I am wondering if an irrational number can be represented as a fraction in this way:
For example (to represent $\pi$):
$$\pi= 3.14159265359...=\frac{314159265359...}{100000000000...}$$
In the fraction $\frac{314159265359...}{100000000000...}$, the numerator is an integer whose digits have the same order as digits of $\pi$, and the denominator is simply $10\,^{(\#\,of\,digits \,of \,numerator\,-\,1)}$. Isn't an irrational number represented as a fraction in this way? Probably I misunderstand the concept of the irrational number. Thanks in advance.
$\endgroup$ 33 Answers
$\begingroup$If the numerator and denominator contains a finite number of digits each (and both are integers), then it's a rational number, but it won't equal $\pi$. If they have infinitely many digits, then I don't know what the expression even means, but even if one could make sense of it, and the value would be $\pi$, the numerator and denominator wouldn't be integers, and thus you wouldn't have a rational number, at least not a priori.
$\endgroup$ $\begingroup$An irrational number is a number that can't be represented as a ratio (i.e., a fraction) of two integers.
Since the digits of pi go on forever, your numerator is an infinite sequence of digits. That isn't an integer; only a finite sequence of digits defines an integer.
Additionally, it's hard to know what your fraction means. You say the denominator is $10^{\text{# of digits of numerator}-1}$, but the number of digits in the numerator is infinite, so is your denominator $10^{\infty-1}$? That isn't well-defined; it certainly isn't an integer.
$\endgroup$ $\begingroup$Actually, the fact that it's irrational means that you cannot represent it this way! This is precisely the definition.
A rational number can be represented as a fraction $a/b$, with $a,b$ integers and $b \neq 0$.
An irrational number cannot.
$\endgroup$
