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Suppose you start with the number 1 (and 0 because why not) and the set is closed so that you may add, subtract (smaller from bigger), multiply, and divide (except 0) any elements of the set. It is easy to see all non-negative rational numbers are elements of this set. If we now also allowed for Euclidean distance as well, what would this set look like? That is, $a,b\in S \Longrightarrow \sqrt{a^2+b^2}\in S.$ For starters, all square roots of rational numbers will be in the set (simple to prove for integers, and the rationals follow by closure under division).

However, I wonder if I can take the square root of any element of the set. For example, are the higher (even) roots of rationals included? What about the root of one plus some other root? Basically, does $x \in S \Longrightarrow \sqrt{x} \in S$?

It has proven difficult to give this set an explicit representation. That's why my attempts to solve equations or find a counter-example have ended without success so far. If anyone could help, I would gladly appreciate it!

EDIT: While this apparently is a Pythagorean field, the reverse does not seem to be true. For example, $\sqrt{\frac{1}{2}}$ does not have a representation under Pythagorean fields at all. So, the question whether Pythagorean fields are Euclidean Fields seems easily answered by the counterexample $\sqrt{\sqrt{2}+1}$. However, this does not seem applicable for the case $a\sqrt{1+b^2}$ which it seems I must pay attention to.

Lyde
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  • I think this is what is called the (positive) constructible numbers. – Jeppe Stig Nielsen Feb 20 '22 at 20:21
  • What does "closure under Euclidean distance" mean? Do you mean that $(a,b),(c,d)\in S\times S\implies \sqrt {(a-c)^2+(b-d)^2}\in S$ ? Something else? – lulu Feb 20 '22 at 20:28
  • What I mean is this: $a,b \in S \Longrightarrow \sqrt{a^2+b^2}\in S$. – Lyde Feb 20 '22 at 20:39
  • How is that a Euclidean distance? The distance from $a$ to itself would be $\sqrt 2 ,|a|$, right? In any case, I suggest editing your post to reflect your definition. – lulu Feb 20 '22 at 20:42
  • In any case, the standard construction of a square root can be found, e.g., here. It's a short exercise in the Pythagorean Theorem to confirm that it works (and another exercise to confirm that this is compatible with your, non-standard, definition). – lulu Feb 20 '22 at 20:43
  • I appreciate it, but the problem is exactly that I have a restriction in place, not allowing for circles. In a geometry where constructed line segments may be translated and divided into $n$ parts, all of the above can be constructed. I am now trying to prove that $\sqrt{x}$ is not always an element of this set, since I have not been able to find a construction. – Lyde Feb 20 '22 at 20:50
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    the most detail on such fields is in Hartshorne, Geometry: Euclid and Beyond. – Will Jagy Feb 21 '22 at 01:32
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    Lyde, your question is a "Project 2" in page 158, 4th edition, Euclidean and Non-Euclidean Geometries by Marvin Greenberg. My work is mostly pages 520-528, then part of it is Project 4 on page 538. As I mentioned above, there is considerable detail on ordered fields in Hartshorne, Geometry: Euclid and Beyond. My impression is that Pythagorean fields are the best fit for Hilbert's axioms. – Will Jagy Feb 21 '22 at 02:04
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    See https://math.stackexchange.com/q/116716/72031 – Paramanand Singh Feb 22 '22 at 16:42

2 Answers2

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An ordered field $P$ such that, given any element $x \in P,$ we also have $ \sqrt{1+x^2} \in P.$ Such a field is called Pythagorean. This is your example: given field elements $a,b$ with nonzero $a >0,$ we see that $\frac{b}{a} \in P, \; \; $ and $ \sqrt{1 + \frac{b^2}{a^2} } \in P . \; \; $ Then $$ \sqrt{a^2+b^2} \; = \; \; \; \; a \; \sqrt{1 + \frac{b^2}{a^2} } \; \in \; P \; \; .$$

Will Jagy
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  • Thanks a lot for your answer! I will check out the sources you recommended. I don't think this suffices as an explicit representation of S, as for example $\sqrt{1+x^2}\sqrt{1+y^2}=\sqrt{1+x^2+y^2+x^2y^2}$ must not necessarily have a representation of the form $\sqrt{1+z^2}$, right? – Lyde Feb 21 '22 at 09:19
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    @Lyde agreed. However, it is $$ (1+xy) \sqrt{ 1 + \left( \frac{x-y}{1+xy} \right)^2 } $$ if we demand $xy > -1 ; ; ; $ – Will Jagy Feb 21 '22 at 16:14
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After quite a bit of research, this is difficult to prove properly! It relies on finding some kind of invariant that stays the same regardless which of these operations is applied. Here are some ideas from Auckly, D. (1995). Totally Real Origami and Impossible Paper Folding. The American Mathematical Monthly.

  • Algebraic numbers $\alpha$ are the zeros of polynomials with rational coefficients. Moreover, there is a unique minimal (irreducible) polynomial in $\mathbb{Q}[x]$ for which $\alpha$ is a zero. The other roots of this polynomial are the conjugates of $\alpha$. We call algebraic numbers totally real iff all its conjugates are real.
  • To show: The sum, difference, product, quotient, Euclidean distance of totally real numbers is totally real.
  • This is done using symmetric polynomials. They remain unchanged when swapping terms (such as $x^2+y^2$ but not $x^2-y^2$). Also, every symmetric polynomial is a linear combination of products of three so-called elementary symmetric polynomials: $s=\sum_{i=1}^{n}x_i$, $p=\prod_{i=1}^{n}x_i$ and $s'$, the sum of all products of $k$ distinct $x_i$.
  • We eventually find $\sqrt{x}\in S \Longrightarrow \sqrt{x}$ is totally real.
  • Then not all conjugates, for example, of $\sqrt{1+\sqrt{2}}$ are real. Hence, $\sqrt{1+\sqrt{2}} \not\in S$.

The quoted source has the most detailed proof I could find online. Thanks again for everyone's help!

Lyde
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