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Let $\pi(x)$ denote the number of prime number not exceeding $x$. I will show that the function $\pi(x)$ can not be expressed as quotient form of polynomial like that $\frac{P(x)}{Q(x)}$ where $P(x)$, $Q(x)$ are polynomials.

My simple solution is the following. On the contrary, Suppose that $\pi(x) = \frac{P(x)}{Q(x)}$ and $degP(x) = n$, $degQ(x) = m$ Prime Number Theorem says that $\pi(x)$ is asymptotic to $\frac{x}{log x}$ as $x \to \infty$ and $\frac{P(x)}{Q(x)}$ is asymptotic to $x^{m-n}$ as $x \to \infty$ .

$\frac{P(x)}{Q(x)}$ and $\pi(x)$ are same But they are not asymptotic to each other. This is a contradiction that completes a proof.

What about my solution?

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    It is quite fine, provided you can show that $x/\log x$ and $x^a$ are not asymptotic, for any integer $a$. (You have an absolute value in the exponent, but that is not correct) – 2017-01-19
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    @MarianoSuárez-Álvarez Can you tell me the point that is not correct? I just want to express for the exponent to be positive. – 2017-01-19
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    If P is equal to X and Q to X^2, then P/Q is clearly not asymptotic to anything of the form X^n with n a non negative integer. – 2017-01-19

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