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

reek

New member
Jul 21, 2012
1
We can build 1/33 like this, .0303... (03 repeats). .0303.... 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?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If a decimal never repeats and never terminates, then we cannot express it as the ratio of two integers, and so it is called irrational.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
If a decimal never repeats and never terminates, then we cannot express it as the ratio of two integers, and so it is called irrational.
On the other hand, an irrational number can be approximated to an arbitrary number of digits by a rational number. For instance, given any irrational number $\pi$, we can trivially approximate it to $n$ decimals as:

$$\overset{\approx}{\pi_n} = \frac{\lfloor 10^n \pi \rceil}{10^n}$$

However, the numerator and denominator will endlessly grow as $n \to \infty$.

In fact, assuming the digits of $\pi$ are randomly distributed, then the numerator is uniform in $0 \leq \lfloor 10^n \pi \rceil \leq 10^n \pi$. Note this is generally not true (I don't even think it is ever true) but it is a good enough approximation for our purposes.

Then, the probability that $\lfloor 10^n \pi \rceil$ is divisible by $10$ is basically $\frac{1}{10}$, which is subcritical, therefore it is clear that the numerator and denominator will grow exponentially with $n$, largely unsimplified.