Transcendental a infinitely close to rationals?

Question

Apologies that this question is rather vague, but I do not know how to state it more precisely. Is, say pi, infinitely "close" to some rational number? More importantly, are all transcendental numbers infinitely close to some rational number? Again, I apologise for my vagueness.

Edit: is the answer the same if we replace the transcendental with the irrationals?

Answer 1

Since $\pi$ is irrational, every rational number $q \in \mathbb{Q}$ is not equal to $\pi$ (this is a bit tautological). Therefore the distance $|\pi - q|$ is strictly greater than zero, so $\pi$ is not "infinitely close" to any particular rational number.

On the other hand, the set $\mathbb{Q}$ of rational numbers is dense in the set of real number, so for any distance $\epsilon > 0$ you can find a rational $q$ such that its distance to $\pi$ is less than $\epsilon$. In other words, $|\pi - q| < \epsilon$. So in this sense, $\pi$ (and any real number, really) is "infinitely close" to the set of rational numbers.

A similar example: the number $0$ is not "infinitely close" to any particular number in the interval $(0,1]$, but it is infinitely close to the interval itself.

Hopefully this will clear up your confusion.

Answer 2

Consider specifically the following sequence of rational numbers:

$$3\\3.1\\3.14\\3.141\\3.1415\\ \vdots$$

These are all rational (for example, the third one is $\frac{314}{100}$), and approximate $\pi$ as closely as desired.

Answer 3

You can't find rationals infinitely close to $\pi$, but you can find hyperrationals infinitely close to $\pi$. In other words, an affirmative answer to your question requires an extension of the number system from the reals to the hyperreals: $\mathbb R\subset\mathbb R^\ast$. To see that one can indeed find a hyperrational infinitely close to $\pi$, consider the extended decimal expansion of $\pi\in\mathbb R^\ast$: $$\pi = 3.141\cdots a_{n-1}a_n a_{n+1}\cdots a_{H-1}a_H a_{H+1}\cdots$$ where $a_n$ is a typical digit at a finite rank $n$ whereas $a_H$ is a typical digit at an infinite (hyperfinite) rank $H$. Then truncating the decimal expansion at rank $H$ one obtains a hyperrational $$3.141\cdots{}\cdots a_H 000\cdots$$ infinitely close to $\pi$. The reason this is a hyperrational is because it is the ratio of two hypernatural numbers, namely $3141\cdots a_H$ and $10^H$.

Of course, the same goes for any irrational number. In fact, this approach allows one to build the real number system starting with hyperrationals; see

Davis, Martin Applied nonstandard analysis. Pure and Applied Mathematics. Wiley-Interscience [John Wiley & Sons], New York-London-Sydney, 1977.