Are integers real? - MathOverflow [closed]

Are integers real?

2013-01-28T08:43:07Z

Do you think that $\mathbb Z \subset \mathbb R$? On one hand this inclusion is quite handy. We like to write things like: $$ \sqrt{n} \quad \text{for $n\in \mathbb Z$} $$ which requires the number $n$ to be a real number (where $\sqrt\cdot$ is defined). On the other hand it is difficult to obtain such an inclusion when comes to definition. One would like to be able to define whole numbers without the need to define real numbers. This becomes more tricky when one notices that there are other inclusions which one would like to satisfy. For example I would like $1$ to be a polynomial with whole coefficients, or maybe a polynomial with complex coefficients, or maybe a real function of one variable...

I would say that it is not possible to satisfy all these inclusions. So maybe we must not insist on saying that $\mathbb Z \subset \mathbb R$ in the first place. Are there alternatives?

One possibility I see is that of having many different sets isomorphic to $\mathbb Z$. We should use the name $\mathbb Z$ for integers as we use $V$ for vector spaces. We should say: let $\mathbb Z$ be any set of integers. Or: let $\mathbb R$ be a set of reals and let$\mathbb Z$ be the set of real integers $\mathbb Z \subset \mathbb R$. And so on...

Another possibility I see (but I'm not sure if it can be really founded) is that of redefining the meaning of equality $=$ and distinguish between equality and identity. We could try to take an "object-oriented" approach where equality could be defined like any other operation. So one could define $1 = 1/1$ i.e. the integer $1$ is the same as the rational $1/1$ and one should define the sum of integers and rational by converting the integer to a rational and then performing the sum between rationals. This also modifies the concept of 'set' since the set ${1, 1/1 }$ is equal to ${1/1}$ and hence has a single element. This is, more or less, how types work in computer languages. Can this approach be made rigorous?

Answer by Gian Maria Dall'Ara for Are integers real?

2013-01-28T09:26:25Z

Expanding on the first part of Daniel Spector's comment, in all the cases you mentioned there is a canonical inclusion $\mathbb{Z}\hookrightarrow X$ for a certain set $X$ (reals, polynomials with complex coefficients, etc. ). When we write $\sqrt{n}$ for $n\in\mathbb{Z}$ (or any analogous expression), you are doing a little abuse of notation, identifying $n$ with its image under the canonical embedding of $\mathbb{Z}$ into the reals (or the appropriate set $X$). This usually doesn't cause any ambiguity and that's why we do it all the time.

Answer by Spice the Bird for Are integers real?

2013-01-28T09:26:44Z

The answer to this question depends on the type of set theory that you are working in, and the way you decide to code the integers and real numbers inside set theory. For instance in material set theory, we often (it is not mandatory) define the natural numbers to be finite Von Neumann ordinals,integers as certain classes of ordered pairs of naturals the rationals as certain equivalence classes of certain ordered pairs of integers, then the reals as classes of Cauchy sequences. In this kind of set theory, in no way is the integers a subset of the reals. But their is a cannonical inclusion of the integers into the reals.

In a structural set theory what is in the set does not matter so much as how different sets relate to one another. This is inherently a categorical discussion of what set theory is. Now in category theory, a sub-object is an equivelence class of monomorphisms. In this kind of set theory the integers are a subset of the reals.

For the differences between material and structural set theory see http://ncatlab.org/nlab/show/set+theory .

A subobject is defined here: http://ncatlab.org/nlab/show/subobject

Also a nice exposition of structural set theory is here: http://golem.ph.utexas.edu/category/2012/12/rethinking_set_theory.html

Also relevant to material and structural set theory and relations to type theory are here: http://golem.ph.utexas.edu/category/2013/01/from_set_theory_to_type_theory.html

Answer by arsmath for Are integers real?

2013-01-28T09:34:09Z

We do in fact have many different sets isomophic to $\mathbb{Z}$. The correct definition of $\mathbb{Z}$ is a set with some operations on it that satisfy some axioms. This is easier to see with $\mathbb{N}$, which is a set with a 0, a "plus one" operations, that satisfies mathematical induction. There is only one such set up to oeration-preserving isomorphism.

$\mathbb{R}$ can be axiomatized similarly, as a complete Archimedean ordered field (see the synthetic approach section at Wikipedia). Again there is only one such set up to operation-preserving isomorphism. You can identify in that set a copy of $\mathbb{Z}$, so you can treat $\mathbb{Z}$ as a subset. But this is a matter of notation. If you had a compelling reason to make it disjoint, you could require that, instead.

We say "the integers" or "the reals" because there is only one, up to isomorphism. Any theorem you prove about $\mathbb{Z}$ or $\mathbb{R}$ that doesn't use the internal representation will transfer to any isomorphic copy, so we don't need to know which one.

There is a whole mathematical notion of types that actually underpins languages with more-complicated type systems, such as ML or Haskell. You can think of $\mathbb{Z}$ and $\mathbb{R}$ as a type in a type system. Some typing systems have a notion of "subtype", but let's suppose that you don't. Then $\mathbb{Z}$ and $\mathbb{R}$ are types, and there's a designated monomorphism $i$ from $\mathbb{Z}$ to $\mathbb{R}$. Then 2 + 3.5 is an overloaded operation that is syntactic sugar for $i(2) + 3.5$. I don't know if anyone has ever worked out a clear account of what mathematicians do from this point of view, but what they do is not very complicated.

If you do want to allow subtypes, there's a notion of order-sorted algebra that allows the kind of overloading you probably have in mind. I don't know of a canonical link, but introductions are easy to find.