How do we know that the number $1$ is not equal to the number $-1$? How do we know that the number $1$ is not equal to the number $-1$? (I am not talking about the multiplicative inverse of an arbitrary field, but the integer/rational, real or complex number $1$.)
Is that an axiom?
Since someone in the comments asked for more background/motivation: I'm just a naive bachelor student and I wanted to prove that for any number $x$, $x=-x$ implies $x=0$ (which is true if and only if $1\neq -1$, see my last question).
Why is this question different to the question in my last post?
In my last post, I wanted to know if you can prove that $x=-x$ implies $x=0$ for an element $x$ of an arbitary field. It turned out that you can not (there are fields where this is not the case). In addition, it turned out that this is true if and only if $1\neq -1$. ($1$ stands for the multiplicative identity, not the number one.)
Since we normally assume that $x=-x$ implies $x=0$ if $x$ for some number $x$ (at least in my lectures), I wanted to know wheter you can prove that $1\neq -1$ or not. (To be honest, I didn't think that you have to distinguish e.g. between the integer $1$ and the natural number $1$...)
To put it into a nutshell: My first question was wheter you can prove $1\neq-1$ where $1$ stands for the multiplicative identity of an arbitrary field (the answer is no.) But we know that the answer is yes if $1$ stands for the integer/real number/complex number one, which led to this post.
That depends on the context, e.g. $1=-1$ in $\mathbb{Z}/2\mathbb{Z}$. Mathematics is a very strict discipline, you should always be precise about the meaning of symbols.
However judging by your tags you are asking about integers. To answer that question we first need to know what those symbols are and what integers are. One way to construct the set of integers $\mathbb{Z}$ is via $(\mathbb{N}\times\mathbb{N})/\sim$ where $(a,b)\sim (a',b')$ iff $a+b'=a'+b$. In that situation our $1$ is formally $[1,0]_\sim$ while $-1$ is $[0,1]_\sim$. For them to be equal we would have to answer the question whether $1+1=0+0$ over naturals?
Now Peano arithmetic jumps in. First of all $1$ is defined as the successor of zero: $1=S(0)$ by definition. The successor function $S$ and the zero $0$ are primitives in the Peano axioms. The addition is then defined recursively by $a+0=a$ and $a+S(b)=S(a+b)$.
Thus $0+0=0$. On the other hand $1+1=0$ has no chance of happening because
$$1+1=S(1+0)=S(1)$$
by definition, while $0$ is not a successor of anything by a Peano axiom.
Note that similar reasoning can be applied to any non-zero integer $x\in\mathbb{Z}$ since either $x$ or $-x$ can be represented by $[n,0]_\sim$ for some non-zero $n\in\mathbb{N}$. Non-zero is essential because the only natural that is not a successor is zero.
In any ring we have $1=-1\iff 1+1=0$ (just add/subtract $1$ from both sides).
Though, in $\Bbb Q$ [or $\Bbb R$ or $\Bbb C$] we have $1+1=2\ne 0$.