# 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 ?

• $$101^2-100^2=10201-10000=201>100$$ So, the possibility goes to $0$. – Takahiro Waki Dec 12 '16 at 18:40
• @TakahiroWaki: this explanation is quite right, but probably a little terse to allow the OP to generalize. – Yves Daoust Dec 13 '16 at 11:29
• By the way, I was aware it does not mean that the limit is always 0. For example, by exchange natural sequence, $a_n=n^2$ are always square, so it's 100 percent. – Takahiro Waki Dec 18 '16 at 17:08

## 3 Answers

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.

"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

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

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

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