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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.

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    $\begingroup$ Whatever that thing in the numerator is, it's not an integer. $\endgroup$
    – littleO
    Apr 20, 2018 at 16:46
  • $\begingroup$ Obligatory SMBC comic: smbc-comics.com/?id=2208 Which highlights a semantic distinction: you can use irrational numbers in fractions, but an irrational number can't be represented by a fraction itself. $\endgroup$
    – Andy Walls
    Apr 20, 2018 at 16:50
  • $\begingroup$ The definition of an irrational number is that it can't be represented as the $ratio$ of two integers. Hence irrational. What you have, at beast, is a way of making a sequence that approaches the value of an irrational number. $\endgroup$
    – Joffan
    Apr 20, 2018 at 16:50

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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.

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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.

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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.

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