Language Regular or not

Question

Is it regular?

$\left \{ \left ( 0^{n} \right )^{m}|n<m,n,m\geq 1 \right \}$

Comments

Put n=1 and then m=2,3,4........... , We will get all the strings of length greater than or equal to 2 already, Hence Regular.

For other values of n, we will get strings which are already there in the language because of n=1.

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@Deepak poonia (Dee)  Can i take m =1 then I have to take n=0 so epsilon shouldn't be there? ?because m>=1 is given

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You can't take m=1 because it is given that $n \geq 1$ and $m > n$

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Okk got it thanks, didn't see that.

Answer 1

Yes it is regular we can draw a DFA.

Comments

@srestha : $000^{*}$ will not produce $n=m=1$

if $n=m=1$ then $L=((0)^{1})^{1}=0  \not\equiv  000^{*}$

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Always take $n=1$ and $m>1$,

$L=(0^{1})^{2}=00,n=1,m=2$

$L=(0^{1})^{3}=000,n=1,m=3$

$L=(0^{1})^{4}=0000,n=1,m=4$

 

and so on...

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no it was prev. one

I got it