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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 form2012-11-15

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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.


Edit: Regarding Nishant's questions. We have by the definition of complement and intersection \begin{align*} L &= \{a^n b^m \mid nm \ge 10\}\\ &= \{a^n b^m \mid \neg(nm < 10)\}\\ &= \{w \in \Sigma^* \mid \exists n,m\ge 1 : w = a^nb^m \wedge \neg(nm < 10)\}\\ &= \{w \in \Sigma^* \mid \exists n,m \ge 1 : w = a^nb^m\} \cap \{w \in \Sigma^* \mid \neg\exists n,m \ge 1: w=a^nb^m, nm < 10\}\\ &= L_2 \cap \complement L_1 \end{align*}

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6434/discussion-between-nishant-and-martini)2012-11-15