Let \(A=\{a, b\}\) and \(L=A^*\). Let \(x=\{a^nb^n,n>0\}\). The languages \(L…
2023
Let \(A=\{a, b\}\) and \(L=A^*\). Let \(x=\{a^nb^n,n>0\}\). The languages \(L \ U \ X\) and \(X\) are respectively :
- A.
Not Regular, Regular
- B.
Regular, Regular
- C.
Regular, Not Regular
- D.
Not Regular, Not Regular
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Correct answer: C
Answer: Regular, Not Regular
Reason: The union L ∪ X equals L because L = A* contains every string over the alphabet; since X ⊆ L, L ∪ X = L. A* is a regular language, so L ∪ X is regular.
Why X is not regular:
Assume for contradiction that X = {a^n b^n | n > 0} is regular. Let p be the pumping length from the pumping lemma.
Take the string s = a^p b^p in X. By the pumping lemma, s can be split as s = xyz with |xy| ≤ p and |y| > 0, so y consists only of a's.
Pumping down with i = 0 gives xz = a^{p-|y|} b^p, which is not of the form a^n b^n (the number of a's is less than the number of b's), contradicting that all pumped strings remain in X.
Therefore X is not regular.
Conclusion: L ∪ X is regular (equals A*), and X is not regular.