Let $1=0$ we know that $x \cdot y=0$ if $x=0$ or $y=0$ so

$1(1-1)=0 $ if $1=0$ or $1=1$ but $1=1$ is contradiction since we assume that $1=0$ so $1 \neq 0$

Is my prof correct ? If not how should it look like ?

  • $\begingroup$ Not a contradiction. I don't know what axioms you are using, but the field ones (i.e. the ones with multiplication, addition, identities, etc) aren't going to give you any contradictions since $\{0\}$ is a valid field. Edit: I guess actually a zero ring is not a field. My mistake! $\endgroup$ – Bruce Zheng Oct 23 '14 at 16:52
  • $\begingroup$ I don't think you can prove it. It's an assumption for $\mathbb{R}$ to be a nontrivial field. $\endgroup$ – John Oct 23 '14 at 16:52
  • $\begingroup$ @BruceZheng The axioms I take for a field do not have $\{0\}$ as a valid field... Sure, a commutative division ring, but not a field. To be complete, the definitions I use are to say that $(F,+,\cdot,0,1)$ is a field if both $(F,+,0)$ and $(F\setminus\{0\},\cdot,1)$ are abelian groups and distributivity holds. Since there is no group structure on $\emptyset$, we must have $F\neq \{0\}$. $\endgroup$ – Hayden Oct 23 '14 at 16:55
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    $\begingroup$ @mariuszCzarnecki it is important that you include your axioms since there are clearly different accepted definitions. $\endgroup$ – Hayden Oct 23 '14 at 16:56
  • $\begingroup$ possible duplicate of Is $\{0\}$ a field? $\endgroup$ – Adriano Oct 23 '14 at 17:04

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