Tips to proof a language is not regular using Pumping Lemma

Question

I am trying to prove that the following language is not regular using the pumping lemma

L = {aⁱ b^j | i = 2j for some j ≥ 0}

I have decided to choose s = a^2p b^p, in this way |s| ≥ p and I can split it in three pieces xyz where for every i ≥ 0, xyⁱz ∈ L.

Any tips for continuing the proof?

Thanks!

Apply the lemma! you're almost done. You know that the lemma holds, so there exist such xy'z , choose i+1, show that the word is not in the language so the language is not regular. — Benjamin Gruenbaum
One of the conditions of the pumping lemma says that |xy| <= p. In this case p < 2p so |y| < 2p. If I chose i = 2, the string would be xyyz. The length is |xyyz| = |xyz| + |y| < 2p+p + p. So the number of a's can't be twice the number of b's. Is it correct? — colis
Right, just explain why the number of _a_s can't be twice as much as the number of _b_s and you're done — Benjamin Gruenbaum
@colis read my this answer and this answer In which I have explained that in pumping lemma we don't know where looping part so you need to break strings in L in all possible ways. And If still its a problem let me know. — Grijesh Chauhan

Pascal NoPascensor Pascal NoPascensor · Accepted Answer · 2014-02-27T14:57:12

Choose s = a^2p b^p is right!

As said by Grijesh Chauhan we must break strings in L in all possible ways.

So you can split s in:

where |xy|≥ 0 and |y|>0.

Taking i=2, you have xy²z:

that is:

Since l contains at least one 'a' (because |y|>0). You can say L is not regular