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+25 votes
On the set $N$ of non-negative integers, the binary operation ______ is associative and non-commutative.
asked in Set Theory & Algebra by Veteran (59.9k points)
edited by | 1k views
Associative but not commutative ? Division , subtraction are not possible as these are not associative as well. Any answer ?

3 Answers

+31 votes
Best answer
Define Binary operation $*$ on $(a,b)$ as : $a*b = a$

1. It is associative : $(a*b)*c = a*c = a$, and $a*(b*c) = a*b = a$

2. It is not commutative : $a*b = a$, whereas $b*a = b$.
answered by Boss (11.3k points)
selected by
Can you explain a little more?
@arjun sir :what is the proper operation ...the operation in the ans is assumption only
mod operation?
no, thats not associative. Not getting any standard operators.
so what can be the ans
division also associative but not commutative
Division is not associative because in general (A/B)/C is not equal to A/(B/C).  
For example: (8/4)/2 is not equal to 8/(4/2).
yes ,thanks :)
subtraction operation?


subtraction will voilate closer property, it is said that "on a set of non negative integers"...but subtraction could lead to negative integers.....

for eg: 2*4 = 2-4 = -2// not a non negative integer.

+1 are right.Thanks for correction

@rahul+sharma+5, subtraction by default is not associative either.

+26 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 

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.

answered by Boss (23.6k points)

nice example (y)
If $f∘g$ exists then $g∘f$ might not even exists.

  • Matrix multiplication is one more.
Thanks, Rajesh Pradhan and Sachin :) for eye opener example.
@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 vote
Set Intersection operator is also possible.
answered by Active (1.6k points)
Set intersection is commutative also

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