Math Challenge!

[Surlethe]

Prove that the rational numbers are exactly the repeating or terminating decimals! For extra credit, characterize the nature of the repeat in terms of the associated rational number.
Spoiler

Use the division algorithm, long division that we all learned in elementary school.

(This came up in my college algebra class the other day and it was a cute little exercise.)

[LionElJonson]

It's because there's only so many numbers that can be left over from any given subtraction, so sooner or later you'll wind up cycling through them, for any given possible rational number (2-digit numbers for any given single-digit divisor, 3-digit numbers for any given 2-digit divisor, et cetera), since each stage removes the single-largest single-digit number from the remaining amount. This means that you'll eventually get a repeating number or a precise termination (the remainder is zero), so irrational numbers are impossible to get.

Dunno how to formally write that up as a proof, though.

[Grog]

It's because there's only so many numbers that can be left over from any given subtraction, so sooner or later you'll wind up cycling through them, for any given possible rational number (2-digit numbers for any given single-digit divisor, 3-digit numbers for any given 2-digit divisor, et cetera), since each stage removes the single-largest single-digit number from the remaining amount. This means that you'll eventually get a repeating number or a precise termination (the remainder is zero), so irrational numbers are impossible to get.

Dunno how to formally write that up as a proof, though.

That is as good as any proof of the fact that rational numbers have repeating or terminating decimals.
However you forgot to prove that all repeating or terminating decimals are rational.

[LionElJonson]

It's because there's only so many numbers that can be left over from any given subtraction, so sooner or later you'll wind up cycling through them, for any given possible rational number (2-digit numbers for any given single-digit divisor, 3-digit numbers for any given 2-digit divisor, et cetera), since each stage removes the single-largest single-digit number from the remaining amount. This means that you'll eventually get a repeating number or a precise termination (the remainder is zero), so irrational numbers are impossible to get.

Dunno how to formally write that up as a proof, though.

That is as good as any proof of the fact that rational numbers have repeating or terminating decimals.
However you forgot to prove that all repeating or terminating decimals are rational.

They're not irrational numbers (which are defined as non-repeating, non-terminating numbers), so therefore they're rational numbers, right?

[Surlethe]

Irrational numbers are defined as (real) numbers which cannot be written as the ratio of two integers. The challenge in this thread is to justify their characterization as non-repeating, non-terminating decimals.

[LionElJonson]

All terminating numbers can be written as (significant digits)/10^n, where n is the number of significant digits after the decimal point, and therefore rational. The repeating portion of all repeating numbers can be written as (repeating digits)/(10^n+10^(n-1)...+1), where n is equal to the number of repeating digits, and are therefore rational. They can be shifted to the left or right by multiplying or dividing by 10^x, where x is the number of digits you're shifting them, and since they can be added to terminating numbers to create things like 15.2446446446..., all terminating or repeating numbers are therefore rational.

[Grog]

All terminating numbers can be written as (significant digits)/10^n, where n is the number of significant digits after the decimal point, and therefore rational. The repeating portion of all repeating numbers can be written as (repeating digits)/(10^n+10^(n-1)...+1), where n is equal to the number of repeating digits, and are therefore rational. They can be shifted to the left or right by multiplying or dividing by 10^x, where x is the number of digits you're shifting them, and since they can be added to terminating numbers to create things like 15.2446446446..., all terminating or repeating numbers are therefore rational.

This argument does not work. For example 0.3333.... is not 3/(10+1), but it is might be close to one way of doing it. Spoiler

There is a number you can divide by to get it right but your expression is wrong.

[Kuroneko]

It's wrong, but it essentially works. Let M have m digits. The division algorithm in base 10m can easily verify that M/[10m-1] is the decimal formed by repeating the digits of M. Thus any repeated decimal x can be written in the form 10-n[N + M/[10m-1]].

[Grog]

It's wrong, but it essentially works. Let M have m digits. The division algorithm in base 10m can easily verify that M/[10m-1] is the decimal formed by repeating the digits of M. Thus any repeated decimal x can be written in the form 10-n[N + M/[10m-1]].

That was what I meant by the hint.

[LionElJonson]

All terminating numbers can be written as (significant digits)/10^n, where n is the number of significant digits after the decimal point, and therefore rational. The repeating portion of all repeating numbers can be written as (repeating digits)/(10^n+10^(n-1)...+1), where n is equal to the number of repeating digits, and are therefore rational. They can be shifted to the left or right by multiplying or dividing by 10^x, where x is the number of digits you're shifting them, and since they can be added to terminating numbers to create things like 15.2446446446..., all terminating or repeating numbers are therefore rational.

This argument does not work. For example 0.3333.... is not 3/(10+1), but it is might be close to one way of doing it. Spoiler

There is a number you can divide by to get it right but your expression is wrong.

*facepalm*