# (Ordered) field's multiplicative identity added to itself $n$ times isn't its additive inverse

Let $(K,+,*)$ be a field. Let $1_K$ be the field's multiplicative identity. Let $(-1_K)$ be its additive inverse. Let $0_K$ be the field's additive identity.

I'm trying to prove that for every $n>0_{\mathbb{N}}$, $f=\underbrace{1_K+...+1_K}_{n\, times\,+}$ isn't equal to $0_K$.

Since $+:K{\times}K{\rightarrow}K$, I know that I can write $f$ as $1_K+k$. I can also prove that for any two field elements $x$ and $y$, $x+y=0{\implies}x+y+(-x)=0+(-x){\implies}y=-x$ and so $f=0$ only iff $k=(-1_K)$.

And here we arrive at my original question: how to show that $k$ can't be equal to $(-1_K)$?

The statement is not true. Consider $\mathbb Z/3\mathbb Z$. This is a finite field, and $1+1+1=0$.
• What's $\mathbb{Z}/3\mathbb{Z}$? – asdasdfsss Jan 1 '17 at 17:49
• Yes, I believe that is true, since then $x+1>x$ for any $x$. – Tim Raczkowski Jan 1 '17 at 18:09
• $1>0$ is a consequence of the ordered field axioms btw. – Tim Raczkowski Jan 1 '17 at 18:22
• Consequence of $x^2\ge 0$ for all $x$. – Tim Raczkowski Jan 1 '17 at 18:30