Irrational numbers and repeating patterns

[Nick89]

Hi,

I want to show that an irrational number (let's say pi) can never have an (infinitely) repeating pattern (such as 0.12347 12347 12347 ...).

Is it possible to 'proof' (or just make it more acceptable, I don't need a 100% rigorous proof) this easily, without using too much complicated math?

Obviously the fact that no pattern has been found yet is not a convincing argument; the digits we know currently might not even have ended their first 'repeating string'.

Note; I know and accept that an irrational number cannot have an infinitely repeating string of digits (although I've never seen the proof). I just want to show (to someone else, who does not know a lot of higher math) that it is, since he won't accept the 'because it has been proven' argument.

I have read the wikipedia article on irrational numbers, and they try to explain it using long division, where a rational number n/m can be shown to always have either repeating remainders, or a zero remainder. But I don't fully understand it and I can't explain it in more accessible terms to my friend...

Thanks!

 

[yyat]

The rational numbers are exactly those with repeating decimal expansion:

Given a decimal number with repeating digits, there is a very easy way to find a representation as a rational number (see also http://en.wikipedia.org/wiki/Repeating_decimal#Fraction_from_repeating_decimal"): For example, 1.78121212... is 178/100+12/9900.

In the other direction, quoting http://en.wikipedia.org/wiki/Repeat...a_repeating_or_terminating_decimal_expansion":

...For example, consider the rational number 5⁄74...

Only finitely many different remainders — in the example above, 74 possible remainders: 0, 1, 2, ..., 73 — can occur. If the remainder is 0, then the expansion terminates. If 0 never occurs as a remainder, then only finitely many other possible remainders exist — in the example above they are 1, 2, ,3, ..., 73. Therefore eventually a remainder must occur that has occurred before. The same remainder implies the same new digit in the result and the same new remainder. Therefore the whole sequence repeats itself.

To show that $$\pi$$ is irrational requires calculus (makes sense, since it is already used in the definition of $$\pi$$), see for example http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational"

 

[Georgepowell]

To prove it, all you would need to do:

Show that any repeating decimal can be written in the form a/b where a and b are integers.

a.bcdefgrrrrrrrr

where a.bcd... is the beginning (non-repeating) part of the fraction, and r is the repeating part of the fraction.

it can always be written as a.bcdefg00000000 (obviously rational) + 10^-n*r/K (where K is something like 99999).

eg.

1.829347 91819181918191819181918191...

would be equal to:

1.8293470000+10^-6*9181/9999 (this can be simplified into a single fraction)

 

[Nick89]

Ok, so by proving that a repeated decimal corresponds to a rational number, I can conclude that a non-repeating decimal corresponds to a irrational number, and thus that an irrational number can never have a repeating decimal. True?

 

[CRGreathouse]

Ok, so by proving that a repeated decimal corresponds to a rational number, I can conclude that a non-repeating decimal corresponds to a irrational number, and thus that an irrational number can never have a repeating decimal. True?

1. A repeated decimal corresponds to a rational number
2. A non-repeating decimal corresponds to a irrational number
3. An irrational number can never have a repeating decimal

You said: From (1), I can conclude that (2), and thus (3).
Should be: From (1) I conclude (3).

(2) is correct but is not proved by (1) and is not needed to prove (3).

 

[Nick89]

Thanks, that makes sense.