Math 20 - Lesson 66

Uploaded by PCCvideos on 16.09.2009

A Portland Community College mathematics telecourse.
A Course in Arithmetic Review.
Produced at Portland Community College.
On this lesson let's just jump right in,
because the product of signed numbers
is easier than addition or subtraction.
Okay, the rules are quite simple.
The product of two like signed numbers is always positive.
Positive times positive is positive, [ (+) x (+) = (+) ]
negative times negative is also positive. [ (-) x (-) = (+) ]
The "why" of this we will have to leave for an algebra course,
but later in this lesson
we will try to make this seem a little bit reasonable.
But actually the reason is a logical one,
which is the scope of algebra.
For the time being, the product of like signs is always positive.
Simply remember that and things go smoothly.
So when you're multiplying two negatives,
simply remember that a negative times negative is positive.
Negative times negative is positive.
And again, two negatives being multiplied is positive.
Then at this stage, all you need to do is
to multiply their absolute values,
which is nothing more than an arithmetic review.
So, 1.5 times .4 is .6 [ (1.5)(.4) = .6 ]
And on the next one, once we dispose of the sign,
simply a matter of 3/4 times 5/6, [ (-3/4)(-5/6) ]
which after, using the cancellation law,
gives me 5/8 [ 3/4 · 5/6 = 5/8 ]
And the last one, after deciding
that the product will be positive [ (-2)(-1 1/3) = + ]
we multiply their absolute values [ |-2|·|-1 1/3| = + ]
and get 8/3, [ |-2|·|-1 1/3| = +8/3 ]
or if we wish, 2 2/3. [ 8/3 = 2 2/3 ]
But the algebraic part of the signed numbers is simple.
A negative times negative is positive.
The old-fashioned arithmetic
is actually the harder part to the problem.
Now what if the two signed numbers are of different signs?
Our rule there is that
the product of two unlike signed numbers is always negative.
So a negative 3 times a positive 4, [ (-3)(4) ]
unlike signs, product-wise is always negative.
The product of their absolute values.
On the other hand,
if I have positive 3 times negative 4 [ (3)(-4) ]
as long as the signs are different, [ (3)(-4) ]
the product will be negative. [ (3)(-4) = - ]
And of course, 3 times 4 is 12. [ (3)(-4) = -12 ]
So basically, if the signs are the same the product is positive,
if the signs are different, the product is negative.
And that's it for multiplication.
It's really that simple.
So in all three cases here we are multiplying unlike signs.
In that case, the product is always negative.
Then you can simply look at their absolute values and multiply.
If you need scratch paper, then that's okay, or a calculator.
So that gives me .2
And in this case, doing the whole number arithmetic
of 11/5 times 11/5, getting 121/25, [ (11/5)(11/5) = 121/25 ]
converting that to 4 and 21/25. [ 121/25 = 4 21/25 ]
Now I did this to indicate to you
that the algebra is really quite simple;
that it's the arithmetic that takes time.
The algebraic portion of this problem
is simply the determination of the sign of the answer,
and when you are multiplying,
if the signs are different, the product sign is always negative.
Then from that point on, you are back to old-fashioned arithmetic.
Summarizing, when multiplying two like signs,
the answer is positive,
and two unlike signs the answer is negative.
Why? Students always ask "why?"
Primarily because in their daily lives
they haven't seen something,
at least they think they haven't, that works this way.
Of course, the answer to the "why" is primarily mathematical,
but let's try to illustrate it with non-mathematics for a moment
and we might show you
that this really does seem sort-of reasonable.
Consider this example:
Let's presume that you have a reel,
you might be reeling out video tape or film,
or just a hose, a garden hose.
Now it seems reasonable that the amount that is reeled out
is the reeling rate times the time it's going to reel.
Let's symbolize that in a simple formula,
the amount that's out is the rate,
the reeling rate [ R] times the time [T] [ R · T ]
Now using that formula, if the reel winds out at 4 feet per minute,
what will be the status in 5 minutes?
So the formula says the amount that's out is the rate,
which is 4 feet per minute, [ R = 4 ft/min ]
times the time, which is 5 minutes, [ 4 ft/min · 5 min ]
and using what we learned in our measurement chapter,
we can cancel these measurements
and we get 4 times 5 is 20 feet. [ (4 ft)(5) = 20 ft ]
And of course, this is very reasonable.
If it is reeling out 5 minutes
at 4 feet per minute, [ (4 ft/ min)(5 min) ]
common sense says it will be 20 feet in 5 minutes.
But now consider this next example.
Now if the reel is winding in at 4 feet per minute,
what will be the status in 5 minutes? Well it's now winding in.
So it doesn't take too much imagination to see
it seems reasonable to call that
minus 4 feet per minute [ -4 ft/min ] because it is winding in,
there is 4 feet less out every minute.
So minus 4 feet per minute. [ -4 ft/min ]
It's still going to be in 5 minutes from now,
so negative times positive is negative [ (-4 ft/min)(5 min) = - ]
and 4 times 5 is 20 feet. [ (-4 ft/min)(5 min) = -20 ft ]
And now if we interpret negative to mean "20 less feet,"
that is reasonable, isn't it?
If it's pulling in 4 feet every minute,
in 5 minutes there'll be 20 less feet.
So we realize reeling out we will call positive,
reeling in, we will call negative.
And in that sense,
a negative times a positive indeed is negative.
But let's look at another application of this.
Now, if the reel winds out at the rate of 4 feet per minute,
we've agreed that 'out' is 'plus' 4 feet per minute.
What would have been the status 5 minutes ago?
So 5 minutes from now we called plus 5,
it seems reasonable that 5 minutes ago is minus 5 minutes.
And of course, if positive times negative really is negative,
we'll get negative 20 feet.
Now, of course, if it's reeling out 4 feet per minute,
5 minutes ago, there indeed was 20 feet less out.
So you see, our model of using that reel
is fitting what the algebra says it should be.
Now if it reels in at 4 feet per minute,
we've agreed that's minus 4,
what would have been the status 5 minutes ago?
And time past we've agreed should probably be a negative number.
Now, if it's reeling in,
5 minutes ago, there was 20 more feet out, which is plus 20.
But see our rule says that negative times negative is positive
and that's what our common sense says it should be.
Now I hope we didn't go too fast,
but if you think about this just for a moment,
it appears that our rules for multiplying signed numbers
does indeed work like things that are reeling in and out
or things going in a cycle, a wheel, a dynamo, a planet in orbit,
time going forward and backwards, seasons, cycles,
and indeed, that's one of the reasons why this rule was developed.
Actually the real reason though, is a mathematical, logical one.
And you'll find that developed more thoroughly
in a full-blown algebra course.
Now as it turns out,
one can find situations which work like that
from every kind of field from business to astronomy.
Once you become used to the operations on signed numbers,
you will find it easy to work problems
which formerly you found difficult to even talk or think about.
Such is the power of mathematics.
And so it was with that reel.
It was a little bit difficult to think fast
about what you're going to have in 5 minutes or 5 minutes ago,
under the different conditions.
But by using the rules of signed numbers,
all we had to do was to remember
that reeling out is positive, reeling in was negative,
time in the future is positive, time in the past is negative,
and then just remember the algebra.
When multiplying unlike signs the answer is negative.
When multiplying like signs, the answer is positive.
When multiplying unlike signs, the answer is negative.
So you see the algebraic part of it is really, truly quite simple.
Let me use this next problem to make a special point.
Many of you as you begin to approach algebra
approach it with a little bit of apprehension
because unfortunately,
algebra has a bit of a reputation of being hard.
Let me use this example to show you
that that's really not the case.
Let's look at the algebra part of this multiplication.
The algebra part is simply this:
Once we are prepared to begin to get the answer,
our first question
when we are adding, subtracting, multiplying or dividing
is ask what the sign is?
Well, we know if we are multiplying two positives,
two negatives or two positives, we get a positive.
So this is positive.
And a positive times a negative is negative.
That's the algebra part.
Now I ask you, is that at all hard?
Now after we have determined the sign, the algebra part,
we have to actually multiply the absolute values,
which means forget the signs,
which means old-fashioned, grade school arithmetic.
In this case, of fractions.
That's where the messy work begins.
And unfortunately, algebra gets a little bit of blame for that.
But you see the algebra was very easy, wasn't it?
Now we have to simply go back and recall our arithmetic.
And if you recall, when you are working with fractions,
and in particular multiplying fractions,
we had what was called:
the Cancellation Law of Fraction Multiplication
which simply said this: If I'm multiplying fractions,
I may divide any top multiplier anything I want.
So I see, I can divide this [ 25 ] by 5.
And the law says I may do that
providing I find one bottom multiplier I can also divide by 5.
So in this case I have two of them,
either this [ 35 ] or this [ 15 ]
So let me pick on this [ 35 ],
5 goes into [ 35 ] here 7. [ 35 ÷ 5 = 7 ]
And since we cross out after we've divided,
that's where the word 'cancellation' comes in,
but please remember,
'cancellation' really is another way of saying
'division under special conditions'
and the 'special conditions' is I have to be multiplying.
If I do this at any other time, I'll get erroneous results.
So let's see if we can cancel anywhere else.
Well, I see I can divide
both of these top [ 5 ] and bottom [ 15 ] multipliers
once again by 5.
Then I see I can divide this [ 3, 3 ]
top and bottom multipliers by 3, that gives me 1,
then I see I can divide each of these [ 7, 14 ] by 7,
then each of these [ 2, 4 ] by 2,
and now I multiply the remaining multipliers on top [ 1 · 1 · 1 ]
and get 1, [ (-1/1)·(-1/1)·(1/2) = 1/ ]
and the remaining multipliers on the bottom, [ 1 · 1 · 2 ]
and get 2, [ (-1/1)·(-1/1)·(1/2) = -1/2 ]
so this whole thing, [ (-25/35)·(-14/15)·(3/4) = -1/2 ]
after I would have multiplied and reduced would be negative 1/2.
But again, please note.
The algebra part of the problem is remarkably simple,
it's the simplest part of it.
The arithmetic itself is hard or easy
depending upon you and your recall of past lessons,
stemming from perhaps, in some cases, many years ago.
So it's remembering something from the past
that is necessary to algebra
that frequently causes the difficulty, not the algebra itself.
Keep that in mind as you go further into algebra
and that will put aside much apprehension
and let you see it for really the easy subject that it should be.
Now let's develop a special multiplication rule
that will provide a short-cut for much multiplication.
Very frequently in algebra and its applications
when you need to multiply,
you don't just settle for multiplying for two at a time,
you're multiplying a long string.
So let's look at that specifically.
Let's not look at their absolute values,
because that's just plain-old arithmetic, let's look at the signs.
Now if they were all positives,
there would be no problem, would there?
The answer would, of course, be positive.
But what happens if they're all negative
or there's a mixed stream of negative and positive?
Let's consider first the case
where there's an odd number of factors
and let's presume that they're all negative.
Notice if I multiply a negative times a negative,
that's always positive. [ (-)(-) = + ]
Then the next negative times a negative [ (-)(-) = + ]
[ (-)(-) = + ] is also positive,
and of course, positive times positive [ (+)(+) = + ]
[ (+)(+) = + ] will always be positive.
So again, negative times negative [ (-)(-) = + ]
is again positive, and so on,
clear down the stream, even if there were a zillion factors.
But because there is an odd number of them,
there were always be one left over
after all the other pairs have produced positive products.
So this negative, the one lone one left over, times this positive
of course, will always be negative.
So we can make a special note:
If I'm multiplying a long string of factors,
and if there is an odd number of negatives,
the product will be negative too.
And then all I have to do is to multiply the absolute values.
Well then, what about the other case?
If I'm multiplying an even number of factors,
and here again, recall in algebra,
the word 'factor' means 'to multiply' or 'multiplier,'
if there's an even number of them,
then that means if I sub-divide them into pairs,
each of those pairs producing a positive product,
there is no odd one left over.
And of course, all of these positives produced by these
even negatives, when multiplied is, of course, positive.
So we make a special note:
that if I'm multiplying a long string of factors,
and if there's an even number of negatives,
the results will be positive.
So once again, an odd number of negatives is negative,
an even number of negatives is positive, when multiplying.
Of course, when adding or subtracting
that's an entirely different game, isn't it?
Let's apply this to a situation.
First we note that I'm multiplying everywhere.
By the way in algebra,
this is always the first thing you look at.
Never the numbers or the signs,
the first thing you look at are the operations.
What kind do I have? Are they all the same?
If they are the all the same, what is it?
In this case, they're all multiplications.
Then that, all being multiplications,
tells me I may use a special short-cut rule.
Then I simply want to count the number of negatives.
So there's 1,
that's an even number so I know my product will be positive,
and the algebra bit is over.
Isn't that easy?
Now we can just begin to do the multiplications.
And in standard grade-school arithmetic
we take 5 times 7 times 2 times 3 times 1 times 2 [ 5·7·2·3·1·2 ]
that's rather involved,
but we can remember from our rules in previous lessons
that multiplication, too, is commutative and associative,
which tells me I can multiply in any order that I want.
So notice this. I will do this first.
5 times 2 is 10, and hold that in my mind. [ 5 · 2 = 10 ]
Then 2 times 3 is 6 [ 2 · 3 = 6 ]
times 7 is 42, [ 6 · 7 = 42 ]
and 42 times 10 is 420. [ 42 · 10 = 420 ]
So you see, the algebra isn't making things harder,
indeed here's a case where it's making things easier.
It's giving me permission to jump around as I multiply
in order to find the easiest combinations.
And that's the two algebra laws,
the Associative and Commutative Laws of Multiplication.
And again the algebra tells me to determine the sign,
because I'm only multiplying, I count the number of negatives.
If it's even the product is positive.
So again, the algebra part is simple.
And the algebra gives us additional rules
that will simplify our arithmetic.
Here's another special law
you might want to tuck at the back of your mind
which will simplify a lot of situations when you get to algebra.
And that is when you have the opposite of a number,
sometimes it's convenient to think that
that's equal to -1 times that number. [ -a = (-1)·(a) ]
Now let's look at this
in the light of our positive and negative numbers.
Let's say I had a negative 4, [ -4 ]
but recall we also stated that that could be thought of
as the opposite of 4,
and this law says I can treat that as -1 times 4 [ (-1)(4) ]
And of course, we know that
a negative times a positive is negative,
so this [ (-1)(4) = -4 ] is negative 4,
which is where I started from. [ -4 = (-1)(4) ]
So you might say Hey, what's the big deal then?
But this way of looking at it
sometimes will help explain a certain situation.
Let's look at this.
If I were to take, let's say, negative 3/4.
Now in this case, [ - 3/4 ] it is negative, the fraction, 3/4.
But let's look at it
as -1 times the fraction 3/4. [ (-1) · 3/4 ]
Now let's recall, on a fraction
forgetting this [ -1 ] for just a moment.
I can always multiply top and bottom by the same number.
And specifically that will also work
if that top, that number is a negative number,
let's say, negative 1. [ 3/4 · (-1)/(-1) ]
Now, 3 times -1 is -3 on top, [ 3/4 · (-1)/(-1) = -3/ ]
and 4 times -1 is -4 on the bottom. [ 3/4 · (-1)/(-1) = -3/-4 ]
But I still have that times -1. [ (-1) · (-3)/-4 ]
But, recall from elementary school
that if I'm taking a whole number times a fraction,
I simply take this whole number times the top
to get the new top, over that denominator.
So taking this whole number [ -1 ] times this top [ -3 ]
and negative 1 times negative 3 is positive 3, [ (-1)(-3) = 3 ]
and on the bottom we still have negative 4.
Playing around a bit, which we will further in the next lesson,
we have division of signed numbers,
we find that this [ - 3/4 ]
can also be written [ -3/4 ] as negative 3 over positive 4.
So note, negative ¾ [ - 3/4 ]
is the same thing as negative 3 over 4, [ - 3/4 = (-3)/4 ]
is the same thing as 3 over negative 4. [ - 3/4 = (-3)/4 = 3/-4 ]
Later in algebra this will be really convenient
to be able to go from this form [ - (3/4) ] to this form [ -3/4 ]
or to this form [ 3/-4 ]
So think of this tentatively, just informally,
as though the fraction has three signs,
the sign of the whole thing,
the sign of the top and the sign of the bottom.
And that I may opposite any two of them.
So if I opposite these two [ - 3]
this [ 4 ] becomes plus and that [ 3 ] becomes minus
and I get this [ -3/4 ] But on the other hand,
if I opposite these two [ - 4] this [ 3 ] becomes plus
and this [4] becomes minus and I get this [ 3/-4 ]
Actually, I said "three forms" there is a fourth here, isn't there?
Because I could have written this as 'the opposite of'
that's negative 1, times,
then minus 3 over minus 4 [ - -3/-4 ]
And of course, I would have gotten here [ - -3/-4 ]
by changing two signs, leaving this [ - ] one alone,
and oppositing the top and the bottom to give me this [ -3/-4 ]
So you see when we come to working with fractional expressions,
we're going to have four different ways of expressing this.
Bear this in mind.
We'll come back on this later.
But let's just sort of tickle our fancy with it now
so that when we get it seriously in a later lesson,
this will help us tie into it more thoroughly.
So at this point, it is quite sufficient
if you simply notice that this [ - 5/6 ]
could be written as minus 5, or negative 5 over 6,
or 5 over negative 6 [ 5/-6 ] or, leaving this [ - ] alone,
writing it as negative 5 on top and negative 6 on the bottom.
Four forms of saying the same fraction.
Now one might ask:
Are any of these forms preferred over any others?
At the beginning, this [ - 5/6 ] form,
simply because it keeps you thinking
in terms of rules that are fairly simple
as far as multiplication, addition, and subtraction.
This [ -5/6 ] when you get into algebra proper
will be considered the more preferred form.
These [ 5/-6, - -5/-6] will be useful now and then
but generally not preferred as a concluding process.
Usually we like to keep the bottom positive if possible.
So let's conclude this lesson by going back to its basics.
And that is, if I'm multiplying like signs of two factors,
the answer is always positive,
and if I'm multiplying two unlike signs,
the answer is always negative.
That's the algebra part.
Then the arithmetic part is: multiply the absolute values.
And of course,
that's just elementary-school arithmetic, isn't it?
But see, actually the algebra part is easy.
Two like signs, positive.
Two unlike signs is negative.
Now please note, some students in algebra
will try to short-cut this rule and say:
"Two negatives make a positive." That is not true.
Because if I am adding two negatives,
the results is not positive,
in this case it's negative, isn't it?
So the rule is:
If I'm multiplying two negatives, the results is positive,
as opposed to: adding two negatives, the answer is negative.
So don't give yourself short-cut rules that aren't true.
So this is our basic lesson and that's remarkably simple.
And our short-cut rule,
if we're strictly multiplying a long stream of factors,
then we simply count the number of negatives.
So 1, 2, 3, 4, 5, that's an odd number,
so because it's an odd number, the answer is negative.
If it were even, the answer would be positive.
Then we simply multiply the absolute values,
knowing that now the Commutative and Associative Laws
give me the freedom to skip around.
So I see that this [ 2, 5 ] is 10, this [ 9, 7 ] is 63,
10 times 63 is 630, [ 10 · 63 = 630 ]
times 2, [ 630 · 2 ]
so my answer is negative 1860.