Skip to content
Exercise 12.1 · Q2

Q.On Z\mathbb{Z}, define ∗* by (m∗n)=mn+nm, ∀ m,n∈Z(m*n) = m^n + n^m,\ \forall\, m,n \in \mathbb{Z}. Is ∗* binary on Z\mathbb{Z}?

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
2% · 2/84 Questions
✓ Free question

We look for one ordered pair of integers where mn+nmm^n+n^m fails to be an integer -- a single counterexample is enough to disprove closure.

Step 1. Try a pair with a negative exponent. Take m=2, n=−1∈Zm=2,\ n=-1\in\mathbb Z.

Step 2. Compute mn+nmm^n+n^m. mn=2−1=12m^n=2^{-1}=\dfrac12, and nm=(−1)2=1n^m=(-1)^2=1. So m∗n=12+1=32m*n=\dfrac12+1=\dfrac32.

Step 3. Check membership. 32∉Z\dfrac32\notin\mathbb Z, so the output leaves Z\mathbb Z for this pair -- condition (ii) of Definition 12.1 fails (the operation is not even defined as an integer here, let alone unique-in-Z\mathbb Z).

Conclusion. Since there exists at least one pair (m,n)∈Z×Z(m,n)\in\mathbb Z\times\mathbb Z for which m∗n∉Zm*n\notin\mathbb Z, ∗* is not a binary operation on Z\mathbb Z.

✓Final answer

No, ∗* is not binary on Z\mathbb Z. Counterexample: m=2,n=−1⇒2−1+(−1)2=32∉Zm=2,n=-1\Rightarrow 2^{-1}+(-1)^2=\tfrac32\notin\mathbb Z.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.