New operator definitions Getting comfortable with evaluating newly defined function operators
New operator definitions
- 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 are going to do in this video is a little bit of fun.
- We are actually going to define our own operators.
- And what's neat about this it kind of shows how broad
- mathematics can be. And in a more practical sense it is something
- that you actually might see on some standardized test.
- And the reason why they do that is so that you can appreciate
- that these are not the only operators out there, plus
- exponentiation and all those, that in mathematics you can define whole
- new set of operators. Let's just do that. So let me just define x diamond
- x diamond y and I'm going to define that as
- I'm just gonna define that as 5x minus y. So you
- could view this kind of defining a function
- but we are defining it using an operator
- so if i have x diamond y by definition we've defined
- this operator, that means that's going to be equal to 5x minus y.
- So given that definition, what would-what would
- 7 diamond 11 be?
- Well, you just go to the definition. 7 diamond 11
- for instead of a x we have 7, so it's going to be
- 5 times 7. Just let me do 5 times 7 minus, and instead
- of a y, we have an 11.
- So one way to think about it is every--in our definition
- every place you saw a x you can replace with 7
- and every place you saw y, you replace with 11.
- So you have minus 11 over here. So this is the 7
- This 7 is this 7
- and this 11--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 star--or not, this isn't a star--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-- a -- write it this way
- a star b--let's say that is the same thing as-- I don't know
- a over a plus b. So it's the same idea
- What would,
- 5 star --5 star 6 be?
- Well you go back to the definition. By definition
- every place you see a you would now replace with 5
- every time you'd see b, you'd now replace with 6
- 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
- 5/11. And then you can compound them.
- We haven't defined and order of operations for these particular
- operations we just have defined.
- So we are going to be careful to use parentheses
- when we do, when we put some of these together.
- But you could do something like, something interesting
- like -1 diamond--negative 1 diamond 0 star--0 star 5
- and once again we just focus on parentheses, because
- that is the only thing that is telling us what to start on first
- because we haven't, we haven't figured out
- we haven't defined whether diamond takes precedence over
- star or star takes precedence over diamond.
- The way we have that thing saying hey, you do multiplication before you do addition
- we haven't defined it for those operations.
- But that's what parentheses help us do.
- So we want to evaluate the parentheses first.
- 0 star 5, that is 0, because so now this you can zero as the a
- and the 5 as the b
- this is going to be 0 over 0 plus 5. Over 0 plus 5.
- which is just going to be 0. So this is 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 first, the first number
- in our operator. The first term we are giving the operator
- I guess you can think of it that way.
- So 5 times that, so that will be 5 times negative 1
- x is negative 1
- minus y. Well y here is the zero. Minus 0
- So 5 times negative 1. 5 times negative 1 is negative 5
- And you will see and the idea is just to make you feel comfortable
- defining new operators like this, and not feel being daunted
- if you all of a sudden see a diamond. And they are
- defined the diamond for you
- and you are like you never saw a diamond. They are actually defining it for you
- So you shouldn't say I never saw a diamond.
- You'll just say, oh well they have defined diamond for me
- this is how I use that operator. And sometimes you'll see even wackier things.
- You'll see things like, let me draw
- So they'll define. So this is, I don't know if you'd even consider this as an operator.
- By definiton, if somebody writes a symbol like this
- and they put an a b c--let me write 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
- but they are just defining how (do) you evaluate this crazy expression.
- And so if someone were to give you, say evaluate this diamond
- 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 five, a three and a two. And we would just use the definition of how we evaluate the diamond.
- And we'd say every time we see an a, we say that's negative 1
- so we have negative 1 times d, well d. Well d is whatever is in the bottom right section of this
- diamond or this kite. So d is going to be 2. Let me write this this way.
- This is a.
- This is b.
- This is c.
- and this is d.
- So this is going to be negative 1
- times 2, minus b
- Well b is 5
- minus five
- 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. Negative 7 over 3.
- And you can go crazy like this. And it might be a fun thing actually
- if you have some spare time.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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This is great, I finally understand quadratic functions!
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At 2:33, Sal said "single bonds" but meant "covalent bonds."
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