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## Algebra (all content)

### Course: Algebra (all content)>Unit 7

Lesson 30: Two-variable functions

# New operator definitions 2

Worked examples of working with more newly defined operators. Created by Sal Khan.

## Want to join the conversation?

• what is there was a qustion like this where the X and the + dont match If x⊗y=(8−x)(y) and x⊕y=x2−4y2, find 3⊗(4⊕−2).
• The expressions don't have to match in any way. The ⊕ would give you an expression of x and y, and so will the ⊗. Suppose you don't divide by zero or raise zero to its own exponent anywhere there, you'll get a number as a result (or an expression if you use parameters). Otherwise, the expression would simply be undefined, just as one over zero is undefined (although some people would say it's infinity, but they won't be able to answer if I told them to calculate infinity over infinity).

There is an exponent for negative numbers, if that's what you were referring to.

As for your example:

3⊗(4⊕−2) = 3⊗[4²-4*(-2)²]
3⊗(4⊕−2) = 3⊗[16-4*4]
3⊗(4⊕−2) = 3⊗[16-16]
3⊗(4⊕−2) = 3⊗[0]
3⊗(4⊕−2) = (8-3)*0
3⊗(4⊕−2) = 0

The answer is 0. I hope that helps.
• What in the world is this!/!/!/!?/?/?/?
• Essentially Sal is defining a few new operators (that is, symbols just like the ones you're accustomed to using, like +, -, or ÷ ), which are indicated by the two circle symbols in the video. We already know what +, -, and ÷ do to two terms, we're familiar with those. We don't know what those new symbols do, though, so Sal is just attaching some rules to them.

To clarify a bit more, we could define an operator, indicated by @ (this can be any symbol you like!). We could say that x@y = x+y. What we've just done is defined a new operator @ which does the exact same thing as a plus sign. x and y are variables, they can be any value! So for this example:
1@1 = 1+1 = 2
2@4 = 2+4 = 6
2@7 = 2+7 = 9
and so on.

What's going on in the video is that Sal is just using a more complicated version of this. If x@y = x+3y (again, the symbol can be anything, we're just inventing it to play around), that means we can enter any values for x and y, and find out what the solution is based on that definition. For example:
1@1 = 1+3*1 = 4
2@3 = 2+3*3 = 11
5@3 = 5+3*3 = 14
and so on.

The video's a slightly more intricate problem, as you've defined not one but TWO new operators. In order to find the answer, it's just important to make sure you follow the brackets to find the order of operations. We can't use PEMDAS here, because we aren't working with the traditional operators of +, -, and so on. So we have to follow the brackets: do the operations INSIDE the brackets first, then the ones outside.

For more info Sal explains the process in more detail here: https://www.khanacademy.org/math/algebra/algebra-functions/new_operators/v/new-operator-definitions
• How did he know (or decide) which equation to use where? He solves the second half of the whole equation using the first one. Then plugs that solution into the second one given. But ... How did he know (or decide) to do that? Would it have worked either way?
• He just uses the operator definition when encountered an operator. For example, if you encounter the operator *, then follow the definition of x*y and solve that part. After that, if you encounter the operator #, follow the definition of x#y and solve it. That's it.
(1 vote)
• What if a operator is defined multiple times? Is this undefined behavior?
(1 vote)
• yes, if you have the exact same operator defined in 2 different ways, then this means it is undefined
(1 vote)
• Question 2 on this video is this: Why are both operators defined with x and y. Why not have one operator defined with x and y and one defined with a and b. Or even xa and xb or something to differentiate the two sets a little more? What effect would labeling them differently have on the final product? So for instance...

X & Y = 2X+3Y and x @ y = 7x + 2 find 7 & (4 @ 2) Hopefully my use of the & and @ are clear as replacements for other operator symbols. Wouldn't this have the same answer as x & y = 2x+3y and x @ y =7x+2 find 7 & (4 @ 2)? Or am I misunderstanding something intrinsic to this concept?
(1 vote)
• Yes it indeed does have exactly the same answer think of variables as a cloak before it is revealed (solved for) it doesn't matter what kind of cloak is concealing it as it is the same
(1 vote)
• In this video at he writes down 4(-2)^2. If in that example the -2 were instead a 0, or in any similar situation, would the exponent go inside the parenthesis or outside. Even though the answer is the same. 4(0)^2 = 4(0^2). And more importantly where should I go to learn the reasons why, and how to implement the rule later?
(1 vote)
• one for zero its doesn't matter as this is all multiplying 0 by 0 or by the 4 so it equals zero but to the first question . think of it like this 0^2+0^2+0^2+0^2 which is multiplying now pemdas exponents first whats 0^2? 0 so now you got 0+0+0+0=0 it can go in it or not as its the same as (0)^2 is that same as (0^2) think of parenthesis as a section you have as a whole, and the whole thing is ^2 but the whole is 0 and the whole is ^2 or putting it inside is still making that specific whole ^2 so it is still he same but with binomials in parenthesis don't as you can as (for example) (x+y)^2 is (again the wholes) (x+y)(x+y) not (x^2+y^2) so only for 1 term you can do this. check quadratics to explain binomials and beyond etc and check out a section teaching parenthesis to help
(1 vote)
• we know,
i^2= -1 [by the of definition of 'i']
=> i.i= -1
=> underroot(-1).underroot(-1)= -1
=> underroot{(-1).(-1)}= -1
=> underroot(+1)= -1
=> +1= -1 ......why
(1 vote)
• As you said, the definition says `i² = -1`. Therefore you can say `sqrt(-1)` can be written as `+- i`, but strictly it's not defined and your proof not valid. ;-)
(1 vote)
• y'all should go watch my new movie https://www.youtube.com/watch?v=TLJJbqQBRUo
(1 vote)
• yall should go watch my new movie