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L = {a^n: n ≥ 2, is a prime number}.

This is not a regular language.

What about L*?

Is it regular? Please explain.

asked in Theory of Computation by Junior (679 points) | 72 views

1 Answer

+1 vote

Let ∑ = {a} and L1 = ∑* 

Then we can say that L1 is regular.

Now take any arbitrary string x belongs to ∑*

then x = am for some integer m

Now, m must have a prime factor p

Hence m = p*k (for some integer k)

Hence, am = ap . ap . ap . ap .....(concatenating the string ap k times)

Now, ap belongs to L (given in the question )  which means (ap )k  belongs to L*

Hence am belongs to L*  which means x belongs to L*

So, for any arbitrary string x belongs to ∑* we get that x belongs to L*

Hence ∑* = L* 

Thus L* is a regular language.

answered by Loyal (8.4k points)
edited by
0

@https://gateoverflow.in/user/Kushagra+Chatterjee 

L* contains Language L which is not prime. So How can you say that L* is regular?



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