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What is the contraposition of the statement "If $x\in \mathbb{R}$ and $x\neq 0$, then $\exists x^{-1}\in \mathbb{R}$ such that $x\cdot x^{-1}=1$"? Where $1$ is de multiplicative identity that satisfies: $\exists 1\in \mathbb{R}$ and $1\neq 0$ such that if $x\in \mathbb{R}$, then $x\cdot 1 = x$.

Context:

I'm trying to prove that if $a\neq 0$ then $a^{-1}\neq 0$.

I can do this by using the fact $a\cdot 0 = 0$ as it was suggested in Correct Logical equivalent of the statement: "Every real number except zero has multiplicative inverse.", or suppose $a^{-1}=0$ as it was suggested in Multiplicative inverse of $0$

However, it seems to me that just the contraposition of the axiom will give an immediate proof, but I don't know how to express such contraposition.

Darvid
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    $$\forall x ~ \bigg((x \in \mathbb R \land x \ne 0) \implies (\exists x^{-1} ~ x^{-1} \in \mathbb R \land x \cdot x^{-1} = 1)\bigg)$$ can you do contraposition on that? – DanielV Oct 01 '22 at 23:06
  • Is this correct? $$\forall x \big( \forall x^{-1}\in \mathbb{R} \lor x\cdot x^{-1} \neq 1 \Longrightarrow x\notin \mathbb{R} \lor x=0 \big)$$ – Darvid Oct 01 '22 at 23:16
  • Very close, the $\forall x^{-1} \in \mathbb R \lor$ part is wrong – DanielV Oct 01 '22 at 23:26
  • How about now? $$\forall x (\forall x^{-1}\notin \mathbb{R} \lor x\cdot x^{-1} \neq 1 \Longrightarrow x\notin \mathbb{R} \lor x = 0)$$ – Darvid Oct 01 '22 at 23:31
  • The expression $\forall (P(z) \lor Q(z))$ is grammatically wrong and meaningless. – DanielV Oct 01 '22 at 23:34
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    @Darvid I think what you meant to write is $\forall x (\forall x^{-1}(x^{-1}\notin \mathbb{R} \lor x\cdot x^{-1} \neq 1) \Longrightarrow x\notin \mathbb{R} \lor x = 0)$. This would be the correct answer. – Dark Rebellion Oct 02 '22 at 16:29
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    A simpler form, if you just assume everything to be a real number, would be $\forall x\in\mathbb{R} (\forall x^{-1}\in\mathbb{R}(x\cdot x^{-1} \neq 1) \Longrightarrow x = 0)$ – Dark Rebellion Oct 02 '22 at 16:30

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What is the contraposition of the statement

"If $x\in \mathbb{R}$ and $x\neq 0$, then $\exists x^{-1}\in \mathbb{R}$ such that $x\cdot x^{-1}=1$" ?

Spelling out its implicit universal quantification, your original statement is equivalent to $$\forall x\in(-\infty,0)\cup(0,\infty) \;\;\exists y{\in}\mathbb R\quad x\cdot y=1.$$ In full: $$\forall x\;\Big( x\in(-\infty,0)\cup(0,\infty) \implies\exists y \;\big(y\in\mathbb R\:\land\: x\cdot y=1\big)\Big).$$ As such, its contrapositive is $$\forall x\;\Big( \forall y \;\big(y\in\mathbb R\implies x\cdot y\ne1\big)\implies x\not\in(-\infty,0)\cup(0,\infty) \Big).$$ Restricting the domain of discourse to $\mathbb R,$ this becomes $$\forall x\;\Big( \big(\forall y \:\:x\cdot y\ne1\big)\implies x=0 \Big),$$ which is equivalent to $$\forall x\;\exists y\;\big( x\cdot y\ne1\implies x=0 \big).$$

ryang
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