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For the set $N$ of natural numbers and a binary operation $f : N \times N \to N,$ an element $z \in N$ is called an identity for $f,$ if $f (a, z) = a = f(z, a),$ for all $a \in N.$ Which of the following binary operations have an identity?

  1. $f (x, y) = x + y - 3$
  2. $f (x, y) = \max(x, y)$
  3. $f (x, y) = x^y$
  1. I and II only
  2. II and III only
  3. I and III only
  4. None of these
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Answer : A

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An Element $z \in N$ is called an Identity element for $f$ if $f(a,z) = f(z,a) = a, \forall a \in N$ 

$\text{I}$ : $f(x,y) = x + y -3$

We will say that $z$ is an identity element if $f(x,z) = x = f(z,x)$, Hence, $x + z -3 = x = z + x -3$

So, We have $z = 3$. 

$\text{II}$ : $f(x,y) = \max(x,y)$

We will say that $z$ is an identity element if $f(x,z) = x = f(z,x)$, hence, $\max(x,z) = x = \max(z,x) , \forall x \in N$.

So, We have $z = 0$ (if $N$ consists of $0$)

Or We have $z = 1$ (if $N$ does not consist of $0$)

$\text{NOTE}$ : Whether $0$ belongs to Set of natural numbers or not, is an ambiguous thing. In Number theory, generally we don't consider $0$ as a Natural number. But in Computer Science, $0$ is considered as a Natural number by many authors. 

In this question, It doesn't matter whether or not $0$ belong to $N$ or not. We can say that the Set $N$ will have an Identity element in both the cases.  (Also See below NOTE 3)

$\text{III}$ : $f(x,y) = x^y$

We will say that $z$ is an identity element if $f(x,z) = x = f(z,x)$, hence, $x^z = x = z^x , \forall x \in N$.

But, we know that $x^y $ is Not necessarily equal to $y^x$. (See below NOTE 2) 

Hence, there is No identity element for this Binary Operation.


$\textbf{Some Extra Points}$ : 

$\text{NOTE 1: }$

For any Set  $A$.

Saying that a Binary Operation $f : A \times A \rightarrow A$, is as good as saying that the Given Set $A$ is Closed under the Binary Operation $*$. 

$\text{NOTE 2 : Solutions of the Equation } \bf{x^y = y^x}$

Obvious partial solution is  $X = Y$, but there are  other solutions like $2,4$ and $4,2$.

 In fact, these solutions and $X = Y$ are the only solutions in positive integers.  And the only integer solutions are $X = Y$ and $(2, 4), (4, 2), (-2,-4), (-4, -2)$

Refer Proof : http://mathforum.org/library/drmath/view/66166.html

$\text{NOTE 3 : }$

I had said in the Second Binary Operation that $0$ or $1$ could be the Identity element. But after seeing the Third Binary Operation, I can say that Whoever Professor set this Question, must have thought of $0$ as a Non-Natural Number because If $0 \in N$, then $f(0,0) $ will be Indeterminate and $f(0,0) = 0 \,\,or \,\,1$. Which will lead to $f$ Not being an Binary Operation. 

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Answer: A

  1. Identity element is $3.$
  2. Identity element is $1.$
  3. There is no identity element. $(x^y ≠ y^x,$ when $x ≠ y)$
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I-   f(x,y) = x+y-3 = x= y+x-3  =>  y=3 Here identity elements is 3


II-  f(x,y) = max(x,y)=x=max(y,x)  => y=1 Here identity elements is 1


III- f(x,y) =x^y is not same as f(y,x) = y^x. So no identity element.

so A is correct answer

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–6 votes

Answer is (D)

All of them have identity element.

(1) has 3 as identity element.

(2) has 1 as identity element.

(3) has 1 as identity.

Answer:

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