Math 20 - Lesson 5

A Portland Community College mathematics telecourse.
A Course in Arithmetic Review.
Produced at Portland Community College.
Again, this lesson will presume
that you have your multiplication tables memorized.
If you don't, then seek a specialist at your school
for special help.
It's not a difficult task, but all of our following work
will assume that you can recall these individual facts
very, very rapidly and accurately.
First, and as usual, let's build our vocabulary.
When two or more whole numbers are multiplied,
each of your multipliers is called a 'factor.'
The answer is called a 'product.'
So in this double multiplication problem we have down here
there are three numbers involved in the multiplication,
so each of them we'll call 'factors.'
And the result in this case [ 6 x 8 x 2 = ]
of 6 times 8 is 48 [ 6 x 8 = 48 ]
times 2 is 96. [ 48 x 2 = 96 ]
The answer [96] we call a 'product.'
So 'product' is the results of having multiplied.
We cannot possibly underestimate
the importance of knowing these vocabularies we are building.
All future explanations in all future math courses
will be done with this terminology.
And along with a new vocabulary is a new symbology.
The use of this small 'x' is a usual way
most people are familiar with an indicated product.
Other ways that you will encounter in the future
is if we put the numerals in parentheses [ (3)(8) ]
and shove the parentheses against each other [ (3)(8) ]
with nothing in between.
That will mean multiplication,
so always between here is an implied multiplication,
or we might just put one of the numbers [ 3(8) ] in parentheses.
It might be sometimes the first [ (3)8 ] and not the second.
Usually, this is not used so much, but it legally is possible.
And you probably are familiar with
using a dot raised to the middle [·] to indicate multiplication.
You can see this might be confusing
with a decimal point in the future,
so we'll try to avoid this [ 3 · 8 ]
except where the meaning is very obvious.
Where we're using a letter [m] to mean 'some number,'
and we shove another letter or a constant against it,
[ 3m ] it means to multiply.
So this [ 3m ] will mean 3 times some number.
If I had something like this, [ a · b ]
that will mean 'some number' times 'some other number,'
but usually we will not write that dot.
We'll simply show it that way. [ ab ]
So as you can see, there are many ways of showing multiplications
As you move on, these two right here
or three, really, will be the most common one.
So do become very familiar with those.
As you can see,
the use of the parentheses shoved against each other
is much neater than this approach right here,
so more and more we're going to replace this: [ 298 x 476 x 529 ]
by this, in our notation. [ (398)(476)(529) ]
Do you recall the standard setup
learned by most of us for multiplication?
Without too much ado,
let me simply work one with you watching over my shoulder.
First, use guidelines if it will help you.
With practice, of course, you won't bother with that.
Then we started with the end digit of our bottom factor, 4,
and we thought 4 times 7, from memory, is 28. [ 4 x 7 = 28 ]
We could have written 28 here.
20|8
But traditionally we write the units digit here [ |8 ]
And 20 goes above the next column as an addition carry.
Then we think 4 times 2 is 8, [ 4 x 2 = 8 ]
plus 2 is 10. [ 8 + 2 = 10 ]
No ones and the 10 goes up here to the next column as a carry.
4 times 5 is 20, [ 4 x 5 = 20 ]
plus 1 is 21. [ 20 + 1 = 21 ]
Now we're through with these carries
and we're through with this ones digit [4]
so we go next to the tens digit [6]
6 times 7 is 42. [ 6 x 7 = 42 ]
We always start under our bottom multiplier, so 42
we'll put 2 here
and the 4 goes to the next column as a carry.
Now we think 6 times the next one, 2, is 12.
Plus that 4 as a carry is 16.
So 6 and the 1 to the next column as a carry.
Now we go to the next 6 times 5 is 30. [ 6 x 5 = 30 ]
Plus the 1 is 31. [ 30 + 1 = 31 ]
Now we're done.
For each digit in our bottom multiplier
we have a line of digits here.
If I have 3 digits here, I would have 3 lines and so forth.
Now we add these digits. We get 8,
2,
7,
3,
3.
And this, our answer, [ 33,728 ]
which we now call the 'product'
of these two factors. [ 527 x 64 ]
Quit an involved process. Isn't it?
But much, much simpler than was used in the world
say 600, 800 years ago.
There are many other setups that can be used to multiply.
Some of them are rather clever, and cute
perhaps your teacher will show you one or two of the others,
and in our extended lesson, I'll show you yet another one.
But let's stay with this one,
which is the basic one with which most people are familiar.
You should practice quite a few of these
until this becomes a very automatic procedure.
So it recalls multiplying
which means you need to re-memorize your table.
Additions, multiply, additions and additions,
so there's a lot involved here.
For the sake of future discussions,
it might be well here to remember
that multiplying whole numbers
is actually a shortcut for repeated additions.
Originally 3 x 5 was a symbol [x] invented
to mean take three of those and add them together [ 5 + 5 + 5 ]
And you can see by that definition,
5 times 3 means something different from that.
It means take five 3s [ 3 + 3 + 3 + 3 + 3 ] and add them together
And in both cases, we get 15. [ 3 + 3 + 3 + 3 + 3 = 15 ]
An interesting fact that surprises as children,
but in time we learn to live with it,
that 3 times 5 and 5 times 3 [ 3 x 5 = 5 x 3 ]
are exactly the same process. [ 5 + 5 + 5 = 3 + 3 + 3 + 3 + 3 ]
Not the same process, exactly the same value.
In fact, they are different processes. Aren't they?
This fact that I can multiply
this way [ 3 x 5 ] or this way [ 5 x 3 ]
and get the same results later in algebra
you refer to as the "commutative law of multiplication."
But can you see by this repeated addition law
a problem like this: [ (495)(236) = ]
by addition, gets completely out of hand.
Can you even imagine having to take 495 of those [ 236 ]
and adding them together? [ 236 + 236 + 236...]
That would take forever.
So as complicated as our multiplication setup was,
it's far simpler than having to go back to basics.
Implied in this definition is another word
that we'd best pause and consider more formally.
The term 'whole number'
simply refers to the numbers 0, 1, 2, 3, 4
and so on and so on and so on forever. [ 0, 1, 2, 3, 4...]
What we're basically ruling out here
are fractions, square roots, pi, decimals, etc.
So those [ 0, 1, 2, 3, 4...] are my 'whole numbers.'
No matter how we look at it,
multiplication is an involved process,
and none of us get overly thrilled of that.
So wherever possible, we prefer doing it with a calculator.
Do learn to do it by hand though because interesting enough,
that cuts down on the errors on calculators.
You wouldn't think so, but that is fact.
So on a calculator, each of these [x] becomes a button.
And of course, after punching each of those buttons,
we want to know what the product is equal to.
And so the equal [=] also becomes a button,
so on our calculator we simply punch those as buttons.
[4][7][x][8] [2][x][3][0][=]
And our answer, a very involved one,
1, 1,
5,
6.
115,620
Now always remember we might have made a mistake,
so always clear, and redo it, in reverse order if you wish.
[3][0][x][8][2][x][4][7][=] 115,620.
So apparently we're correct.
And, of course, most of our problems,
in that mythical 'real world,'
comes to us in verbal forms.
A carpenter ordered 17 roof truss assemblies at $56 each.
Can you see where this, that original definition is?
17 roof trusses at $56 each.
That would be 56, plus 56, plus 56, [ 56 + 56 + 56...] 17 times.
Rather than doing that addition, we do a multiplication instead.
Now a good practice at this point throughout your lesson
is to do the problem first by hand
and then check it with a calculator rather than the answer book.
And if they don't agree, then use the answer page
to check which one of those two is correct,
hand arithmetic or calculator.
So let me go through this very quickly.
7 sixes is 42. [ 7 x 6 = 42 ]
Carry the 4.
7 fives is 35, [ 7 x 5 = 35 ]
and 4 is 39. [ 35 + 4 = 39 ]
I'm done with that.
1 times 6 is 6, [ 1 x 6 = 6 ]
1 times 5 is 5. [ 1 x 5 = 5 ]
Now adding 2,
15,
9.
And I'm done. [ 56 x 17 = 952 ]
Now, checking it with a calculator.
And we got the same number. [ 952 ]
So do work by hand, always by hand.
Then check with a calculator.
It would be nice in this course
if you became good at both processes
rather than using a calculator to avoid this.
That will definitely hurt you in the long run.
Avoid that.
Let's look at another method of long multiplication.
Just for fun, really.
We had just finished doing this particular problem
by the traditional approach.
And sometimes
when we work a problem by the same method every time,
we pretty soon get the feeling
that that's the only way it can be done.
Let's demonstrate that we could have done this multiplication
by an entirely different setup.
First, let's draw a rectangular box,
a column for each digit in one of our multipliers [ 527 ]
[ 527 ] so we have a column for the 5,
[ 527 ] the 2,
[ 527 ] and the 7.
Then one row for each digit in our second multiplier [ 64 ]
so we have a row [ 64 ] for the 6,
and a row [ 64 ] for the 4,
and going from the top, down.
You see how we did that?
Now, in each of these boxes let's divide it by a diagonal.
Each small box within the larger one
can be located or denoted by two numbers.
You could say this is the 2, 6 box.
Whereas this box is the 2, 4 box.
This box, that would be the 5, 4 box.
This box 7, 6 box.
So once you have set up such a box,
let's take the name for each box.
This is the 7, 6 box.
Multiply those two mentally.
7 times 6 is 42. [ 7 x 6 = 42 ]
Put the tens digit in the upper portion,
and the units digit in the lower portion.
Now in the same manner, fill in every box.
The interesting thing about this approach is
it doesn't make any difference what order you work.
Whereas here, order was all important.
So this box is the 5, 6 box.
5 times 6 is 30. [ 5 x 6 = 30 ]
This box is a 2, 6 box.
2 times 6 is 12. [ 2 x 6 = 12 ]
This box is the 5, 4 box.
Five 4s is 20. [ 5 x 4 = 20 ]
Here 2 times 4 is 8, 0 here if you wish. [ 2 x 4 = 8 ]
This box, 7 times 4 is 28. [ 7 x 4 = 28 ]
And here you don't have to worry about carries or adding on.
It's just a flat recall of your multiplication tables.
And if you had forgotten a multiplication tables,
let's say you had forgotten that 7 times 6 is 42,
you can stop, look it up, and write it down
until you do learn it.
Once you have filled in this box,
these outside numbers, the multipliers, are no longer needed.
Now notice we have diagonals [/] going through this box.
Let's call those diagonals 'number paths.'
So for our next step, let's simply ask ourselves
to add all the numbers along each number path.
So adding those gives me 8, [ 8 ]
which I'll write down below it.
Adding the next number paths we have
2 and 2 and 8 is 12. [ 2 + 2 + 8 = 12 ]
And if we have any carries, take it up to the next number path.
So going to the next number path
we have 5 and 2 is 7. [ 5 + 2 = 7 ]
Then up to the next number path.
1 and 2 is 3, [ 1 + 2 = 3 ]
and then the next number path, 3.
Now let's just read what we get here backwards, so to speak.
3, 3, 7, 2, 8.
33,728.
That was the answer to our original problem.
Isn't this neat?
Now I show this to you
not to necessarily show you a better method
but to get across the point that there are other methods.
So if you're in a classroom or talking to a math tutor,
be open for yet other ways
of approaching a problem in mathematics.
There are always many, many ways of approach.
So let's do this once again
to see if it really firms up in our mind.
It's a very simple technique really.
Let's multiply this problem.
Now by that the way, you might show this to a friend.
People seem to rather like this.
First you want a box.
Then you want as many columns
as you have digits in one of your multipliers.
Then you want as many rows, in this case three of them,
as you have digits in your second multiplier.
Then you draw a diagonal through each one of these sub boxes.
Then we recall that each sub box,
let's say this one right here is named by two digits.
This is the 8, 3 box
and 8 times 3 is 24. [ 8 x 3 = 24 ]
So now simply recalling our multiplication tables,
we fill in every box.
This box is a 5, 3 box.
That's 15. [ 5 x 3 = 15 ]
In any order you want it.
If you wanted to jump down to this box,
4 times 7 is 28. [ 4 x 7 = 28 ]
And in that manner fill in all of these sub boxes
as we've done here, and once the box is filled,
the original multipliers, you no longer need.
Then you simply add all the digits in each number path,
and if you get a carry from each any one number path,
you carry it over to the next number path.
So adding down here it's 5,
here is 9. [ 0 + 3 + 6 = 0 ]
Here is [ 5 + 2 + 2 + 5 + 8 ]
7, [ 5 + 2 = 7 ]
9, [ 7 + 2 = 9 ]
14, [ 9 + 5 = 14 ]
22. [ 14 + 8 = 22 ]
I had to carry a 2 [ (2) + 1 + 4 + 3 + 8 ]
so it's 3, [ (2) + 1 = 3 ]
7, [ 3 + 4 = 7 ]
10, [ 7 + 3 = 10 ]
18. [ 10 + 8 = 18 ]
Carry one more. [ (1) + 2 + 2 + 1 ]
3, and 3 is 6.
No carry.
And 1.
And our claim is that this 1, 6, 8, 2, 9, 5 is our answer.
Let's check that [ 168,295 ] on a calculator.
So 485 times 347 equals 168295, [4][8][5][x][3][4][7][=] 168295
168,295
this particular method is a handy one
if you are trying to do long multiplication
and you don't yet have your multiplication tables memorized.
This will allow you to do the long multiplication
while in the process of memorizing your table.
If you need to fill in the table
and you forget one of your multiplication facts,
let's say you had forgotten what 8 times 4 was,
then simply look it up on your multiplication tables.
8 times 4 is 32, [ 8 x 4 = 32 ]
and simply say 8 times 4 is 32
and write it over two or three times
saying it out loud each time you write it in
and then get on with the business of finishing the problem.
It's a nice method,
but there are several other methods equally as good as this,
and you might ask your instructor if she or he knows of any.
It's rather fun sometimes to take a fresh approach on a problem
that might have been bothering you somewhat.
While reviewing our multiplication facts,
let's simultaneously get ready for algebra.
Now let's see what we mean by this phrase:
"let's get ready for algebra."
Here's a simple algebraic sentence: [ m/3 = 6 ]
The important thing about this at first
is not being able to work it, whatever 'work' means,
but, like any sentence, to be able to read it.
This is a math sentence. [ m/3 = 6 ]
Remember, a variable, in this case 'm' simply means 'some number'
This little dash here [/] means 'divided by.'
3 is to be 6.
So reading this, this says
'Some number,' when divided by 3, is 6. [ m/3 = 6 ]
And solving the equation means
what number could have replaced this variable and made it true?
And of course our number facts simply tells us it's 18.
We know that without doing any work.
We just know from memory
that 18 divided by 3 is 6. [ 18/3 = 6 ]
However, this one [m/45 = 56 ] is not so obvious. Is it?
What number divided by 45 gives me 56? [m/45 = 56 ]
So quite frequently a text book will start with easy problem
where the answer's obvious right from the start,
and unfortunately, sometimes that can mislead you
because the process is not to know that 18 is the answer,
but what can we do to get rid of this [ /3 ]
not just the three, but the division [ /3 ] as well.
So the sentence simply says that m is something.
So the fact we want to bear in mind from this lesson is this.
That this mark [/] is a division sign, some number divided by 3.
So we ask ourselves what process undoes dividing?
So we note first that a multiplication always undoes division.
So in this case if I were to multiply this by 3,
3 times undoes division by 3. [ 3 x m/3 = m ]
It's very important to realize that our focus point
is on this, that this [x] undoes this [/]
and the three is almost incidental.
But then also recall one of the fundamental laws of algebra,
in algebra, one can multiply both sides of the equation,
and those sides are called 'members' of the equation in algebra
by the same amount.
So in this equation, [ m/3 = 6 ]
this is one side [ m/3 ] one 'member' of the equation
And this is the other 'member.' [ 6 ]
So if I multiplied this member by three [ 3 x (m/3) = ]
to undo that division by three, [ 3 x (m/3) = ]
then I must also multiply
the other side by three. [ = (6)(3) ]
So this [ 3 x ] undoes this [ /3 ]
leaving me with my symbol for 'some number' [ m = ]
And the other side is simply 18. [ (6)(3) = 18 ]
So this says, "Look, your number called 'm' is 18." [ m = 18 ]
So now this problem, which is not quite as obvious,
is simply a matter of asking how do we get rid of this
or more specifically, how do we undo this division?
So you undo division by multiplication,
and since I'm undoing division by 45, I will multiply by 45,
but then our multiplication law of algebra says
that's okay providing I multiply the other side by 45 as well.
So this [ 45 x ] undoes this [ /45 ]
And my math sentence now reads:
"My number 'm' is 56 times 45." [ m = 56 x 45 ]
And if we do that arithmetic, this [ 56 x 45 ]
would give us 2,520. [ m = 2,520 ]
So by undoing this [ 45 x m/45 = ]
Then multiplying the other side by the same amount [ = 56 x 45 ]
this new sentence actually tells us [ 45 x m/45 = 56 x 45 ]
how to find the value of that unknown number called 'm.'
So in algebra, if we see a simple equation [ s/148 = 75 ]
where the number that's operating on the variable, the unknown,
is dividing into it, [ s/148 ] we simply think
to undo dividing I multiply,
in this case by 148, [ 148 x s/148 ]
then algebraically I recall
that I have to multiply
the other side by the same amount. [ (75)(148) ]
Then this [ 45 x ] undoes this [ /45 ]
and this says my s is 75 times 148. [ s = (75)(148) ]
So in fact my algebra sentence
is now telling me what I must do to find the value of my variable
Isn't that simple?
And if you wish we even now have
another way to go about our multiplication.
And we have an answer
by this new box method of multiplication of 11,100.
So however you do it, use this lesson as an opportunity
to once again become strong and comfortable in multiplication.
This is your host, Bob Finnell.