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Is it possible to represent $e=2.71\ldots$ and $\pi=3.14\ldots$ without using the infinity sign nor using infinitely-many numbers, and algebraically, like representing the golden ratio, which is $1:\dfrac{\sqrt{5}+1}{2}$?

Blue
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Math Lover
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    Yes, use the symbols “$e$” and “$\pi$” ;) – MPW Jan 04 '19 at 23:23
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    @Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals. – Cheerful Parsnip Jan 04 '19 at 23:25
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    Not really, since $\pi$ and $e$ are transcendental https://en.wikipedia.org/wiki/Transcendental_number – SmileyCraft Jan 04 '19 at 23:25
  • So can you prove that pi and e are transcendental? – Math Lover Jan 04 '19 at 23:28
  • Within the comments section here? Not efficiently. You can however read other questions on this site on the topic. – JMoravitz Jan 04 '19 at 23:29
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    This article proves both $e$ and $\pi$ are transcendental: http://sixthform.info/maths/files/pitrans.pdf – Noble Mushtak Jan 04 '19 at 23:29
  • $\mbox{Arccos}(-1)$ – José Alejandro Aburto Araneda Jan 04 '19 at 23:32
  • @José I think the answer can also be 3pi, or something like that. – Math Lover Jan 04 '19 at 23:35
  • Within the real numbers $e$ and $\pi$ are definite numbers, clearly defined and unambiguously so. They can be approximated arbitrarily accurately by rationals, decimals or algebraic numbers, but have been proven not to belong to any of these classes of numbers. – Mark Bennet Jan 04 '19 at 23:41
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    @MathLover in your most recent comment, referring to $\arccos(-1)$ as possibly also being $3\pi$, not only $\pi$, that is incorrect. The way $\arccos$ is defined is as a function and functions only have one output at a time, not several. $\arccos(-1)$ is equal to $\pi$ and only $\pi$. – JMoravitz Jan 04 '19 at 23:41
  • $\mbox{Arccos}$ is a function, choosing a domain and codimain – José Alejandro Aburto Araneda Jan 04 '19 at 23:43
  • I don't think $\pi$ and $e$ being transcendental prevents the asked for representation. Algebraic numbers are much more limited than the restrictions posed in this question. For example, $(-1)^{-i}=e^\pi$ is transcendental. – Todor Markov Jan 04 '19 at 23:56
  • $\pi$ can be represented in 54 symbols "the ratio of a circle's circumference to its diameter". And it can be represented in 1 symbol by "$\pi$". And although $\cos (3\pi) = -1$, the definition of $\arccos(n)$ is the unique number between $0$ and $\pi$ inclusively. And$\pi$ is the only answer. I think you need to think about what exactly you are asking and what exactly you mean be "representing". If you mean as a digital decimal the answer is "no". But if we allow symbols such as $\sqrt{}$ then, we can invent symbols for trig concepts as well. in the end "$\pi$" ITSELF is a "representation". – fleablood Jan 05 '19 at 00:01
  • $\arccos(-1)$ and $\cos(3\pi) = -1$ is mostly an irrelevant red herring. $\arccos$ is not an algebraic expression. Now I still maintain there is nothing magical or "better" about writing numbers algebraically but the simple fact is $\pi$ and $e$ aren't algebraic. Which is hardly surprising. Most numbers aren't. – fleablood Jan 05 '19 at 00:14

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Yes. You can represent $e$ in one symbol "$e$" and you can represent $\pi$ in one symbol by using the single symbol "$\pi$".

However you can't express them as integers because.... they are not integers. You can't express $\frac 35$ as an integer either.

You can't express them as a terminating decimal because they are not an integer scaled to a power of $10$.

Nor can you express them as a ratio between two finite integers because the are not rational.

Only difference between them and numbers such as $\sqrt 2$ or the golden ratio which are also not rational, is that they are not roots to any polynomial with integer coefficients. $\sqrt 2$ is the one of the roots to $x^2 -2 = 0$ and the golden ratio is solution to $x^2 - 2x -4= 0$. Neither $\pi$ nor $e$ are a solution to any polynomials with integer coefficients.

Numbers that are solutions to polynomials are called algebraic numbers and the irrational algebraic numbers can't be expressed as finite decimals. But if we invent the notation, and we are mathematicians--- we can invent anything we like as like as it makes sense---, $\sqrt[k]{m}$ to mean a $k$-th root of $m$, then we can express any algebraic number as some combination of roots.

Numbers, such as $e$ and $\pi$ which are not algebraic are called transcendental. And, no, they can not be expressed via a digital decimal system. Nor can they be expressed as a series of root signs.

But so what? Using digital decimals to represent numbers is completely arbitrary and does not in any way make a number more "real" than any other. (The main reason we use digital decimals is because the are convergent and thus every thing can be approximate and accurately expressed through and infinite number of decimals if necessary.)

But to "represent" a number means nothing more than having a symbol for it. And if a number can by defined --- in a way that makes sense so that we know such a number exists--- we can define any symbol we want for them.

So I stand by my answer. You can represent $e$ and $\pi$ with one symbol each: "$e$" and "$\pi$"

fleablood
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To summarize (some) of the comments:

The numbers that you can write algebraically, like the golden ratio--using any finite sequence of operations consisting of addition, subtraction, multiplication, division, square roots, cube roots, and/or higher-order roots applied to integers--is a subset of the algebraic numbers.

Transcendental numbers are real numbers that are not algebraic numbers. It can be shown that $\pi$ and $e$ are transcendental; for example, see How hard is the proof of $\pi$ or $e$ being transcendental?.

Hence neither $\pi$ nor $e$ can be written algebraically. So the answer is "no."

David K
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  • +1 for bringing up the concept of "written algebraically". However maybe we should define what "written algebraically" and explain to the OP that this is actually what s/he was asking. After all. We can always represent $\pi$ and $e$ as.... $\pi$ and $e$. – fleablood Jan 05 '19 at 00:04
  • Never mind. I see the OP has editted the post to mean "written algebraically" – fleablood Jan 05 '19 at 00:07