39 votes 39 votes On the set $N$ of non-negative integers, the binary operation ______ is associative and non-commutative. Set Theory & Algebra gate1994 set-theory&algebra normal group-theory binary-operation fill-in-the-blanks + – Kathleen asked Oct 4, 2014 recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 5.7k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply worst_engineer commented Jul 15, 2015 reply Follow Share Associative but not commutative ? Division , subtraction are not possible as these are not associative as well. Any answer ? 5 votes 5 votes Deepak Poonia commented May 14, 2022 i edited by Deepak Poonia May 29, 2022 reply Follow Share $\color{red}{\text{Find Video Solution Below:}}$Detailed Video Solution$\color{darkred}{\text{Find More Variations here:}}$Variations on Associative, Commutative Property 5 votes 5 votes Please log in or register to add a comment.
52 votes 52 votes Define Binary operation $\ast$ on $(a,b)$ as : $a\ast b = a$ It is associative : $(a\ast b)\ast c = a\ast c = a$, and $a\ast(b\ast c) = a\ast b = a$ t is not commutative : $a\ast b = a$, whereas $b\ast a = b$. Happy Mittal answered Sep 27, 2015 edited Apr 23, 2021 by Lakshman Bhaiya Happy Mittal comment Share Follow See all 20 Comments See all 20 20 Comments reply Show 17 previous comments Kiyoshi commented Dec 20, 2021 reply Follow Share How about matrix multiplication of two matrices A and B..?? It is associative but not commutative.. 1 votes 1 votes Deepak Poonia commented Dec 21, 2021 reply Follow Share @ASNR1010 Is set of matrices same as set of non negative integers? 1 votes 1 votes Kiyoshi commented Dec 21, 2021 reply Follow Share No sir, It is not same… But here I mean by matrix multiplication of two matrices both containing the non-negative integers. So, result after binary operation has to be a matrix containing non-negative integer.(closure satisfied) and matrix multiplication we already know associative and not commutative.. 1 votes 1 votes Please log in or register to add a comment.
47 votes 47 votes The most important associative operation that's not commutative is function composition. If you have two functions f and g, their composition, usually denoted f∘g, is defined by (f∘g)(x)=f(g(x)). It is associative, (f∘g)∘h=f∘(g∘h), but it's usually not commutative. f∘g is usually not equal to g∘f. For our case suppose $\forall$x $\in$ N of non-negative integers, if f(x)=x2 and g(x)=x+1, then (f∘g)(x)=(x+1)2 while (g∘f)(x)=x2+1, and they're different functions. Rajesh Pradhan answered Jan 15, 2017 Rajesh Pradhan comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments HeadShot commented Aug 30, 2018 i edited by Mk Utkarsh Sep 24, 2018 reply Follow Share @Sachin Mittal 1 What would be the count of such operations ? Is there any way to do so ? For eg. For commutative operations on set of n elements , possible ways are $n^{\frac{(n^{2}+n)}{2}}$ 1 votes 1 votes Deepak Poonia commented Apr 10, 2020 reply Follow Share "Composition" operation can only be the answer when set consists of functions (on suitably defined set of functions). For this question, this won't be a viable answer as the given base set is set of all natural numbers. 8 votes 8 votes HitechGa commented Jul 12, 2021 reply Follow Share this is fine as an example of an operation that is associative but not commutative. But this does not satisfy what the question demands. Suppose $a,b\in \mathbb{N}$ and $\star$ is the binary operation. You need to define something for: $a\star b$. Using this example of function, how to define $a\star b$? Your example of $f \circ g$ takes only one parameter as input from $\mathbb{N}$ set, but $a\star b$ requires us to take two values $a$ and $b$. 2 votes 2 votes Please log in or register to add a comment.
0 votes 0 votes @Arjun sir @Lakshman Patel RJIT ‘<’(smaller than) is possible. it is associative. e.ge. 1<(2<3) = (1<2)<3 it is not commutative. e.g. 1<2 != 2<1 Overflow04 answered Sep 9, 2022 Overflow04 comment Share Follow See all 0 reply Please log in or register to add a comment.
–1 votes –1 votes Set Intersection operator is also possible. Mehak Sharma 1 answered Dec 21, 2016 Mehak Sharma 1 comment Share Follow See 1 comment See all 1 1 comment reply Rajesh Pradhan commented Jan 15, 2017 reply Follow Share Set intersection is commutative also 3 votes 3 votes Please log in or register to add a comment.