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A Portland Community College mathematics telecourse.
A Course in Arithmetic Review.
Produced at Portland Community College.
This very short lesson simply wishes
to reverse the process we had on the last tape.
We have a mixed number, let's go back to its improper form.
In this case, let's simply jump
right on top of the technique
without worrying about the logic behind it.
It is simply this.
Let's multiply the denominator [ 7 ] times the whole number [ 5 ]
which is 35. [ 7 · 5 = 35 ]
Then let's add that [ 35 ] to the top [ 2 ]
and 35 plus 2 is 37. [ 35 + 2 = 37 ]
And write that [ 37 ] over that denominator, which is 7.
And we claim that this [ 37/7 ] is the improper fraction form
to this mixed number. [ 5 2/7 = 37/7 ]
Now, we can prove it
even though we didn't go through the logic
of getting from here to here, it's fairly simple.
But let's work backwards.
From the last lesson, this reads 37 divided by 7.
37 divided by 7. [ 37 ÷ 7 ]
And indeed 7 goes into 37, 5 times. [ 37 ÷ 7 ]
Five sevens is 35. [ 7 · 5 = 35 ]
Subtract. [ 37 - 35 = 2 ]
That leaves a remainder of 2, which we write over the divisor,
and see, it is exactly where we started.
So once again, the denominator [ 7 ] times the whole number [ 5 ]
That product added to the top [ 35 + 2 = 37 ]
gives me my numerator [ 37 ] of my improper fraction.
And the denominator [ 7 ] stayed the same. [ 37/7 ]
Let's restate that formally for our notes.
And this is a case where in your note
a diagram probably can speak much better
than a paragraph of words.
Starting here [ 8 ]
multiply those two [ 8 · 3 ] my first step, that's 24.
Adding that product [ 24 ] to this [ 5 ] [ 24 + 5 ]
gives me 29 [ 24 + 5 = 29 ]
over the denominator [ 8 ] which stays the same [ 29/8 ]
Okay. Denominator times whole number,
that product added to the numerator,
that becomes the numerator of my improper form.
Denominator stays the same.
I repeat the technical verbiage primarily
so that you can get more and more used to it as we go along.
A verbal example:
Joseph has 2 3/10 dollars.
How many dimes is that?
Now this says he has 2 whole dollars
and another dollar
which has been divided into 10 equal parts
which we call dimes, he has 3 of those.
So now using our procedures
for changing a mixed number to an improper fraction,
we multiply the denominator [ 10 · 2 ] times the whole number: 20
add the sum [ 20 + 3 ] to the top: 23 and put it over the bottom.
So he has 23 of those parts
where the whole has been divided into 10 pieces called dimes.
Or in short, he has 23 dimes.
Another problem.
If 1/4 inch represent 5 feet on a scale drawing,
how many feet do 2 3/4 inch represent?
So each one of these 1/4 pieces of an inch represents 5 feet.
So our question is:
How many of these 1/4 inch pieces does he have?
Then I multiply it by 5 to find out how many feet that is.
So 2 3/4 changed to a an improper fraction.
Multiply these two is 8, [ 4 · 2 = 8 ]
add to that [ 8 + 3 ] is 11 over 4. [ 11/4 ]
So I have 11 1/4 inch pieces.
But each one of those is 5 feet.
So 5 times those 11/4 inches [ 5 · 11/4 in ]
gives me 55 feet. [ 5 · 11 fourth inches = 55 ft ]
So we can state here that 2 3/4 inch
represents the same distance as 55 feet,
apparently, on a map scale.
There it is. Our task is just that simple.
To change a mixed number [ 6 5/7 ] to improper fraction,
we simply multiply these two, [ 7 · 6 ]
the denominator [ 7 ] times the whole number [ 6 ]
that's 42, [ 7 · 6 = 42 ]
add to the numerator [ 5 ], 47, [ 42 + 5 = 47 ]
put over the original denominator [ 7 ] and you're done [ 47/7 ]
Let's make this a short lesson and leave it right there,
easy as it is.
So let's jump from that very simple lesson
to our continuing reviewing
so that towards the end of this chapter,
we will be able to use this prime factorization task
and other reviews to be able to do a problem like this
very, very easily.
So to prime factor we need to first have in memory or before us
a list of prime numbers.
Here we have such a list.
Then we begin a long, it appears in this case,
a long series of divisions by prime numbers.
Now, we can cleanly see that 2 will go into this, [ 12,152 ]
so dividing it by 2, we get 6076. [ 12,152 ÷ 2 = 6076 ]
And this [ 6076 ] being even tells me
that 2 will still go in at least once more.
So dividing this [ 6076 ] by 2, we get 3038, [ 6076 ÷ 2 = 3038 ]
which is still divisible by yet another 2.
So dividing a third time, [ 3038 ÷ 2 = 1519 ] we get 1519,
and we have carried the 2s as far as we can.
So now we go to the next prime number [ 3 ]
and we can see if 3 will divide into it
by actually trying to divide by longhand or by calculator
or by using the divisibility test.
So 1 and 5 is 6 and 1 is 7 and 9 is 16. [ 1 + 5 + 1 + 9 = 16 ]
3 will not go into 16 [ 16 ÷ 3 ] so will not go into 1519.
So 3 is not a prime divisor.
We can see by inspection alone,
because this [ 9 ] is not 0 or 5, that 5 won't.
7, we'll just try.
So if you try to divide 7 into this number, [ 1519 ÷ 7 ]
you find that it does go 217 times. [ 1519 ÷ 7 = 217 ]
Now to see if 7 will go in again, we have to simply try it again.
And if we do, it goes in 31 times. [ 217 ÷ 7 = 31 ]
But now 31 becomes a prime number, and that's when I quit.
So, the prime factorization of 12,152
is one,
two,
three 2s
being multiplied together, called 2 cubed, [ 2³ ]
times 7 times 7, which is 7 squared, [ 2³ · 7² ]
times 31 to the first if you wish. [ 2³ · 7² · 31¹ ]
So here is the prime factorization of this. [ 2³ · 7² · 31¹ ]
Again, review this
until you're smooth and very, very comfortable.
So that you hardly ever make a mistake
or even think seriously about what you're doing.
And always keep at the back of your mind
the reason we're doing it
is to be able to perform a problem like this
within another half dozen lessons or so.
So let's just while we're here concentrate on this denominator
and write it in prime factored form.
We've done 32 so many times
that perhaps some of you have it memorized already.
2 goes into 32, 16 times. [ 32 ÷ 2 = 16 ]
2 goes into that 8. [ 16 ÷ 2 = 8 ]
2 goes into that 4. [ 8 ÷ 2 = 4 ]
2 goes into that 2. [ 4 ÷ 2 = 2 ]
This [ 2 ] is prime.
So 2 times 2 times 2 times 2 times 2, or 2 to the 5th, [ 2⁵ ]
is simply another way of writing 32. [ 2⁵ = 32 ]
So we have prime factored this denominator to 2 to the 5th.
Now, let's extend our review
of the last chapter a bit further.
Do you recall what is meant by the phrase
'the least common multiple' of say 3, 4, and 6?
Recall what the phrase was asking us for?
Recall that the word 'multiple' or a multiple is a number
which a given number will divide into evenly.
Example:
We can say that 15 is a multiple of 3
because 3 divides evenly into 15.
But 3 has lots of multiples.
15 is just one of them.
And so is 18 a multiple of 3 because 3 divides evenly into 18.
Or 48 because 3 divides evenly into 48.
So this list of multiples of 3,
that is, numbers that 3 will divide into, goes on forever.
And of course you can see that 3 is a multiple of itself
because 3 certainly goes evenly into 3 once.
But say a number like 28 is not a multiple of 3
because 3 will not divide into 28 evenly.
are you beginning to feel comfortable with this word 'multiple?'
So going back to our question of just a moment ago,
multiples of 3 are all the numbers 3 will go into evenly.
And multiples of 4 are all the numbers
that 4 will divide into evenly.
And multiples of 6 are all the numbers that 6 will go into evenly
Let's look at a list of all the multiples of 3, 4, and 6
each if that's possible.
You see, to list the multiples of 3, 3 is the first one,
it's really just a matter if you will of counting by 3s.
3,
6,
3 more is 9,
3 more is 12,
3 more is 15.
And so on, and so on, and so on.
And of course we eventually realize,
hey, this list goes on forever.
So there is an infinite number of multiples of 3.
Another way we could have looked at this
rather than saying start with the multiple itself, 3,
and then add 3, add 3 more, add 3 more, add 3 more, add 3 more,
we could have begun with our set of natural numbers.
There, up here on the bright white spots.
We could have said to find the multiples of 3,
simply repeatedly take 3 times each of the natural numbers.
And so on.
3 times 14 is 5 [3 · 14 = 5 ] 42,
3 times 15 is 45, [3 · 15 = 45 ]
3 times 16 is 48, [3 · 16 = 48 ]
and of course this process goes on forever.
so if now I wish to find the multiples of 4,
I would simply take 4 times 1, [ 4 · 1 ]
4 times 2, [ 4 · 2 ]
4 times 3 is 12, [ 4 · 3 = 12 ]
4 times 4 is 16, [ 4 · 4 = 16 ]
and so on forever.
So now I have a list of the multiples of 4.
So you see, multiples is a list, and the list goes on... forever.
So now if we also want the multiples of 6,
6 times 1, [ 6 · 1 ] 6 times 2, [ 6 · 2 ] etc.
Is all of this coming back from the last chapter?
It's very, very important.
So we do have lists of the multiples of 3, 4, and 6.
But our original question was not
what are the multiples of those 3 numbers, we have that,
that's 3 lists, but What is the least common multiple?
Well, before we answer that, let's back up once more
and ask: What do we mean by common multiples?
Well, if we begin to look at all three of these lists
and look for common numbers notice this:
12 is a multiple of 3 because 3 goes into it. [ 12 ÷ 3 ]
12 is a multiple of 4 because 4 goes into it. [ 12 ÷ 4 ]
And 12 is a multiple of 6
because 6 goes evenly into it. [ 12 ÷ 6 ]
Now let's see if there are any more common multiples.
See, each of these is a multiple of those three numbers
so it's a common multiple.
Whereas 9 is a multiple of that [ 3 ]
but not of this [ 4 ] or this [ 6 ] so 9 is not a common multiple
So if we search a bit,
we will find that 24 is common to all
each [ 3, 4, and 6 ] of my list.
If we search again,
we would find that 36 is common to each of the lists.
If we search some more,
we find that 48 is also common to each list.
So so far we have one, [ 48 ]
two, [ 36 ]
three, [ 24 ]
four [ 12 ] common multiples of the numbers 3, 4, and 6.
And the list of common multiples
turns out to be a list that is also infinite, it goes on forever.
There is no end of numbers, whole numbers,
that 3, 4, and 6 will divide into.
So now we can can move up one word
and cleanly see what we mean by least common multiple.
Of that infinite list of common multiples, 12 is the least.
So 12 is the least common multiple of these three numbers.
Another way of stating that would be:
What's the one number, the smallest number, that exists
that each of these will divide into evenly?
And of course in this case 12 is it.
3 goes into 12 four times, [ 12 ÷ 3 = 4 ]
4 goes into 12 three times, [ 12 ÷ 4 = 3 ]
6 goes into 12 twice. [ 12 ÷ 6 = 2 ]
So now do you have perhaps a clearer and a more comfortable feel
for what we mean by the phrase 'least common multiple,'
which we will abbreviate as LCM for future lessons?
Therefore, after this brief review,
you should at least be very comfortable
with what is meant by a statement like this:
What is the least common multiple
of any two numbers or three or four?
What are we really asking for?
Well, we're asking for the smallest number in existence
that 6 and 10 will each divide into.
Now, please note that 6 and 10 will both divide into 60,
will they not?
6 goes into here 10 times, [ 60 ÷ 6 = 10 ]
10 goes into here 6 times. [ 60 ÷ 10 = 6 ]
They'll both go into 90.
6 goes into 90, 15 times. [ 90 ÷ 6 = 15 ]
10 goes into 90, nine times. [ 90 ÷ 10 = 9 ]
They'll both go into 120.
There's an infinite number of numbers
that these two will go into evenly.
But we're asking: What's the least of them?
And without too much thinking with two simple numbers like this
and without any work, you'll probably realize it's 30.
30 is the smallest number that each of these will divide into.
Most of the numbers you've worked with in the past
are such that you can simply inspect the numbers
and simply know the least common multiple without doing anything.
Just look, know, write it down, or use it.
However, here the answer to the question is not so obvious.
Not at all obvious, is it?
What is the lowest number that each of these will divide into?
But at least we can understand what the question is asking.
Find a number that both of these will divide into somehow.
And it's not at all obvious, is it,
that this [ 38,808 ] is going to be that number?
So within the next one or two lessons following this,
as part of our extended review,
assuming then that you at least know what we mean
by this phrase "least common multiple,"
and that you also know how to prime factor any number
including rather large ones,
we will show you or remind you
that finding the least common multiple
of these two or any number of numbers, no matter how large,
is really quite routine.
So while you're doing the next few lessons,
you might be reviewing Chapter 2 and asking yourself:
How did we do this?
Now, can you nail it down
so that you're very comfortable in that process?
You must be within three or four more lessons.
Now, as important as these reviews will prove to be,
let's not in each lesson forget the main topic of that lesson.
Therefore, in this lesson our principal considerations
is with problems like this, [ 3 5/7 ]
to change from a mixed number form to improper form.
Mixed number means I have a whole number with a fraction.
See, we say it in mixed number manner,
"Three 'and' five sevenths" [ 3 5/7 ]
And we wish to change this
to what we've unfortunately learned to call improper fractions.
What we really mean is a single, pure fraction.
Which simply means I want to know what the new top will be,
what the new bottom will be, two questions.
Therefore, from a few moments ago,
can you see the single point being made in this lesson?
It was simple enough.
Simply this.
To form our new top or numerator,
we first multiplied these two, [ 7 · 3 ]
the denominator times the whole number.
That's 21. [ 7 · 3 = 21 ]
Then added that [ 21 ] to the numerator [ 5 ]
So that gives me 26, [ 21 + 5 = 26 ]
which becomes the numerator of my improper fraction.
And then the denominator of my so-called improper fraction
simply stays the same as the original fraction. [ 26/7 ]
And we remind ourselves
that 26/7 is just another way of saying the number 3 5/7.
In short these are equivalent, [ 26/7 = 3 5/7 ]
which means they look different but they have the same value.
So once again, bottom times whole number
added to top becomes new top.
And bottom stays the same.
Just that simple.
And to encourage you to begin to think more and more verbally,
each lesson always gives you some verbal or story problems
to try to encourage you to translate from these words
to an arithmetic statement.
So let's see how we would handle this one.
Pies at a corner coffee shop
are cut into 6 pieces for each serving.
See, we're talking about fractions now.
We're dividing a whole, pie in this case, into 6 pieces each pie.
How many orders can be filled if they have 12 5/6 pies?
Well, this [ 6 ] tells me how many pieces are in each pie.
This tells me one pie has only 5 of those 6, original 6, left.
Then in addition to that, 'and,' I have 12 full pies.
Well you can see that if I convert this to a mixed number
leaving the denominator the same,
then bottom [ 6 ] times whole number [ 12 ] [ 6 · 12 ]
is 72. [ 6 · 12 = 72 ]
Plus 5 is 77. [ 72 + 5 = 77 ]
So we have 77 of those 1/6 sized piece of pies.
So we have 77 servings of those 1/6 of the pie left.
Later you'll find that it helps to think of it in this way:
I have 77 of, or times, 1/6 of a pie available.
Is this beginning to make sense?
I trust so.
And frequently towards the end of the lesson,
you'll see what appears to be the same problem
but the parts, the numeral parts,
are becoming messier or more involved.
That's to cause us to realize
that just because the numbers get larger,
the math procedure does not change,
it simply becomes a little bit messier.
To change this to improper fraction,
the bottom still remains the same, that's simple enough.
And to find the new top,
we take the bottom [ 47 ] times the whole number [ 208 ]
and then that, added to this top [ 13 ] to get to the new top.
So see, you can know what to do,
in fact you must know what to do before you even do it.
So this simply says the arithmetic is going to be somewhat messy.
I'm going to have to multiply to
and then add that product to that.
Well, we simply will multiply by longhand
if that's the way we must do it
and be very, very careful and to double-check our work,
or use a calculator if it's permitted and you have one.
So all the calculator does
is help us to do whole number arithmetic fast.
See, it's not really doing the fraction for us, is it?
So we take the bottom [ 47 ] times the whole number [ 208 ]
which gives me, whoops.
The bottom [ 47 ] times the top number [ 208 ]
which gives us a very large number.
[ 9776 ] Well, so be it.
Then we must add that product [ 9776 + 13 ] to the top.
So plus 13 equals, [+][1][3][=]
and we have the top [ 9789 ] of our improper fraction.
The important point we're trying to make
is because the numbers get messy, the math doesn't change.
Our arithmetic simply becomes somewhat involved or messy,
and we simply have to do the same thing
more slowly, more carefully,
or if you're in the circumstances where it's permitted,
use a calculator to do the arithmetic itself of whole numbers.
But the fractions, you have to have that procedure in your mind.
Most of the problems you'll encounter, however,
are so simple you can do the work mentally.
Again, when you're changing a mixed number
to pure improper fraction if you will,
the denominator [ 7 ] stays the same.
Then denominator [ 7 ] times the whole number [ 7 · 10 ] is 70,
add it to the top [ 70 + 5 ] is 75, and you're done.
It's just that simple. [ 75/7 ]
So, until your next lesson, this is your host, Bob Finnell.