Welcome to a second lesson in multiplying integers
with two-color counters. In this lesson we'll look at examples
of a negative times a negative, and a negative times a positive.
We'll define a yellow counter as positive one, and a red counter as negative one.
So we we can model positive two, using two yellow counters,
negative three using three red counters, and zero
using one yellow and one red counter. Before we take a look at our new examples though,
I do want to review one example from the last lesson. When we had a positive times a
negative,
for example, positive four times negative two. We view this as starting with zero or starting
with nothing.
Then because of the positive four here, we were given,
or we were adding four sets, or four copies of negative two.
And since we can model negative two using two red counters,
we were given, or adding four copies
of two red counters, which look like this.
And notice how the result is eight red counters,
which equals negative eight. Four times negative two
equals negative eight. So now for the new examples,
let's take a look at negative four times negative two.
Well now because the first factor is negative four,
we'll view this as, we're taking away, or removing
four sets, or four copies of negative two.
So because in this case we're removing four copies
of two red counters, we'll have to start with a bunch of zeros
so that we have four sets of two red counters to remove.
So notice how if we start with a sea of zeros, or a bunch of zeros
like we see here, this is still equivalent or equal to zero, like we see here.
But now in this form, we can remove four copies
of negative two. The four copies of negative two
would be these four copies of two red counters.
And now if we remove these and simplify, we can find the product
of negative four and negative two. So now we'll remove our four sets, or four copies of
negative two. And notice how we still have some zeros here, so let's go ahead and simplify that.
These four red counters and four yellow counters
would simplify to zero, or equivalent to zero,
leaving us with a total of eight yellow counters
which is equal to positive eight. Negative four times negative two
equals positive eight. Now let's take a look at
negative four times positive two. Again this means we're going to be removing,
or taking away four copies, or four sets of positive two this time.
So we're going to start with a sea of zeros, or a bunch of zeros.
and now in this form we can go ahead remove four copies of positive two.
Here are the four copies of positive two.
If we remove these, we'd be left with these counters.
Again notice how we have some zeroes here. These four red counters and four yellow counters
are equivalent to zero or simplified to zero.
Which would leave us with eight red counters,
which is equal to negative eight. Negative four times negative two
equals negative eight. Let's look at this example one more time
with a slightly different approach. Remember before,
we started with a whole bunch of zeros, or a sea of zeros so that for negative
four times positive two,
we could remove four copies of positive two.
What we could do though, is recognize that we have to remove
four copies of positive two, which means it could stop with just
eight zeros. Which will require less simplifying. If we start with eight, zeros, now we
can easily identify the
four copies of positive two, that we want to remove,
which would be these four copies here. Notice once we remove these,
there's no extra simplifying required, because we're left exactly
eight red counters, which again is equal to negative eight.
So negative four times positive two
equals negative eight. So there is more than one way to do these problems.
But I think when first starting out, starting with the sea of zeros
is probably a bit more straightforward. I hope you found this helpful.