# Regular Language : $\{a^m b^n \mid mn \ge 10\}$

I am little bit confuse here about below language is it regular language $$L= \{a^m b^n \mid mn \ge 10 \}.$$

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 Some body will edit my question I am not able to write it properly in mathematics form – Nishant Nov 15 '12 at 13:44

As the language $$L_1 = \{ a^ib^j \mid 1 \le i, 1\le j, \;ij < 10\}$$ is finite, it is regular. Moreover $$L_2 = \{a^ib^j \mid i,j \ge 1 \}$$ is regular (in has $aa^*bb^*$ as regular expression). But so $L = L_2 \setminus L_1 = L_2 \cap \complement L_1$ as the set of regular languages is closed under taking complements and finite intersections.
 martini you are correct according to above example you have given even But in question mn which is product >= 10...As I think this will not going to be regular..if mn <= 10 then it will dfenitely going to be regular correct me if i am wrong – Nishant Nov 15 '12 at 14:25 But, as I wrote, your $L$ (product $\ge 10$) is the complement of my $L_1$ (product $<10$) intersected with $aa^*bb^*$, and hence, regular. – martini Nov 15 '12 at 14:28 Ok acording to you {a^nb^nc^n where n<= 10 } is regular then it's complement L={ a^nb^nc^n n>10 will be regular} is it fine as we know { a^nb^nc^n n>10 will be regular} is not regular.... – Nishant Nov 15 '12 at 14:35 No, as this is not the complement. The complement in $a^+b^+c^+$ is $L = \{a^nb^kc^l \mid n\ne k \vee n \ne l \vee k \ne l \vee n > 10\}$ – martini Nov 15 '12 at 14:40 martini: Is {aibj∣1≤i,1≤j,ij<10} it complement of my problem I don't think so Might be we are interpreting it wrong martini can you check again my above problem is it regular or not In my point of view not looking regular.. – Nishant Nov 15 '12 at 14:54