Is there a way to represent the relation $<$ on the real numbers without using multiplication?

Question

I know for $x,y\in\mathbb R$ we have $x<y$ iff there exists a $z\in\mathbb R$, so that $y=x+z^2$. Is there a similar way to „prove“ $x<y$ using only $+$ and $=$? And if not, is there a more or less simple proof for that?

Answer 1

You can't do that.

Your question, in other words, is whether or not $<$ is definable in $(\Bbb R,+)$, and the answer is no. The simplest way to do that is to show that there's an automorphism of $(\Bbb R,+)$ which is not order preserving.

For example, $x\mapsto -x$.