If \(L_1 = \{a^n | n \geq 0 \}\) and \(L_2 = \{b^n | n \geq 0 \}\), consider…
2014
If \(L_1 = \{a^n | n \geq 0 \}\) and \(L_2 = \{b^n | n \geq 0 \}\), consider
(I) \(L_1⋅L_2\) is a regular language
(II) L1⋅L2 = \(\{a^n b^n|n \geq 0\}\).
Which one of the following is CORRECT?
- A.
Only (I)
- B.
Only (II)
- C.
Both (I) and (II)
- D.
Neither (I) nor (II)
Attempted by 103 students.
Show answer & explanation
Correct answer: A
Key insight: L1 = a* and L2 = b*, so L1⋅L2 = a*b*.
Regularity: a* and b* are regular, and regular languages are closed under concatenation, so a*b* is regular.
Equality check: a*b* = {a^i b^j | i, j ≥ 0}, which is not the same as {a^n b^n | n ≥ 0}. For example, aab (a^2 b^1) belongs to a*b* but not to {a^n b^n}.
Conclusion: The statement that the concatenation is a regular language is true, while the statement that the concatenation equals {a^n b^n | n ≥ 0} is false. Therefore, only the claim that the concatenation is regular is correct.
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