Wikipedia mentioned the limitation of Gödel's theorems. According to it, Euclidean geometry doesn't satisfy the hypotheses of Gödel's incompleteness theorems, that is, it cannot define natural numbers. Why is that? Isn't the number of points, lines, or polygons natural number? I thought it would be easy to define natural numbers in Euclidean geometry.


3 Answers 3


No, the natural numbers cannot be defined in Euclidean geometry.
In what sense is the number of points, lines, or polygons a natural number? In Euclidean geometry there are infinitely many such objects.

Tarski proved that Euclidean geometry (or rather, his axiomatization of Euclidean geometry) is decidable, while Goedel showed that no reasonable theory of natural numbers is decidable. This alone shows that the natural numbers cannot be defined in geometric terms.

It might help to say something more about what definability means in this situation. Euclidean geometry is in some precise sense just the theory of real closed fields, that is, the theory of the real numbers (as an ordered field).
And by theory I mean the set of all first order statements in the language of ordered fields that are true in $\mathbb R$. Now, you might think that the natural numbers can be defined in $\mathbb R$ by saying that the natural numbers are all numbers that can be obtained by iteratively adding 1's. Or you might say that the natural numbers are the smallest subset of $\mathbb R$ that contains $1$ and is closed under addition.

The problem with these two "definitions" is that none of them can be expressed in first order logic, where you can only quantify over elements of the structure (in this case $\mathbb R$) and not over subsets.

I hope this clarifies things a bit.

  • $\begingroup$ Is this a first-order definition of natural numbers in real numbers? $\text{Natural}(a) \leftrightarrow (a=0)\lor(\text{Natural(a-1)})$ $\endgroup$
    – xzhu
    Dec 15, 2011 at 5:43
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    $\begingroup$ No, it is not. In fact, it is not a definition at all. If every real number $x$ satisfies $\text{Natural}(x)$, then $\text{Natural}$ satisfies this equivalence. I guess, what you want to say is this: Let Natural be the smallest set that contains 0 and is closed under taking successors, i.e., the smallest set that satisfies your equivalence. This is not first order since you need quantification over all sets of numbers. $\endgroup$ Dec 16, 2011 at 15:43

Gödel's theorems apply to formal systems. Euclidean geometry in itself is not a formal system. So you have to look to particular formalizations of Euclidean geometry. I suppose there may be many of them. Some will be strong enough that Gödel's theorems apply. But apparently others are not.

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    $\begingroup$ I don't think it's so much an issue of one formalization versus another as an issue of first-order logic versus higher-order logic. $\endgroup$
    – user13618
    Jan 12, 2012 at 20:56
  • $\begingroup$ @BenCrowell Different formalizations are sometimes created explicitly with first-order logic in mind. Tarski's was, for example. That's why it is so crippled it cannot talk about sets of points or define a polygon. No geometer would limit themselves to such a crippled system, just like no number theorist would limit themselves to Pressburger arithmetic, a formalization of arithmetic explicitly designed to be decidable, at the cost of being so crippled it cannot even define divisibility. $\endgroup$
    – Matt
    Nov 17, 2020 at 17:34

To have a Gödel theorem be provable for a system the system must have enough structure to be able to describe a statement that refers to itself as an unprovable (Gödel) statement.

Arithmetic with multiplication, addition, and first order logic is rich enough; Presburger arithmetic (no multiplication, but multiplication can be simulated by additions) and Euclidean Geometry are not rich enough.

Roughly, the two Gödel theorems are (1) Sufficiently rich and consistent systems cannot be complete and (2) The consistency of such systems cannot be proved within the system.

Interestingly, if the Gödel statement were false it could be proved and so must be true; therefore, since the statement says it is unprovable it must be unprovable; and adding it as a theorem does get around the theorems because then another Gödel statement can be found.


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