a^2n b^(2n+1) is a regular language. I am not able to decompose it in xyz so that I can pump any power of y as per pumping lemma. Please help me out.
closed as unclear what you're asking by Raphael♦ Jan 3 '14 at 6:37
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You probably misunderstood the question. For starters, the pumping lemma can't be used to prove that a language is regular, it can only be used to prove that a language isn't regular. The Wikipedia page for the pumping lemma gives a counterexample.
Secondly, as already pointed out by @DavidRicherby, this particular language is not regular, and as it turns out, you can use the Pumping Lemma to prove this statement.
Roughly speaking, the proof goes as follows:
Assume for the sake of contradiction that the language is regular, and thus, the Pumping Lemma holds. The "y" can take one of three forms:
- y contains all "a"s
- y contains all "b"s
- y is of the form $a^ib^j$ for some $i,j > 0$.
Now, can you see how to derive the contradiction?