator
which we'll just call LCD. Now, just repeating this procedure, I want this to look like this.
Well, this has three 2's. This only has one, so I need two more 2's. This has two 7's.
This has only one, so I need one more 7. This has 11. This has none, so I need an 11. This
needs 13. This already has it, so this is what I need to multiply bottom and top by
to make the bottom become this lowest common denominator. So, taking that squared times
that times that times the top, I would get 924 over that messy lowest common denominator.
Now, note, I have a common denominator and an actual value. I don't know what it is except
it's this. Actually, we do because we already multiplied it, but the point is I didn't have
to know what it is. So, now our problem reduces to this. Since I have a common denominator,
I need only add the tops and doing that I would get this problem. Here's my lowest common
denominator just not multiplied out yet. Now, I simply add the numerators to get the numerator
of my answer. 1665 over this factored form of my lowest common denominator. But, even
here, I'm not going to multiply this out because then I just have to reduce. Because in prime
factored form, whatever this number is. . . Now, remember in fact, it was this large number.
In factored form it's easier to see if this is reducible because in the first place, the
only prime numbers that will divide into that huge number is 2, 7, 11 and 13. So, even if
we had a calculator, this prime factorization is going to help me. We can see that 2 won't
divide into the top because it's not even, so we'll take the top which is 1665 and just
check to see if it's divisible by this factor, 7. I get a decimal, so it isn't. So, I try
the same top which was 1665 and see if it'll divide by 11. So, divide by 11. I get a decimal,
so it's not divisible by 11. So, I get the same numerator, try to divide it by 13. I
get a decimal so I know that is not reducible by any of these prime factors. A lot of other
things will go into this, notably 5, but it must be a common divisors and these are the
only ones that divide the bottom. So, we know that this fraction . . and now if we want
we can [multiply] by 56,056... is indeed reduced. Now, this might give you a little bit of a
sense of power to be able to do this because very few people can and yet it's not hard.
The reason they can't, they just have never learned how. It's messy. There's a lot of
arithmetic, but it's very routine. That's a good point to end on to show that even the
messy fractions can now be done. So, let's walk through one of everything we've done
in the last few lessons so that in your mind you can tie it all together. First, if you're
adding fractions, you scan the denominators and ask one question. Are they all the same?
If they are, then that becomes the denominators answer. Then, we simply add the tops either
by hand or by calculator depending upon your teacher's objectives in training you. So,
in this case, we'd have 23 plus 17 plus 16 and we have that the sum is 56/42. Then, we
scan it to see if it's reducible. So, I can divide top and bottom by 2. 56. Divide by
2 by hand or calculator. Then, dividing the bottom as well by 2 in this case. Now, we
have a fraction which at first looks like it's reduced, but then note I can divide top
and bottom yet by 7. 7 goes into 28, 4. 7 goes into 21, 3. So, sometimes you think you're
done reducing, but in fact, you omitted something so always be sure it's reduced. Then, if your
fraction is an improper fraction, you ask whoever you're working this for, do they need
the answer in improper form or do they need it in mixed number form? And again, please
note, we're not saying this is right and this is wrong. They are both good. They are both
the same number. It's always a matter of which is preferred for whatever the use that's going
to be made of the results of this problem. But, the point is, the initial point, to add
fractions the denominators must be the same. It stays the same. You simply add the tops,
reduce, then change to a mixed number if that's what's desired. Then, we moved on to the question
of what happens if we're adding fractions whose denominators are not the same. Well,
if they are not the same as such, we do not have permission to add yet. But, then we ask
ourselves, what's the lowest number both 4 and 6 will each divide into? And that was
12 or stated differently, we were finding the least common multiple of the numbers 4
and 6. But, then we recall I can fraction build each of these into fractions of equivalent
value, but which will now have or then have a denominator of 12. Or, in short, we ask
4 goes into 12 how many times? 3. 6 goes into 12 twice. But, to correctly fraction build,
I must multiply top and bottom by the same amount in each case. Therefore, this fraction
becomes 9/12 which is the same thing as 3/4 just expressed in different form. And 1/6,
similarly, becomes 2/12. but, now they have common denominators which we got first, wasn't
it? And that's what gave us permission to add the tops and I have a fraction which is
improper form and it is reduced. Very frequently when you add fractions, it will be in reduced
form, but we must always check that it is. And very, very frequently in simple everyday
fractions, the denominators will be simple enough you can simply look at them and see
what that lowest common denominator is. But, lets remind ourselves why we wanted the least
common multiple. After all, although it is true that 12 is the smallest number that 4
and 6 will go into, they'll also divide into 24. They'll also divide into 36 as a matter
of fact. In fact, there's an infinite number of numbers that each of 4 and 6 will divide
into and we could build each of these to be in this case, 36. 4 goes into 36 9 times and
of course I have to build top and bottom by the same factor. 6 goes into 36 6 times, so
I build top and bottom by 6. So, this gives me 27/36. and this gives me 6/36 so since
they have common denominators, I certainly have permission to add. So 27 plus 6 is 33.
And that is reducible. I can divide top and bottom by 3 and I get the same answer I had
previously. And of course, it must always come out to be the same results because these
are the same numbers as these just written differently and different than what we had
just preceding this. The point we want to make is that any common multiple would work.
But, if you use anything other than the least common multiple, you will always have to reduce.
Whereas if it's the least common multiple, frequently you will not and always the reducing
will be simpler. So, the least common multiple will give you also the least amount of work
to do. Then, we moved onto addition of fractions where the denominators are not only not alike,
but mentally it's not so easy, possibly impossible, to just see the lowest number that each of
these would divide into. So, at this stage we hope that now you see why the method for
finding the least common multiple has been reviewed so frequently in past lessons because
if it has, then at this point you notice that there's going to be a lot of arithmetic, but
it's not at all difficult, just a little bit time consuming. So, going through the actual
work on this with you from beginning to end, we begin dividing by the lowest prime number
we can think of over and over until we repeated by all the prime numbers we can. And that tells me that 48 in this
case is 2 to the fourth times 3 to the first. Going to the next one and doing the same thing.
And sometimes they are so simple we can do them mentally. We can see that 2 goes into
18 9 times and 9 is 3 times 3 or 3 squared. Okay, same thing here perhaps. 42, well 2
goes into there 21. Perhaps we need some paper to help us. And 21 is 3 times 7. So 42 is
2 to the first times 3 to the first times 7 to the first. And of course, we don't have
to write this little 1 up here, but sometimes it helps us to think. Now, to find the least
common multiple is just this simple. List every base. That's 2, 3, 7. The highest power
used on the 2 was the 4. The highest power used on 3, the highest exponent, was 2. And
the highest exponent on 7 was 1. Now, using this as our guide and using this factored
form instead of the original place value name, we can force each one of these to be this
just by a matter of comparison. I want four 2's. I've got it. I want two 3's. I've only
got one, so I need one more. I want one 7. I give it to myself. So, I'm multiplying the
bottom, in effect, by 21. So, therefore, I must multiply the top by 21 which makes it
become 110 over this bottom whatever it is. Now, I want to build this to look like this,
so I need four 2's. I only have one, so I need three more. I want two 3's. I've got
'em. I want one 7, so I give it to myself and 2 to the third is 8 times 7 is 56, so
the top becomes 7 times 56 or 392. See, it takes a bit of arithmetic here, but the principle
is simple enough. You simply multiply. Okay, now, building this to look like this. I need
the four 2's still, so I need three more. I want two 3's. I give myself one more. Want
one 7. I've got it. So, again, 2 cubed is 8 times 3 is 24, so I've multiplied the bottom
by 24. I multiply the top. If I do that, I get 264. Again, the arithmetic can be involved,
but the principle of what I am to do is simple enough. Now that these are all alike, I don't
even know what they are in place value name, but I do know that they are the alike. I can
actually look and count the factors and see that they are all alike. So, now I simply
add 110 to 392 to 264 and I get 766. Again, a bit of messy arithmetic, but the principle
of what I am to do is simple. I am to add the tops. Is it messy? It's messy, but I still
add the tops. Now, to reduce, I can see the only prime numbers that'll divide the bottom
are 2, 3 and 7. I can see that 2 goes into the top, so I'll do that. 766. Divide by 2
is 383, but I have divided out one of these 2's which will leave me now with only three
2's. But, I can see that no more 2's will go in, so I can try to see if 3 goes in. So,
383 divided by 3. I get a decimal, so although 3 will go into the bottom, in fact, twice,
it won't go into the top. So, then I try 7. 283 or 383 divided by 7. I get a decimal,
so 7 won't go in. So, I know that my denominator is 2 cubed times 3 squared times 7 and furthermore,
this will be reduced before I even get it. So, 2 cubed I know is 8. 3 squared is 9. 8
and 9 is 72 and 72 times 7. . . Well, even without a calculator, 7 times 2 is 14. Carry
the 1. 49, 50. So, here I have my answer and it is reduced. See, how very, very important
and useful that process of prime factoring and that process of finding least common multiple
really is. Without that procedure, this probably would remain an impossible problem for you
as it is for most people. But, now you've moved into the realm of being able to work
a problem that very few people can work. Again, not because it's difficult, but because they
simply haven't been taught and now you have. Become quite comfortable and good at this
addition because you'll find that when we move on to subtraction as we will very shortly,
the principle will remain exactly the same. So, that should remain an easy lesson if this
has become an easy lesson. This is your math host, Bob Finnell. We'll see you at the next
lesson. Good luck.