Welcome to The Nonlinear Library, where we use Text-to-Speech software to convert the best writing from the Rationalist and EA communities into audio. This is: Rational and irrational infinite integers, published by Viliam on March 23, 2022 on LessWrong.
Epistemic status: a fine line between genius and madness
Let's start with the observation that it is much easier to use digits before the decimal point than the digits after it. A six years old child could calculate e.g. 11 + 2, but calculating reliably 0.11 + 0.2 will take a few more years of education.
From this perspective, it is quite ironic that we typically use numbers that only have a few digits before the decimal point, and many, sometimes infinitely many digits after the decimal point, such as 1/3 = 0.333333333... = 0.¯¯¯3.
Obviously, it would be better the other way round.
Technically, we already have infinitely many digits in front of every number. But they are all zeroes; and by convention, we do not write them. Writing a zero or multiple zeroes in front of a number does not change its value; the number 007 mathematically means 7. Therefore, technically, we could write the number 7 with infinitely many digits in front of it, like this: ¯¯¯07.
But it seems like a waste of space to only ever fill those infinitely many pre-decimal places with zeroes. What about using some other digit instead?
Consider the number ¯¯¯9, which consists of infinitely many nines. Written in full, it would be like this: ¯¯¯9 = ...999999999.
The value of this number is -1.
How can we know that? Simply, just calculate ...999999999 + 1 = ...000000000, which is zero. (Zero with an arbitrary number of zeroes in front of it is zero: 0 = 00 = 000 = ¯¯¯0.)
Similarly, ...999999998 = ¯¯¯98 is -2, because ¯¯¯98+2=¯¯¯0. Also ¯¯¯97=−3, ¯¯¯96=−4, and so on.
Seems like we have invented a new syntax for negative numbers. Instead of writing a minus symbol in front of a number, just let the digits underflow. If an odometer in your car showed only nines, you would know what happens if you go the extra mile.
What if we fill the infinite pre-decimal places with digits other than zeroes and nines? For example, how much is ¯¯¯1?
I believe this number is -1/9.
We can easily verify it by multiplying the number by 9 and adding 1.¯¯¯1×9+1=¯¯¯9+1=¯¯¯0.
Okay, this seems strange. How did a fraction smaller than one suddenly become an infinite integer? But as you can clearly see, it did. Remember, our goal was to move the infinite digits behind the decimal point to the front of it. And so far, we are succeeding!
We can similarly get other negative fractions by dividing −1=¯¯¯9 by the denominator, and multiplying the result by the numerator. For example, −1/3=¯¯¯3, −2/3=¯¯¯6, −1/7=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯142857. Check it, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯142857×7+1=¯¯¯9+1=0.
To obtain a positive fraction, we can subtract the negative fraction from zero. For example, 1/9=0−(−1/9)=0−¯¯¯1=¯¯¯89. Check it, ¯¯¯89×9=¯¯¯01. Seems correct.
But there is still problem with fractions like 1/2 or 1/5. There is no integer, not even infinite one, such that if we multiplied it by 2 or by 5, the last digit of the result would be 1. (We can easily check it, because the last digit of the result only depends on the last digit of the original number.) Therefore, some decimal places seem inevitable: 1/2 remains 0.5, and 1/5 remains 0.2. We have not eliminated all decimal places from the fractions, only the infinite sequences thereof.
Using the new syntax, any fraction will have at most finite number of decimal places. To see why it is so, extract the powers of 2 and 5 from the denominator, like this: nd=n2i5jk, then calculate the infinite integer nk and divide it by 2i and 5j. The number of decimal places of the result will be max(i, j). That kinda sucks, but whatever.
By the way, −1/2=¯¯¯9.5, and −1/5=¯¯¯9.8. I assume it is obvious why.
(Having an infinite number of digits both before and after the decimal point would lead to ambiguity. For example, 1/3 could be ...