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given regular language L = {$a^n$b:n≥0}

$L^2$ - L = $L^2$ since there's no string in $L^2$  which is in L.

Am i correct ?
asked in Theory of Computation by Active (1.6k points) | 59 views
0
yes exactly.
0
here $L^{2}=L.L$

only concatenation operation is working here

So, $L^{2}$will contain $a^{n}b.a^{n}b=\left \{ bb,abab,aabaab,aaabaaab,.... \right \}$ where $L=a^{n}b=\left \{ b,ab,aab,aaab,.... \right \}$
+2
@srestha,

the $L^2$ given by you is not correct.

correct will be, $L^2=a^mb.a^nb\ |m,n>=0$
0
yes it will be $a^{*}b.a^{*}b$

and $L$ contains 1 b and $L^{2}$ contains 2 b

that is why $L$ and $L^{2}$ are not same

rt?
0
thanks @nitish

1 Answer

0 votes

Here, L={b,ab,aab,aaab,...................}

Now,

L={bb,bab,baab,..........................}

As we can see there is no common element 

 

answered by Active (3.7k points)


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