It is quite direct to prove that $\lfloor{\sqrt{n}\rfloor} = O(\sqrt{n})$. But how can I prove that $\lfloor{\sqrt{n}\rfloor} = \Omega(\sqrt{n})$?
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It is also pretty straightforward if you write down the definition of $\Omega(\sqrt{n})$. – megas Sep 24 '16 at 00:15
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Welcome to CS.SE! This is a pure math question. Also, What did you try? Where did you get stuck? We're happy to help you understand the concepts but just solving exercises for you is unlikely to achieve that. It looks like the standard techniques in our reference question should make this straightforward. Please read those materials and apply them. If you're still confused, edit the question to show what you've tried and where you got stuck and what specifically you're confused about, and we can consider the question for re-opening. – D.W. Sep 24 '16 at 00:42