GATE CSE 1994 | Question: 2.2

Question

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

Comments

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

Answer 1

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 $b\ast a = b$.

Comments

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

It is associative but not commutative..

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

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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..

Answer 2

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.

Comments

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

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"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.

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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 $b$.

Answer 3

@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

 

Answer 4

Set Intersection operator is also possible.

Comments

Set intersection is commutative also