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If we consider the natural numbers, we have for all $a,b\in\mathbb{N}$ that $a\leq b\Longleftrightarrow (\exists x\in\mathbb{N})[a+x=b]$. (For example if $a=2$ and $b=5$ there exists $3\in\mathbb{N}$ such that $2+3=5$.) Now, as a generalization of this I was trying to prove that if $(\mathbb{F},+,\cdot,\leq)$ is an ordered field, then for all $a,b\in \mathbb{F}$ we have that \begin{align*} a\leq b\Longleftrightarrow (\exists x\geq 0)[a+x=b]. \end{align*}

I've been using the following definitions: A field $(\mathbb{F}, +,\cdot)$ is called an ordered field iff there is an ordering relation $\leq$ on $\mathbb{F}$ such that for all $a,b,c\in \mathbb{F}$ we have \begin{align*} a\leq b\Longrightarrow a+c\leq b+c \end{align*} and \begin{align*} (0\leq a)\wedge (0\leq b)\Longrightarrow 0\leq a\cdot b, \end{align*} where an ordering is defined as a binary relation that's reflexive, anti-symmetric, and transitive.

Now, how do I get from my definitions to what I'm trying to prove?

PAT
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  • I'm aware I didn't properly state quantifiers since I thought it's obvious that $(\exists x)(\forall a,b)$ doesn't make much sense. Ok, inserting $x=b-a$ obviously works, but what about the implication $a+x=b\Rightarrow a\leq b$? Only makes sense for $0\leq x$, though. – PAT Jan 24 '23 at 00:55
  • Well, if we consider the natural numbers, we have for all $a,b\in\mathbb{N}$ that $a\leq b\Longleftrightarrow (\exists x\in\mathbb{N})[a+x=b]$. (For example if $a=2$ and $b=5$ there exists $3\in\mathbb{N}$ such that $2+3=5$.) Now I was trying to prove a generalized version of this for ordered fields. What I meant with $0\leq x$ in my last comment was that if $x$ is negative $a+x=b$ would result in $b\leq a$. – PAT Jan 24 '23 at 01:33
  • $\forall a,\forall b,(a\le b\Leftrightarrow \exists x\in\Bbb F, a+x=b)$ is false because for $\exists x\in \Bbb F,a+x=b$ is true when $b<a$ as well. On the other hand, $\forall a,\forall b,(a\le b\Leftrightarrow\exists x\ge 0, a+x=b)$ is true. – Sassatelli Giulio Jan 24 '23 at 01:44
  • Ok, it clearly makes more sense now then in my initial draft. But how can we formally prove it? – PAT Jan 24 '23 at 02:16
  • Hint: First show $b-a \geq 0$, then use that as your $c$ – Alan Jan 24 '23 at 08:29
  • From $a\leq b\Rightarrow a+c\leq b+c$ we get $a+(-a)\leq b+(-a)$ and therefore $0\leq b-a$ but I cannot follow your second thought. – PAT Jan 24 '23 at 08:36

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