Why this map preserves order?

Question

How to prove that $f:N→\{1,1+1,1+1+1,...\}$ where 1 is an identity element of ordered field, is order-preserving? I guess that maybe property if $a < b$ then $a + c < b + c$ can be useful, but I'm not sure how exactly.

Answer 1

By definition, in your ordered field $F$, $1_F>0_F$, and monotonicity of sum in $F$ helps you to prove by induction your thesis.

Answer 2

Surely some condition needs to be put on $f$, since you can easily construct bad maps that disturb order.

A suitable condition would be for $f$ to be additive.

Suppose additivity: then $f(1)\in\{1,1+1,1+1+1\dots\}$ somewhere. Every other image is now already determined. For example, $f(2)=f(1)+f(1)>f(1)$, where the last inequality follows from the axiom you mentioned in your OP.

By induction, you can show $f(n)=f(n-1)+f(1)>f(n-1)$ holds for all $n$, showing it is order preserving overall.

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In case you meant something even easier (the map $n\mapsto 1+1+\dots+1$ ($n$ times), then you can apply the same reasoning, because this map is additive. In this case it's more transparent that the sums of $1$ are an order-isomorphic copy of $\Bbb N^+$.