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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 ?

UnsinkableSam
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Mario
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  • 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! – Bruce Zheng Oct 23 '14 at 16:52
  • I don't think you can prove it. It's an assumption for $\mathbb{R}$ to be a nontrivial field. – John Oct 23 '14 at 16:52
  • @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 $F\neq {0}$. – Hayden Oct 23 '14 at 16:55
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    @mariuszCzarnecki it is important that you include your axioms since there are clearly different accepted definitions. – Hayden Oct 23 '14 at 16:56
  • possible duplicate of Is ${0}$ a field? – Adriano Oct 23 '14 at 17:04
  • You made a few assumptions. The proof is faulty. Firstly, what does $1-1$ even mean? Is subtraction defined in your axioms? Secondly, do you know that $1-1=0$, because that seems like your second leap? Thirdly, what does $1-1=0$ have to do with $1=1$; there seems to be a disconnect there because I dont see how you went from $1\cdot 0 = 0$ to $1\cdot(1-1)=0$? Is substitution even allowed? Have you defined multiplication in your axioms? Lastly, how do you know that $1\ne 0$ already, because if you concluded $1=1$ but assumed $1=0$ then what is the contradiction? There doesnt seem to be one. – SquishyRhode Feb 14 '20 at 23:02

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