- 14 Oct '11 23:47

One way is to first show the following: every rational number has a terminating or eventually repeating decimal expansion.** Then your claim in question (that a decimal expansion that neither repeats nor terminates is irrational) follows immediately.*Originally posted by tomtom232***how do you prove that a decimal that neither repeats nor terminates is irrational?**

For example:

0.1411411141114 . . . ,

prove this number is irrational.

**This one should not be too hard to show. Here would be the basic idea and it could be made more rigorous:

http://mathlesstraveled.com/2008/09/07/rational-numbers-and-decimal-expansions/ - 15 Oct '11 07:33 / 1 editIf you divide m by n, the fraction left over is some value 0, 1, 2, ... , n-1. If at any point you get a zero, the division terminates there, e.g. dividing 1 by 4 gives a zero leaves 1; then 2 leaves 2, then 5 leaves zero, so 1 / 4 = 0.25 . If you don't get a zero, the division bounces around in values {1, .. , n-1}, until eventually it steps somewhere it has before. If the a'th fraction is the same as b'th, then a+1'th is the same as b+1'th etc, so dividing by n means that either the sequence terminates or then it repeats a cycle that is at most n-1 in length. Actually, if n is a compound number, the maximum length is probably shorter than n-1.. as a hunch, p-1 where p is the biggest prime by which n is divisible and for which lcd(10,p) = 1.

The other half of the proof is to show that all fractions that do terminate or which do have a sequence are rational. If the fraction f terminates and there are d decimals then it can be expressed as f / 1 = (f * 10^d) / 10^d which is an integer over integer, that is, a rational number. If there is a sequence of length l, then

f = f * (10^l - 1) / (10^l -1)

expresses f in a way that terminates, and

f = (f * 10^d * (10^l -1)) / (10^d * (10^l -1))

expresses f as a fraction of two integers - that is, a rational number.

Thus, any repeating or terminating sequence of decimals is a rational number, and every rational number leads to a terminating or repeating sequence of decimals, so non-repeating sequences have to be irrational. - 03 Dec '11 23:28

I believe it*Originally posted by iamatiger***That looks very hard! Are you sure it can be done?***has*been done, but it takes some serious mathematics. You don't just have to prove that e and pi are irrational - which are hard enough, but doable - but also that they're a different kind of irrational. Which is, well, obviously true, but non-obvious to prove.

Richard