Question

Show that L = {w|w = a2n} is not regular, where Σ = {a}.

Answer 1

pumping Lemma - *) If A is a regular Language, then A has a Pumping length p such that any string 'g' where 18129 may be divided into 3 pasos G=XYZ such that The following conditions must te the Дхуі 2 ел Рех елеу j20 2) 140 3) IXY) SP To prove that a longuage is not regular using Pumping Lemma follow the below steps: cwe poove using contradico op 1. Assume that A is regular 2. It has to have a pumping Length (Say D 3 All Songs Longer than p can be Pumped ISIZP 4 wow find a sting &' n A such that 18/21 5. onide s into xyz 6- show that xy/Z&A for some i 7. Then consides all ways that I can te divided into Xyz

L = {w/w aan 1. Assome that t' is Regular 2- Pumping lengon accogling to condioon on let n=1,2,3,- . then 2,418,16. - - - L={2}={aaaaz 3. Divide Language into 3 PSUS XYZ L=şqagaz x zaa IX string lenger (1)=4 Z-a Izlas cose li- lxyl<n - 354 w celea:- 141ZO yoa lyol Izow of 3 - 2=XYZ izo i will take is (=aacaita L-aaaaaaa x failed Our Assumption is using Becase '7' is not Dones of 2, I X Y Z & L for gone i (=an 3 is not desuar a
that is a^2^n not a^2n

a^2n is regular

a^2^n is not regular

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