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With a field axiom concerning a structure of real numbers $\mathbb R$, like the Closure, Associative, Commutative, Distributive law and existence of the identity, inverse, etc. How to prove that $(-1)^2 = 1$? Should proofs of all formulas or statements in this axiom start with "there exists the number $1$, $0$" when we can't use any preceding theorem?

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$(-1)^2-1=^{\color{red}{\vdash}}(-1)(-1+1)=0$. Then $(-1)^2=1$ by the uniqueness of the neutral element.

$\cdot$ in ${\color{red}{\vdash}}$ use distributive.

The existence of $1$ and $0$ are axioms too.

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    $\begingroup$ $(-1)^2 + 1 \neq 0$; you mean $(-1)^2+(-1)$. $\endgroup$ – Arturo Magidin Mar 19 at 6:35
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    $\begingroup$ And by associative I think you mean distributive. $\endgroup$ – awllower Mar 19 at 6:36
  • $\begingroup$ Correct both. I'm sorry. I confused "distributive" when I went to translate. $\endgroup$ – Matheus Nunes Mar 19 at 6:38
  • $\begingroup$ I wonder what is ⊢. I've never seen this. $\endgroup$ – lev.1 code Mar 19 at 6:39
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    $\begingroup$ $\left( -1 \right)^2 = -1$ ? Shouldnt it be $= 1$ ? $\endgroup$ – thinkingeye Mar 19 at 7:18

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