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How to prove that $x\cdot y\neq 0$ when $x\neq 0$ and $y\neq0$ via field axioms?

According to the field axioms, especially the Commutativity of multiplication it is $a\cdot b=b\cdot a$. Is that enough to disprove $x\cdot y=0$ hence proving that $x\cdot y\neq 0$

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  • $\begingroup$ Thank you for your response. To what "definition" are you reffering? To the aforementioned Commutativity? $\endgroup$ – Analysis Apr 23 at 10:00
  • $\begingroup$ Sorry, in my definition of a field it's so that $F\setminus \{0\}$ is an Abelian group with respect to multiplication. You must have some other one. $\endgroup$ – Jakobian Apr 23 at 10:03
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Suppose that $x\cdot y=0$ and $x\neq 0$, then $\frac1x$ exists and $\frac1x\cdot x\cdot y=\frac1x\cdot 0$, what is equivalent to $y=0$.

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  • $\begingroup$ I have to prove: x·y=0 ⇔ (x=0)∨(y=0). I have already proven (x=0)∨(y=0) ⇒ x·y=0. But now I have to prove x·y=0 ⇒ (x=0)∨(y=0). And I decided to do it via Contraposition, thus x≠0 ∧ y≠0 ⇒ x·y≠0. But how am I going to do that? $\endgroup$ – Analysis Apr 23 at 10:06
  • $\begingroup$ Yeah, but since it is $\Longleftrightarrow$ I also have to prove that $(x=0)\vee (y=0) \Longrightarrow x\cdot y=0$. $\endgroup$ – Analysis Apr 23 at 10:12
  • $\begingroup$ @Analysis ok, I understand... Just use the above. I had proven it in my answer, I just skipped the part that $\frac1x\cdot 0=0$ $\endgroup$ – Masacroso Apr 23 at 10:14
  • $\begingroup$ I want to start like that: Suppose that $x\cdot y=0$ with $y\neq 0$ and $x\neq 0$ ..... and than disprove that statement $\endgroup$ – Analysis Apr 23 at 10:16
  • $\begingroup$ @Analysis suppose you start like you want... Then from my answer you get the contradiction that $y=0$, then it is not possible that $x\cdot y=0$ and $x,y\neq 0$ at once, so if $x\cdot y=0$ then $x,y\neq 0$ must be false. $\endgroup$ – Masacroso Apr 23 at 10:17
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If $x\neq0$, then it as an inverse. So$$y=1.y=(x^{-1}.x).y=x^{-1}.(x.y)=x^{-1}.0=0.$$Now, it remains to be proved from the field axioms that you always have $x.0=0$.

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  • $\begingroup$ I have already proved $x\cdot 0=0$ and $-x=(-1)x$. Now I have to prove that $x\cdot y=0 \quad \Longleftrightarrow \quad (x=0)\vee (y=0)$. Proof by cases: I have proven the cases $x\neq 0$ and $y=0$ and vice versa. Now the third case $x\neq 0$ and $y\neq 0$ $\endgroup$ – Analysis Apr 23 at 10:03
  • $\begingroup$ Since you have already proved that you always have $x.0=0$, there is nothing else that you need to prove. The fact that you always have $x.0=0$was all that was needed to complete my proof of the fact that $x\neq0\implies y=0$. $\endgroup$ – José Carlos Santos Apr 23 at 10:06

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