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

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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?
 
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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.
 
MarkFL said:
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.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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