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We've started with $x^2$, saying that there are $\sqrt{10^9}$ numbers that are not $\in 10^9$

i'm thinking that if we add $\sqrt[3]{10^9}$ and $\sqrt[5]{10^9}$ to $\sqrt{10^9}$ and subtract them from $10^9$, we would have gone too far and excluded too many numbers due to repeating numbers such as $2^2$,$2^3$ and $2^5$, for example.

does it even make sense to do so? aren't all numbers in $\sqrt[3]{10^9}$ and $\sqrt[5]{10^9}$ $< 10^9$ already in $\sqrt{10^9}$ ?

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    It is always best when you discover things yourself.2012-06-12

1 Answers 1

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Let $a=10^9$. There are $\lfloor a^{1/2} \rfloor$ squares, $\lfloor a^{1/3}\rfloor$ cubes and $\lfloor a^{1/5}\rfloor$ $5^{th}$ powers within $1$ to $10^9$. So, the P.I.E. gives the answer to be: $a-(\lfloor a^{1/2}\rfloor+\lfloor a^{1/3}\rfloor+\lfloor a^{1/5}\rfloor-\lfloor a^{1/6}\rfloor-\lfloor a^{1/10}\rfloor-\lfloor a^{1/15}\rfloor+\lfloor a^{1/30}\rfloor)$.