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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 $(-1_K)$?

1 Answers 1

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

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    What's $\mathbb{Z}/3\mathbb{Z}$? – 2017-01-01
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    Integers (mod 3). – 2017-01-01
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    And what about ordered fields? Is the theorem true at least for them? – 2017-01-01
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    Yes, I believe that is true, since then $x+1>x$ for any $x$. – 2017-01-01
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    Why does $x – 2017-01-01
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    are you assuming $1>0$? – 2017-01-01
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    I'm assuming only the ordered field axioms, I don't know if this is their consequence. If it isn't, I'm not. I can see how this would prove $x – 2017-01-01
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    $1>0$ implies $x+1>x+0$. – 2017-01-01
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    $1>0$ is a consequence of the ordered field axioms btw. – 2017-01-01
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    In what way? I suppose that it's somehow related to ${\forall}x:0 – 2017-01-01
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    Consequence of $x^2\ge 0$ for all $x$. – 2017-01-01