Define a new operation * on the set of all natural numbers by m*n = m² + n².

(i) Is ℕ closed under *?

(ii) Is * commutative on ℕ?

(iii) Is * associative on ℕ?

(iv) Is there an identity element in ℕ with respect to *?

(i) Yes, ℕ is closed under *. This is because, for any natural number m, m² 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 m² + n² also belongs to ℕ.

(ii) {displaystyle x*y=y*x}Commutative means x*y = y*x where x and y belongs to ℕ

⇒m*n = m² + n²

And n*m = n² + m²

⇒ m² + n² = n² + m²

⇒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 = m² + n²

Further,

(m*n)*o = (m² + n²)*o = (m² + n²)² + o²

Similarly, n*o = n² + o²

Further,

m*(n*o) = m*(n² + o²) = m² + (n²+ o²)²

⇒ (m² + n²)² + o² ≠ m² + (n² + o²)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.

For *, let k be the identity element then m*k = m² + k² = m²

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

Hence, * does not have an identity element.