Q. 94.3( 6 Votes )

# Define a new oper

Answer :

**Consider m*n = m ^{2} + n^{2}.**

(i)

Yes, ℕ is closed under *.

This is because, for any natural number m, m

^{2}is also a natural number.

Further, on adding two natural numbers, we get a natural number only.

So, if m and n belong to ℕ then m

^{2}+ n

^{2}also belongs to ℕ.

**(ii) **Commutative means x*y = y*x

where x and y belongs to ℕ

⇒m*n = m^{2} + n^{2}

And n*m = n^{2} + m^{2}

⇒ m^{2} + n^{2} = n^{2} + m^{2}

⇒m*n = n*m

Hence, * is commutative on ℕ

**(iii)** Associative means (x*y)*z = x*(y*z)

where x, y and z belongs to ℕ

m*n = m^{2} + n^{2}

Further,

(m*n)*o = (m^{2} + n^{2})*o = (m^{2} + n^{2})^{2} + o^{2}

Similarly, n*o = n^{2} + o^{2}

Further,

m*(n*o) = m*(n^{2} + o^{2}) = m^{2} + (n^{2}+ o^{2})^{2}

⇒ (m^{2} + n^{2})^{2} + o^{2} ≠ m^{2} + (n^{2} + o^{2})^{2}

⇒ (m*n)*o ≠ m*(n*o)

Hence, * is not associative.

**(iv)** An identity element is a special type of element of a set with respect to a binary operation on that set,

which leaves other elements unchanged when combined with them.

then b is the identity of a.

For *, let k be the identity element

then m*k = m^{2} + k^{2} = k^{2} + m^{2} = m^{2}

This is possible only when k = 0 but k needs to be a natural number.

Hence, * does not have an identity element.

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