Percentage of natural numbers that are perfect squares?

While reading about the prime number distribution I came across this fact that the percentage of natural numbers that are perfect square is zero. How do I prove this ?

Solution 1:

"Percentage" is a bad word, you mean something like natural density.

For that it's easy, as the limit exists, you just compute

$$\lim_{n\to\infty} {\#\{m^2 < n : m\in\Bbb N\}\over n}$$

The numerator is bounded above by $\sqrt{n}$ so we get

$$0\le\lim_{n\to\infty} {\#\{m^2 < n : m\in\Bbb N\}\over n}\le \lim_{n\to\infty} {1\over\sqrt n}=0$$

Solution 2:

Zero percent of all natural numbers are perfect squares in the sense that the limit of the proportion of the perfect squares to natural numbers is zero.To express it, consider

$$\lim_{x\rightarrow\infty} \frac{\left|\left\{n\in\mathbb{N}\mid n≤x \wedge n\ \mbox{is a perfect square}\right\}\right|}x.$$

Since the numerator is $\lfloor\sqrt{x}\rfloor$, thus approximately $\sqrt x$ and

$$\lim_{x\rightarrow\infty} \frac{\sqrt{x}}x = \lim_{x\rightarrow\infty} \frac{1}{\sqrt{x}} = 0,$$

the statement follows that zero percent of all natural numbers are perfect squares.

Solution 3:

Let $Q_n = \{ x \in \mathbb N : x=y^2, x \le n \}$. Then $\#Q_n \le \sqrt n$.

Then $\displaystyle\lim_{n\to\infty} \frac{\#Q_n}{n} \le \lim_{n\to\infty}\frac{\sqrt n}{n}=\lim_{n\to\infty}\frac{1}{\sqrt n}=0$.