Is a non-repeating and non-terminating decimal always an irrational?

We can build 1/33 like this, .030303..... (03 repeats). .030303.... tends to 1/33 .

So,I was wondering this: In the decimal representation, if we start writing the 10 numerals in such a way that the decimal portion never ends and never repeats; then am I getting closer and closer to some irrational number?

Re: Is a non-repeating and non-terminating decimal always an irrational?

Sorry - I'm not following what you mean by the "10 numerals" - but in general, yes, if in decimal the digit string does not enter a repeating pattern and does not terminate then the number is indeed irrational. So for example:

0.123456789898989... is rational since the "89" digits repeat, and

0.1011011101111011111... is irrational because there is no repeating pattern.

Re: Is a non-repeating and non-terminating decimal always an irrational?

One can show in general that the decimal expansion of every rational number is either a repeating decimal or terminating decimal. So, yes, if a decimal is non-repeating or non-terminating, it must be irrational.

To show the first, imagine, for the fraction, 'm/n', dividing m by n. Doing a long division, at each step, you subtract to bring down the next "remainder". That remainder must be a non-negative integer less than n. After having divided at most n times, you have used up all of those possible remainders and the next must be either 0 or one of the previous divisors. If the remainder is 0, you stop and the decimal is terminates. If it is one of the previous divisors, everything you did before will repeat and you will get a repeating decimal.

Re: Is a non-repeating and non-terminating decimal always an irrational?

you can also prove the converse of what HallsofIvy posted:

if a decimal terminates, or terminates in a repeating cycle of finite length, it is rational.

the first part is easy:

if a decimal d terminates after k digits, then 10^{k}d is an integer, whence:

d = 10^{k}d/10^{k} is rational.

the second part is a little trickier. suppose that after k digits, a decimal d terminates in a cycle of length m.

then 10^{k+m}d - 10^{k}d is an integer (the two cycles parts after the decimal point "cancel").

hence d = (10^{k+m} - 10^{k})d/(10^{k+m}-10^{k}) is rational.

to illustrate this second principle, let's look at:

$\displaystyle d = 0.041\overline{6}$

here k = 3, and m = 1. so we want to multiply d first by 10000 (= 10^{k+m} = 10^{4}), and subtract d times 1000 (= 10^{k} = 10^{3}). this gives us:

$\displaystyle 10000d - 1000d = 416.\overline{6} - 41.\overline{6} = 416 - 41 = 375$ <---this is our numerator.

our denominator is then 9000, so the rational number we obtain is 375/9000.

of course, we can "reduce this". it's obvious we have a common factor of 5, so let's take it out: 375/9000 = 75/1800.

we still have a common factor of 5, so again: 75/1800 = 15/360. and again: 15/360 = 3/72. and a last common factor of 3: 3/72 = 1/24.

so since repeating or terminating decimal <=> rational, if a number is neither terminating nor repeating, it must be irrational (even if there's a "pattern" like:

0.112123123412345123456123456712345678123456789123 456789101234567891011123456789101112........)