Question

For the set $N$ of natural numbers and a binary operation $f:N\times N\to N$, an element $zϵN$ is called an identity for $f$, if $f(a,z)=a=f(z,a)$, for all $aϵN$. Which of the following binary operations have a identity ?
I. $f(x,y)=x+y-3$
II. $f(x,y)=\text{max}(x,y)$
III. $f(x,y)=x^y$

• A. I and II only
• B. II and III only
• C. I and III only
• D. None of these

Answer 1

The correct option is A I and II only

$I:$ $f\left(x\right)=x+y-3$
$x+a-3=x=a+x-3$
so $a=3$
Now 3 is unique, and $3ϵN$
So $I$ has identity.

$II:$ $f\left(x\right)=\text{max}(x,y)$
$\text{max}(x,a)=x=\text{max}(a,x)$
In $N$, the only value of $a$ which will satisfy above equation is $a=1$
Since 3 is unique, and $3ϵN$
So $II$ has identity.

$III:$ $f\left(x\right)=x^y$
$x^a=x=a^x$
Now $x^a=x\Rightarrow a=1$, but $x=a^x$ has no solution for $a$ in the set $N$
So $III$ has no identity.
So only $I$ and $II$ has identity.