Question

\[\Huge \mathsf{\text{Binary Operations}} \]\[\text{- Lesson to be hosted by nbouscal.} \]

Answer 1

yes?no?

Answer 2

Haha alright, why not. Good thing to know.

Answer 3

Basically, a binary operation is a type of map. Say you have sets A, B, and C, a binary operation maps from \(A\times B\to C\). Most commonly, though, the three sets in question will be the same set. So you'll have for example a map from \(\mathbb{R}\times \mathbb{R}\to \mathbb{R}\).

Answer 4

The easy examples that you already know are your usual binary operations from arithmetic. Addition, subtraction, multiplication, division, exponentiation.

Answer 5

I crave for examples.

Answer 6

So, with the easy examples you have like 5+6=11, you see you have three elements from R that are mapped via the binary operation of addition. You map (5,6) to {11}. Or if you have 2*4=8, you're mapping (2,4) to {8}.

Answer 7

But, there are a lot of other binary operations other than those easy examples. For example, composition of functions is a binary operation. If you have f(x) and g(x) then f(g(x)) is a binary operation.

Answer 8

You would write that one \(f\circ g (x)=f(g(x))\) with \(\circ\) as the operator.

Answer 9

So, a binary operation is just something performed on a set to bring it on another? So something like\[7 - 6 = 1 \implies (7,6)\rightarrow \{1 \} \]

Answer 10

Yeah. The key point is that it's binary, which means it's performed on the product of two sets, and it maps to an element from a single set.

Answer 11

You can expand it to more than two things, though, by using a certain property that some binary operations have: the associative property.

Answer 12

Which brings us to an important topic when studying binary operations: properties that they can have. The most important properties are: associative, commutative, identity element, inverses.

Answer 13

Example, please?

Answer 14

Okay we can start with associative property. Associative means that for a binary operation, let's say \(\circ\), you have that \((a\circ b)\circ c=a\circ (b\circ c)\).

Answer 15

:O I didn't know that one.

Answer 16

Easy examples that are associative: addition and multiplication.
Easy examples that are not associative: subtraction and division. Also, exponentiation.

Answer 17

Yes. Yes. I follow.

Answer 18

So is that it?

Answer 19

\[\text{Basic operations}\subset \text{Binary operations} \]If I am not wrong.

Answer 20

After associative you can look at commutative property. Commutative says, for a binary operation \(\circ\), that \(a\circ b=b\circ a\). Again, you can see easy examples with the normal binary operations on scalars. Another good example of noncommutative is matrix multiplication.

Answer 21

What do you mean by basic operations?

Answer 22

Basic operations are \(\times,-,\div, +\)

Answer 23

What is \(a \circ b\)?

Answer 24

Oh, okay. Then yeah, those are a subset of binary operations.

Answer 25

I'm just using \(\circ\) as a placeholder for an unspecified binary operation.

Answer 26

Okay. I see. But \((f\circ g)(x) \ne (g \circ f)(x)\) when they are not inverse functions.

Answer 27

Right, so composition of functions is not commutative.

Answer 28

You can also easily see that subtraction and division are not commutative, but addition and multiplication are.

Answer 29

Oh, I see. :)

Answer 30

Thank you once again! @nbouscal

Answer 31

There's still more :P

Answer 32

:O

Answer 33

Go ahead please.

Answer 34

As I mentioned before, we're also interested in whether binary operations have identity elements, and whether they have inverses.

Answer 35

An identity element is an element \(e\) such that \(e\circ x = x\circ e = x\) for all x in the set. Can you figure out what the identity elements are for addition and multiplication?

Answer 36

Yes.
0 for addition.
1 for multiplication.
1 for division.
0 for subtraction.

Answer 37

Be careful here. Division and subtraction don't actually have identity elements.

Answer 38

Why?\[{x \over 1} = x \]\[x - 0 = x \]

Answer 39

Notice that I said it had to go both ways. e-x=x-e=x, and e/x=x/e=x. There's no e for either of those that fits that definition.

Answer 40

Oh. Yes!

Answer 41

You'll notice from this definition that the identity element has to commute with every element of the set. That is to say, ex=xe for every x. (We often leave the operation out and just right ex when we mean e\(\circ\)x).

Answer 42

write*

Answer 43

Okay :)

Answer 44

Should also point out now that division, exponentiation, and some others are sometimes referred to as "partial binary operations" rather than binary operations proper, because they are partial functions. For example, you can't divide by zero.

Answer 45

So that will explain to some extent why division isn't fitting with any of these properties :P

Answer 46

Haha :)

Answer 47

The last property to talk about for the moment is inverse elements. Basically y is the inverse element of x if xy=yx=e.

Answer 48

You'll find that some operations have inverse elements for every element of the set, others have inverses for some only, and others have no inverses at all other than the trivial: the identity is always its own inverse. Also you'll notice that to have inverses at all, the operation must have an identity.

Answer 49

Some other good binary operations to look at are the ones you find in modular arithmetic.

That covers it pretty well for now, let me know if you have any other questions.

Answer 50

(This is good stuff to learn, btw. If you ever take any higher maths at all you'll need to know this for sure.)