0

I am analysing axiomatic approach to defining real numbers. There are two axioms that postulate existence of $0$ and $1$, namely (according to my notes):

There exists an element $0\in\mathbb{R}$ such that for any $x\in\mathbb{R}$ we have $x+0=x.$

There exists an element $1\in\mathbb{R}$ such that for any $x\in\mathbb{R}$ we have $x\times 1=x$.

Then, I see a really little remark, almost non-existent, that says

We suppose $1\neq 0$.

I searched other sources which provided basically the same axioms but nowhere I could see the sentence $1\neq 0$ as an axiom.

My question is - why isn't it considered as an separate axiom? What it is then? I don't think you can prove it from other axioms of $\mathbb{R}$ and I think it is fairly important for the whole theory to work properly.

Marc van Leeuwen
  • 115,048
mz71
  • 908
  • 1
    You have a typo: "$x \times 0 = x $" should read "$\times1$" – RGS Jan 21 '17 at 11:23
  • 1
    It really makes no sense to talk about (the possible redundancy of) an individual axiom if you don't mention the whole collection of axioms. The condition $0\neq1$ is one of the axioms of fields, but you need more than being a field to be the real numbers. The additional axioms (assuming the real numbers are axiomatised as some particular kind of field) might make $0\neq1$ redundant or not; one cannot tell without knowing what those axioms are. – Marc van Leeuwen Jan 21 '17 at 11:24
  • 2
    Possible duplicate of Reasons behind the field axioms : $1 \ne 0$ and $1/0$ not defined – Asaf Karagila Jan 21 '17 at 11:26
  • Also http://math.stackexchange.com/questions/427078/is-0-a-field – Asaf Karagila Jan 21 '17 at 11:27
  • 1
    @AsafKaragila OP never mentions fields, just real numbers. – Marc van Leeuwen Jan 21 '17 at 11:30
  • @Marc: By talking about additive and multiplicative neutral, the OP talks about fields. – Asaf Karagila Jan 21 '17 at 12:19
  • In linked question, Rudin's book clearly states $0\neq 1$ as part of the (M5) axiom, that approach solves my question. I know now that the question essentially comes down to field axioms. But still, for example here: link $0\neq 1$ isn't taken as an axiom. It should, right? – mz71 Jan 21 '17 at 13:02
  • Yes mathworld is just wrong. I guess they couldn't decide whether $0\neq1$ is an additive or a multiplicative condition. They cheated for distributivity too. – Marc van Leeuwen Jan 21 '17 at 13:14
  • The case of real numbers is an ordered field, so the inequality of $0\neq 1$ is considerably more rooted in the properties of real numbers than it is for fields generally. If $0$ were equal to $1$ all elements of a ring would be equal, which is certainly not the case in the real numbers. Using the ordered property of real numbers, which is important to defining which real numbers are positive, we have $0 \lt 1$. – hardmath Jan 22 '17 at 19:28

1 Answers1

1

Yes, it should be considered an axiom (unless you are able to show it from the other axioms, for we definitely want $1\ne 0$ to hold).

In fact, you can easily present a model of all axioms[1] (except this), namely the set $\Bbb R=\{0\}$ (where $1$ is just another name for $0$) and the obvious operations.

[1] depends on your concrete list of axioms, of course

Hagen von Eitzen
  • 374,180
  • So if we don't specify $0\neq 1$ as an axiom, then $\mathbb{R}$ is not uniquely defined? – mz71 Jan 21 '17 at 12:37
  • @mzg147 As the other comments already say, that depends greatly on what other axioms you have. Some collections of axioms defining $\Bbb R$ includes $1\neq0$, some don't. But if it's not an axiom, then it's provable from the axioms. – Arthur Jan 21 '17 at 13:11
  • @mzg147 If your text says "We suppose that $1\ne 0$" then it sounds to me as if they do not intend to prove $1\ne0$ from the (other) axioms, and once we have such a suspicion, it is unlikely that it is possible to prove it from those. But to be exact, I can only be the umpteenth person to state that it depends on the specific list of given axioms. For example, if one of your axioms reads "$x^2+1=0$ has no solutions", then $1\ne 0$ follows. Or if one axiom reads "$\Bbb R$ has at least two elements" – Hagen von Eitzen Jan 21 '17 at 15:03