How do I put this into an expression "For any nonzero real number $x$, there exist $-x$ real number."? Is it "$\forall x\ne0 \exists(-x)$"?
I don't know how to put "real number $x$" into an equation
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$\begingroup$It looks like what you are trying to do is to formulate the statement that every number $x$ has an additive inverse, denoted $-x$.
A better way to do this is to separate the statement from the definition of the notation, like this:
- $\forall x \in \mathbb R \,\,\exists y \in \mathbb R$ such that $x+y=0$. Denote $-x=y$.
However, this raises a logical issue: how do we know that $-x$ is well-defined?
To answer that, one must also state (and prove) uniqueness:
- $\forall x, y, y' \in \mathbb R$, if $x+y=0$ and $x+y'=0$ then $y=y'$.
Only after that is done is it logically valid to introduce a well-defined notation for the additive inverse.
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