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

We can build $\frac{1}{33}$ like this, $.030303$ $\cdots$ ($03$ repeats).
$.0303$ $\cdots$ tends to $\frac{1}{33}$.

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?

#### Solutions Collecting From Web of "Is a non-repeating and non-terminating decimal always an irrational?"

The decimal expansion of a rational number is always repeating (we can view a finite decimal as a repetition of $0$’s)

If $q$ is rational we may write it as an irreducible fraction $\dfrac{a}{b}$ where $a,b\in\mathbb{Z}$. Consider the Euclidean division of $a$ by $b:$

At each step, there are only finitely many possible remainders $r\;\;(0\leq r< b)$. Hence, at some point, we must hit a remainder which has previously appeared in the algorithm: the decimals cycle from there i.e. we have a repeating pattern.

Since no rational number can be non-repeating, a non-repeating decimal must be irrational.

Closer and closer? If you mean by closer and closer as in approximating an irrational, yes. The idea behind closeness is that there is some destination behind where you are going whilst writing the number, but in an irrational number there is no destination, you simply keep writing. If it would be possible, with perfect information, to keep writing out the decimal expansion of an irrational number, making sure there is absolutely no repetitiveness or patterns, then you would be getting arbitrarily close to the irrational number, (in terms of $\epsilon$ close). An irrational number has a non-terminating, non-repeating decimal expansion. So there is no real idea of closeness here, unless you are talking about distance, ($\epsilon$ close), in which case yes you are getting closer.