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## Two-variable functions

Current time:0:00Total duration:6:50

# New operator definitions 1

## Video transcript

We're all used to the
traditional operators like addition and subtraction
and multiplication and division. And we've seen there's multiple
ways to represent this. But what we're going to do in
this video is a little fun. We're actually going to
define our own operators. And what's neat
about this is it kind of shows how broad
mathematics can be. And on a more practical
sense, it's actually something that you might see on
some standardized tests. And the reason why
they do that is so that you can appreciate
that these aren't the only operators out there--
plus exponentiation and all those-- that in
mathematics, you can define a whole new
set of operators. So let's just do that. So let me just
define x diamond y. And I'm going to define
that as 5x minus y. So you could view this as
defining it a function. But we're defining
it using an operator. So if I have x diamond
y, by definition, we have defined this operator. That means that's going
to be equal to 5x minus y. So given that definition,
what would 7 diamond 11 be? Well, you just go
to the definition. 7 diamond 11. Instead of an x we have a 7. So it's going to be 5 times 7. So let me do it 5 times 7
minus, and instead of a y, we have an 11. So one way to think about
it is, in our definition, every place you saw an x,
you could replace with a 7, every place you saw a y,
you replace with an 11. So you have minus 11 over here. Let me make the number--
So this is the 7. This 7 is this 7. And this 11 is this
11 right over here. And then we just evaluate that. So 5 times 7 is 35. So this is equal to 35 minus
11, which is equal to 24. So 7 diamond 11 is equal to 24. We can define other things. We can define something crazy
like, let me define a-- well, I mentioned a star, let me
use a star-- a star-- let me write it this way-- a star b. Let's say that that
is the same thing as-- I don't know--
a over a plus b. And so same idea. What would 5 star 6 be? Well, you go back
to the definition. By definition, every
place where you see the a, you would now replace with a 5. Every time you saw the b, you
would now replace with a 6. So this is going to be equal
to 5 over 5 plus 6. a plus b. a is 5, b is 6 over 5 plus 6. So this would be 5/11. And then you can compound them. And we haven't defined
any order of operations for these particular operations
that we've just defined. So we're going to be
careful to use parentheses when we put some
of these together, but you can do something
like, something interesting, like negative 1
diamond 0 star 5. And once again, we just
focus on parentheses, because that's the
only thing that's telling us what
to start on first. Because we haven't
figured out, we haven't defined whether diamond
takes precedence over star, or star takes precedence
over diamond the way that we have that saying that,
hey, you do multiplication before you do addition. We haven't defined it
for those operations, but that's what the
parentheses helps us do. So we want to evaluate
these parentheses first. 0 star 5 that is 0-- because
you could view this 0 as the a and the 5 as the b--so
it's going to be 0 over 0 plus 5, which is just going to be 0. So this over here, 0
or 5 just goes to 0. So this whole
expression simplifies to negative 1 diamond--
this diamond right over here-- diamond 0. And now we go to the definition
of the diamond operator. Well, that's five times the
first number in our operator, or the first term that
we're giving the operator. I guess you could
think of it that way. So 5 times that. So it'll be 5 times negative
1, x is negative 1 minus y. Well, y here is the 0. Minus 0. So 5 times negative
1 is negative 5. And you will see--
and the idea here is just to make you feel
comfortable defining new operators like this. And not being daunted if all
of a sudden you see a diamond, and they're defining
the diamond for you. And you're like, wait,
I never saw a diamond. They're actually defining it
for you, so you shouldn't say, I never saw a diamond. You should say, well, they've
defined a diamond for me. This is how I use that operator. And sometimes you'll
see even wackier things. You'll see things like this. Let me draw. So they'll define. I don't know if you would even
consider this an operator. But you'll see something like
this, that by definition, if someone writes
a symbol like this, and they put-- a, b, c-- let me
write it this way-- a, b, c, d. They'll say this is
the same thing as ad minus b, all of that over c. And once again, this
is just a definition. They have this
weird symbolic way of representing these
variables in all this. But they're defining how do you
evaluate this crazy expression. And so, if someone
were to give you, were to say, evaluate
this diamond. Let me evaluate the diamond. So evaluate the diamond
where, in my little sections of the diamond, I have a
negative 1, a 5, a 3, and a 2. We would just use
the definition of how to evaluate this diamond. And we'd say, OK,
every time we see an a, that is going
to be negative 1. So we have a negative 1 times d. Well, d is whatever
is in the bottom right section of this
diameter or this kite. So d is going to be 2. Let me write it this way. This is a. This is b. This is c. And this is d. So it's going to be negative
1 times 2 minus b-- well b is 5-- minus 5, all of
that over c, which is 3. So this is going to be
equal to negative 2 minus 5. So that is negative 7 over 3. And you could go
crazy like this. And it might be a
fun thing, actually, if you have some spare time. Define your own operators
and see how creative you can get with
those operators.