1
$\begingroup$

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 $(-1_K)$?

$\endgroup$
3
$\begingroup$

The statement is not true. Consider $\mathbb Z/3\mathbb Z$. This is a finite field, and $1+1+1=0$.

$\endgroup$
  • $\begingroup$ What's $\mathbb{Z}/3\mathbb{Z}$? $\endgroup$ – asdasdfsss Jan 1 '17 at 17:49
  • 1
    $\begingroup$ Integers (mod 3). $\endgroup$ – Tim Raczkowski Jan 1 '17 at 17:49
  • 1
    $\begingroup$ Yes, I believe that is true, since then $x+1>x$ for any $x$. $\endgroup$ – Tim Raczkowski Jan 1 '17 at 18:09
  • 1
    $\begingroup$ $1>0$ is a consequence of the ordered field axioms btw. $\endgroup$ – Tim Raczkowski Jan 1 '17 at 18:22
  • 1
    $\begingroup$ Consequence of $x^2\ge 0$ for all $x$. $\endgroup$ – Tim Raczkowski Jan 1 '17 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.