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On the set $N$ of non-negative integers, the binary operation ______ is associative and non-commutative.
asked in Set Theory & Algebra by Veteran (59.7k points)
edited by | 919 views
+2
Associative but not commutative ? Division , subtraction are not possible as these are not associative as well. Any answer ?

3 Answers

+28 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
0
Can you explain a little more?
0
@arjun sir :what is the proper operation ...the operation in the ans is assumption only
0
mod operation?
0
how
0
no, thats not associative. Not getting any standard operators.
0
so what can be the ans
0
division also associative but not commutative
+9
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).
0
yes ,thanks :)
+1
subtraction operation?
+4

@rahul

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
yeah.you are right.Thanks for correction
+1

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

+22 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.

answered by Boss (23.1k points)
+9

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

  • Matrix multiplication is one more.
0
Thanks, Rajesh Pradhan and Sachin :) for eye opener example.
0
@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)
+2
Set intersection is commutative also

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