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I have been thinking about this problem

Yes, let's simplify the problem by assuming the set of only positive integers $Z$ I know if I take a range of numbers say $1$ to $10$ , there are only three perfect square existing here; which are $1$ , $4$ and $9$ So the probability is $3/10$ If I increase my range the probability becomes lesser, so I can conclude there are more imperfect-square numbers than perfect squares.

My question Is there any fixed value assign to this, can I know the minimum and maximum value of this probability

Aderinsola Joshua
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    You need to specify more things. Whats a perfect square? Simply x s.t. sqrt(x) is an integer again? What is the probability distribution on lN ? Or are we considering {1, ..., n}? – Stockfish Mar 29 '20 at 13:27
  • A perfect square. X ( a positive integer ) , x s.t. sqrt(X) is a positive integer again. we are considering {1,2,3,.…........,n}, considering all the set of positive integer again – Aderinsola Joshua Mar 29 '20 at 13:35
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    If $n$ tends to $\infty$ , the probability that a random number in the range $[1,n]$ is a perfect square, tends to $0$. So , if we allow the random number to be any positive integer, the probability is $0$. – Peter Mar 29 '20 at 13:38
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    https://en.wikipedia.org/wiki/Natural_density – saulspatz Mar 29 '20 at 13:50

1 Answers1

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The probability that a uniformly chosen integer in $\{1,2,\ldots, n\}$ is a perfect square is given by $$p_n={\lfloor\sqrt{n}\rfloor\over n}\ ,$$ because there are exactly $\lfloor\sqrt{n}\rfloor$ perfect squares in $\{1,2,\ldots, n\}$. As $$\sqrt{n}-1<\lfloor\sqrt{n}\rfloor\leq\sqrt{n}$$ we therefore can say that $${1\over\sqrt{n}}-{1\over n}<p_n\leq{1\over\sqrt{n}}\ ,$$ which proves that $\lim_{n\to\infty} p_n=0$.

Christian Blatter
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