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Since $a$ and $b$ are both integers, there is only a countable amount of numbers $a+b\pi$. Thus not every real number can be expressed as such.

But is there a way to determine if $x$ can be expressed as $a+b\pi$, like a fundamental property of $x$? What about if $a$ and $b$ are rational? Or if instead of $\pi$ you take another irrational number?

Quintium
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    Regarding the second question, nobody yet knows whether $e + \pi$ is rational :-) (click a few links there for more context) – Amateur_Algebraist Feb 16 '24 at 20:25
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    To give a "way to determine" if $x = a + b\pi$, requires you to specify precisely how $x$, $a$ and $b$ are represented. If (for example), they are represented as computable reals, then there is no way to determine any equality. – Rob Arthan Feb 16 '24 at 20:32
  • @RobArthan I meant if given a real number $x$, how can one determine if it's expressible as $a + b\pi$? – Quintium Feb 16 '24 at 20:37
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    But what do you mean by "given"? How is $x$ presented to you? – Rob Arthan Feb 16 '24 at 20:38
  • @RobArthan As a variable, without any specific representation. I thought maybe there'd be a property such as that $x/\pi$ is rational or something like that. – Quintium Feb 16 '24 at 20:43
  • I see. I don't think that there is any useful property that will help. – Rob Arthan Feb 16 '24 at 20:45
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    There is a 'tautological' answer: $x$ is of the form $a+b\pi$ if there are integers $a$ and $b$ such that $(x-a)/b=\pi$. This is what could be called a $\Sigma_1$ definition: it asks for the existence of a pair of integers satisfying a certain condition. (This is the same as asking for a single integer satisfying a certain condition, but we won't go into that). Note that your statement '$x/\pi$ is rational' is also a $\Sigma_1$ statement: it says 'there are integers $a$ and $b$ such that $x\times a=b\times\pi$'. – Steven Stadnicki Feb 16 '24 at 21:02
  • @StevenStadnicki: You haven't said so, but I'll add: when a tautological answer is also so simple, it's not at all clear to me what the question is. – Lee Mosher Feb 17 '24 at 14:20
  • @LeeMosher I've thought about when you asked this, but I wanted to know whether one can somehow link this problem to known categories such as rational, algebraic or transcendental numbers. – Quintium Feb 17 '24 at 18:39

2 Answers2

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The set $ \left\{ a+b\pi, (a,b)\in \mathbb{Z}^2 \right\}$ is dense in $\mathbb{R}$, meaning that given any real number, you can construct a sequence of elements of this set that converges to this number . This is due to $\pi$ being irrational, which can be proved at a college level in a guided exercise. The more general fact that $a\mathbb{Z}+b\mathbb{Z}$ is dense in $\mathbb{R}$ iff $\frac{a}{b} \notin \mathbb{Q}$ can be proved from the density of the additive subgroups of $\mathbb{R}$, usually using the greatest lower bound property.

To answer your question about the fundamental property, I guess more detail is needed on the assumptions you can make on $x$. For other irrational numbers such as square roots, there's a whole theory known as algebraic field extensions.

François Mortier
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    This isn't an answer. It's an observation (that doesn't seem very relevant) and a comment. – Rob Arthan Feb 16 '24 at 21:43
  • @RobArthan Can't say I disagree.Technically, it doesn't answer the question. However I will leave it to those more knowledgeable than I to provide a "definitive" answer. I do get why the observation might not seem relevant to you, but in my view it has the merit of providing some information about the set mentioned in the question. As a general guideline, is providing a pseudo-answer worse than no answer at all ? – François Mortier Feb 16 '24 at 22:20
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Knowledge about this sort of problem is pretty limited. The general relevant area of mathematics is transcendental number theory, which has some interesting and notable results, but for most questions one can think to ask, such as whether $e$ can be a rational linear combination of $1$ and $\pi$, we simply don't know the answer.

A set $A$ of numbers is said to be algebraically independent if for any finite subset $\{\alpha_1, \ldots, \alpha_n\}$ of distinct elements of $S$, no nonzero polynomial $P(x_1, \ldots, x_n)$ with rational coefficients satisfies $P(\alpha_1, \ldots, \alpha_n) = 0$. For example, $\{\pi\}$ is an algebraically independent set (equivalently, $\pi$ is a transcendental number): there is no combination of rational numbers $a_0, \ldots, a_n$ such that

$$a_n \pi^n + a_{n-1}\pi^{n-1} + \ldots + a_0 = 0$$ except for the trivial case where all the $a_i$ are zero. This would mean for instance that no nontrivial rational combination of powers $\pi^k$ with $k \geq 2$ can ever be of the form $a + b\pi$ (here and elsewhere, $a, b$ denote rational numbers).

Relatedly, no algebraic number except $0$ can ever be of the form $a + b\pi$ for rational $a, b$.

If we have a set $S$ of algebraically independent numbers that includes $\pi$, then for any nontrivial rational polynomial $P(x_1, \ldots, x_n)$, the value of $P(\alpha_1, \ldots, \alpha_n)$ for $\alpha_i \in S$ will never be of the form $a + b\pi$ unless all the $\alpha_i$ are equal to $\pi$ (and even there, the possibilities are extremely constrained, as we just saw).

The following sets are known to be algebraically independent:

  • (Nesterenko) $\{\pi, e^\pi, \Gamma(\frac1{4})\}$, where $\Gamma$ denotes the Gamma function.
  • (Baker) $\{\pi, \log 2, \log 3, \log 5, \ldots, \log p, \ldots, \}$ where $p$ ranges over primes ($\log$ is the natural logarithm).

There is a raft of conjectures about algebraically independent sets. One very powerful conjecture is Schanuel's conjecture, and there are various conjectures surrounding the concept of period, but as I say, these are conjectural and we seem to be very far away from settling them.

Of course, all this is about numbers that can't be of the form $a + b\pi$. This is only to be expected; as you say yourself, there are only countably many such $a + b\pi$, a piddingly small subset among all real numbers.

user43208
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