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While reading through Analysis 1 by Vladimir A. Zorich, I encountered this proof which has this 1 step I can't understand. Here is the consequence and the proof:

For every $x\in \mathbb R$ the following is true

$$-x=(-1)\cdot x$$

Proof. $\ \ x+(-1)\cdot x=\underbrace{(1+(-1))\cdot x}_\text{Which of the axioms were used here ?}= 0 \cdot x=x \cdot 0 = 0$. The assumption follows from the uniqueness of the negative of a number.

End of proof.

The underbraced part is what I fail to understand. What addition and multiplication axioms were used in order to make that expression ?

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    $\begingroup$ Distributive law of real numbers $\endgroup$
    – Philipp
    Aug 17 '20 at 16:29
  • $\begingroup$ Please avoid the use of the "analysis" tag. Indeed the tag info explicitly suggest to use a more specific tag. $\endgroup$ Aug 17 '20 at 17:33
  • $\begingroup$ @ArcticChar what is the issue with using the analysis tag? $\endgroup$
    – tau20
    Aug 17 '20 at 17:37
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    $\begingroup$ @GrigoriPerelman Please see here and here for more information. $\endgroup$ Aug 17 '20 at 17:41
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    $\begingroup$ I will try to do that (in general when I edit a post I do not check who's the asker - that's irrelevant). $\endgroup$ Aug 17 '20 at 18:18
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Note that $1\in\Bbb{R}$ is a special element of the set with the property that for every $x\in \Bbb{R}$, $1\cdot x = x\cdot 1 = x$. Next, we also use the distributive law that for all $a,b,c\in\Bbb{R}$, $a\cdot(b+c) = a\cdot b + a \cdot c$. Hence, \begin{align} x+ (-1)\cdot x &= 1 \cdot x + (-1)\cdot x \tag{property of $1$} \\ &= [1 + (-1)]\cdot x \tag{distributive law} \end{align} The rest of the proof follows once you establish that for every $x\in\Bbb{R}$, $0\cdot x = 0$.

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  • $\begingroup$ Many thanks. Completely clear now. $\endgroup$
    – tau20
    Aug 17 '20 at 16:36
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the principal is distribution: $a(b+c) = ab + ac$.

So the proof goes like this:

$x + (-1)x = 1\cdot x + (-1)\cdot x$ (by existence and definition of multiplicative identity)

$=(1+(-1))\cdot x$ (by distribution)

$=0\cdot x$ (by definition of additive inverse)

$=x\cdot 0$ (commutivity of multiplication but I have no idea why he did this)

$= 0$ (This is not an axiom but a proposition can be proven that $0\cdot x = 0$. Have you proven that yet? Does Spivak use that as an axiom?)

Then by definition we have that for every $x$ there exists a unique $-(x)$ so that $x + (-x) = 0$.

If we ever have an $a$ so that $x + a = 0$ it must be that $a=-x$ as the multiplicative inverse is unique. As $x + (-1)x =0$ it must be $(-1)x = -x$.

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Prop: $x\cdot 0 = 0$.

Pf: $x\cdot 0 + (-(x\cdot 0)) = 0$. (Every element $a$, including $x\cdot 0$, has an additive inverse, $-a$, so that $a + (-a) =0$.)

$x\cdot(0 + 0) + (-(x\cdot 0)) = 0$ ($0=0+0$ because $0$ is the additive identity and $a +0 = a$ for all $a$, including when $a$ is $0$.)

$(x\cdot 0 + x\cdot 0) + (-(x\cdot 0)) = 0$ (distributivity)

$x\cdot 0 + (x\cdot 0 + (-(x\cdot 0)) = 0$ (associativity)

$x\cdot 0 + 0 = 0$ (definition of additive identity)

$x\cdot 0 = 0 $ ($a + 0= a$ for all $a$ by definition of additive identity.)

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