Algebra (all content)
New operator definitions 1
Sal solves a few problems where a new operator is defined and Sal evaluates some expression with it. Created by Sal Khan.
Want to join the conversation?
- Are the symbols just like variables, but operations?(33 votes)
- Yes. Think of them as like the f in f(x), the sin in sin(x), the ! in x!, the + in x+y, the √ in √2, and many more things, but defined via operator.(1 vote)
- Can you use symbols like a these? $ % & € £ ¥ ₩ @ #
Or even a drawing?(7 votes)
- Definitely! You can use any symbols that you like or create your own. But to avoid confusion, it's not recommended to use signs, which have already a meaning like #, that has several meaning in mathematics ( cardinality in set theory, primorial in number theory and etc.)(12 votes)
- What does "⊘" mean??(3 votes)
- ø typically means the empty set, in set theory. However, you are probably using it to denote a "new operation".
Hope this helps!(6 votes)
- What is the difference between defining an operator and simply defining a function?(4 votes)
- + and - are actually forms of operators. So at some point they were defined. And f( ) is an operator. It is what you use to say "now use this function"(4 votes)
- What does diamond or star mean .(0 votes)
- it is just a defined operation.
for example # = multiplication(2 votes)
- Is it possible to define an order of operations precedence for operators you define?(3 votes)
- Yes. You can make operators anything you want.(1 vote)
- so are the star and diamond there instead of an adding, subtracting, multiplying, or dividing symbol?(2 votes)
- No. a Star and a Dimond are absolutly NEW operations:not +.-.х,/.
someone TOLD you that 3+4=7. I bet you have never asked why ! Well Sal here tries to explain:
that Dude, who told you that + sign works the way you know, told you this because SOME OTHER dude(the inventor) DEFIEND + sign to do this things, but the inventor of + sign could have invented a STAR(*), for example, IF he NEEDED it. Why he would need a STAR operator? Well, maybe he likede the idea that 3 * 4 would equal 34. Maybe Its good for inventing a warpdrive !:) who knows... You are free to invent your own operators!!!(3 votes)
- What kind of applications does this have in real life?(2 votes)
- at1:54what did he mean?(1 vote)
- how do you know whether to use subtraction, addition, multiplication, or division for the symbols?(1 vote)
- In order to use the symbols, you have to define how to use it, and that is exactly what Sal does.
For instance at0:45he defines x diamond y to be 5x-y.(1 vote)
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.