What are Binary Operations? | Abstract Algebra

What are binary operations? Binary operations are a vital part of the study of abstract algebra, and we'll be introducing them with examples and proofs in this video lesson!

A binary operation on a set S is simply a function f from SxS to S. So a binary operation is a function that takes two elements from the same set and maps that pair to exactly one element also in the same set.

For example, consider addition on the real numbers. Addition takes each pair of real numbers (an element of RxR) and maps it to exactly one element also in R. So if f: RxR to R is defined by addition, then f((3,2)) = 3 + 2 = 5. This is a binary operation. Since any two real numbers add to another real number, we say the real numbers are closed under addition.

Also, addition on the real numbers is commutative, as in the order doesn't matter. For any two real numbers a and b, a + b = b + a. Some binary operations have this property and some do not.

SOLUTION TO PRACTICE PROBLEM:

Addition on {0, 1, 2} is not a binary operation because it is not closed. Notice 2 + 2 = 4 is not in {0, 1, 2). Addition mod 3 on {0, 1, 2} is a binary operation because it is a function and it is closed. By definition the sum of any two integers mod 3 is 0, 1, or 2. Since the mod 3 sum of any two numbers in {0, 1, 2} is equal to exactly one other number in {0, 1, 2}, it is a function from {0, 1, 2}x{0, 1, 2} to {0, 1, 2} and hence is a binary operation.

Division on the rationals (Q) is not a binary operation because division is not defined on all of QxQ. Notice 0 is in Q, so elements of the form (a, 0) are in QxQ, and that is the problem. We can't divide by 0!

Multiplication on the complex numbers (C) is a binary operation. Consider two complex numbers a + bi and c + di, where i is the imaginary unit, and a, b, c, and d are real numbers. Multiplying these complex numbers and simplifying gives us (ac - bd) + (ad + bc)i. By closure of the reals under addition and subtraction, we know ac - bd and ad + bc are reals, so (ac - bd) + (ad + bc)i is a complex number. Thus, since two complex numbers multiply to exactly one other complex number, multiplication on C is a binary operation.

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I hope you find this video helpful, and be sure to ask any questions down in the comments!

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The outro music is by a favorite musician of mine named Vallow, who, upon my request, kindly gave me permission to use his music in my outros. I usually put my own music in the outros, but I love Vallow's music, and wanted to share it with those of you watching. Please check out all of his wonderful work.

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[hengsokgich]

I’m not sure If I can understand what my lecture teaches me, but clicking on your video is the right thing ever.

[benedictding]

So lucky to come across this video before my midterm, you explained better than the textbook!

[samo]

Thanks for this! :D I'm a teacher trying to review past lessons and this video was such a big help in refreshing my brain to binary operations!

[chandrika]

I've been searching for a perfect explanation on this concept and it's damn clear now! Even a lot of paid apps wouldn't have explained better. Tq

[jacobh]

Super useful! Working with some early Real Analysis where the book assumes Binary Operations are familiar, solved my conundrum super quickly!

[JEZZE]

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[fitoremorina]

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Can you make more videos on group theory?
Please, because next week I have a exam and your lectures helped me very much.

[thatfly]

Reading the description, your explanation of why the rationals are not a binary operation under division was unintentionally a bit funny. Because you say “you can’t divide by 0!” Well you can divide by 0!, since it is 1 😆

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[jaafarsoriano]

Please explain about group structures of binary operations with *

[tahiraayub]

Excellent explanation of binary opration

[bennett]

hello. is there a significance in your use of ":" when defining a binary operation as "f: S x S -> S"? I've seen it before in the same context, but I don't understand exactly what it's saying. I understand that every ordered pair in S x S passed through the function f is an element of S, I'm just having trouble translating that meaning to the notation. also, I'm trying to connect this to what i've learned so far. and if f((a, b)) is similar to f(a, b) in this case, would it be okay to say that ∀n(n ϵ S x S -> f(n) ϵ S)? i mean to say that for every element in S x S (can an ordered pair be called an element of S x S?), the function f acted upon that element would be an element of set S? i'm learning this topic by myself, so i don't really have anyone to ask about my questions. thank you for making the video regardless!

[hemchandrathoudam]

Thanks v.much 4 for this explanation on binary operation

[shizukana-gaijin]

Hello. Thanks for the lesson. I have a personal research question: Multiplication is generated by addition. But is there a binary operation that generates the addition?

[muskankushwah]

I didn't get the mod 3 questions answer . how you are exactly adding mod 3 to {0, 1, 2} sets.

Or what the question exactly mean.

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[gqsvtfm]

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[ChocolateMilkCultLeader]

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