0
$\begingroup$

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.

$\endgroup$
9
  • 1
    $\begingroup$ You have a typo: "$x \times 0 = x $" should read "$\times1$" $\endgroup$
    – RGS
    Jan 21, 2017 at 11:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – Marc van Leeuwen
    Jan 21, 2017 at 11:24
  • 2
    $\begingroup$ Possible duplicate of Reasons behind the field axioms : $1 \ne 0$ and $1/0$ not defined $\endgroup$
    – Asaf Karagila ♦
    Jan 21, 2017 at 11:26
  • $\begingroup$ Also math.stackexchange.com/questions/427078/is-0-a-field $\endgroup$
    – Asaf Karagila ♦
    Jan 21, 2017 at 11:27
  • 1
    $\begingroup$ @AsafKaragila OP never mentions fields, just real numbers. $\endgroup$
    – Marc van Leeuwen
    Jan 21, 2017 at 11:30

1 Answer 1

Reset to default
1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ So if we don't specify $0\neq 1$ as an axiom, then $\mathbb{R}$ is not uniquely defined? $\endgroup$
    – mz71
    Jan 21, 2017 at 12:37
  • $\begingroup$ @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. $\endgroup$
    – Arthur
    Jan 21, 2017 at 13:11
  • $\begingroup$ @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" $\endgroup$
    – Hagen von Eitzen
    Jan 21, 2017 at 15:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .