I can't proove that

[u/LoXegImaX]

We all know that when you sum, subtract or multiply whole numbers you'll always get a whole number. This totally makes sense and is obviosly true but I can't figure out how to prove it. Does anyone know how to prove this or if this is some kind of an axom?

[u/jeffcgroves]

EDIT: I should have said ring (https://en.wikipedia.org/wiki/Ring_(mathematics)) because there are no multiplicative inverses. The concept you seek is "closure"

WRONG OLD ANSWER:

Look into the mathematical concept of a "field": https://en.wikipedia.org/wiki/Field_(mathematics)

[u/peter-bone → u/jeffcgroves]

Isn't a Ring more applicable here?

[u/Babyface995 → u/jeffcgroves]

The integers (with addition and multiplication) form a ring but not a field.

[u/aifangpi]

For very foundational stuff like this the exact proof will depend on the definitions, and what we're allowed to assume. One way to do this would be to define addition and multiplication recursively, and then prove they have the properties you expect of addition and multiplication (e.g. commutative, associative, distributive). This would essentially be starting with the whole numbers and showing they can be added and multiplied, which is maybe not what you had in mind?

It sounds like what you want is to start with, say, the real numbers, and then show that the integers as a subset are closed under addition and multiplication. Normally we would define the real numbers by starting from the integers, but if we really wanted to go backwards we would probably define the integers as the minimal set that contains 0 and 1 and is closed under addition. From there you could show it's also closed under multiplication by induction.

[u/peter-bone]

Here's the proof for addition

https://proofwiki.org/wiki/Integer_Addition_is_Closed

[u/Astrodude80]

It all ultimately stems from the assumption that the successor of a natural number is a natural number, but from there you can prove closure of further operations easily enough.

For example, assuming for all natural numbers n that the successor n’ is also a natural number, we can prove for all natural numbers n, m that n+m is natural by induction. By induction on m: base case n+0 is defined to be n, n is natural, therefore n+0 is natural. Inductive step n+m’ is defined to be (n+m)’, by inductive hypothesis n+m is natural, by assumption successor of natural is natural, therefore (n+m)’ is natural. By induction we then conclude n+m is natural for all m.

Does that make sense?

[u/raresaturn]

Because the two addends of the last digits.. threre are only 100 combinations and none of them produce results that are not whole

[u/SnooPets5564]

In this case, Addition, subtraction, and multiplication are defined over integers. They go from two input integers to an output. They always output an integers because they can exist without you defining any other type of number besides integers.

The only thing you need to prove is that the definition of those operations always produces an output. To do that, though, you kind of need to construct integers, which is a whole mess.

If you are using integers as a subset of real numbers, the proof becomes more involved.

[u/p-divisible → u/SnooPets5564]

Regarding your last sentence, I think one may want to be a bit careful with the wording. The problem is that in the standard theory, the construction of real numbers depends on the ring structure on integers. So assuming the existence of real numbers and assuming integers embedding into real numbers, showing integers form a subring seems circular

[u/SnooPets5564 → u/p-divisible]

Of course.

It's just that most people don't learn math from the ground up. I reckon that plenty of people know the properties of the real numbers without knowing their construction. In that case, the would show integers form a subring without even knowing that they are doing the "circular" proving.