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On the set $N$ of non-negative integers, the binary operation ______ is associative and non-commutative.

Associative but not commutative ? Division , subtraction are not possible as these are not associative as well. Any answer ?
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Variations on Associative, Commutative Property

Define Binary operation $\ast$ on $(a,b)$ as : $a\ast b = a$

1. It is associative : $(a\ast b)\ast c = a\ast c = a$, and $a\ast(b\ast c) = a\ast b = a$
2. t is not commutative : $a\ast b = a$, whereas $b\ast a = b$.

How about matrix multiplication of two matrices A and B..??

It is associative but not commutative..

@ASNR1010 Is set of matrices same as set of non negative integers?

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..

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.

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@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 $n^{\frac{(n^{2}+n)}{2}}$
"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.
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 $b$.

@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

Set Intersection operator is also possible.

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Set intersection is commutative also

1
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