# Help with a proof of a consequence from the axioms of addition and multiplication

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 ?

• Distributive law of real numbers Aug 17, 2020 at 16:29
• Please avoid the use of the "analysis" tag. Indeed the tag info explicitly suggest to use a more specific tag. Aug 17, 2020 at 17:33
• @ArcticChar what is the issue with using the analysis tag? Aug 17, 2020 at 17:37
• I will try to do that (in general when I edit a post I do not check who's the asker - that's irrelevant). Aug 17, 2020 at 18:18

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 $$x\in\Bbb{R}$$, $$0\cdot x = 0$$.

• Many thanks. Completely clear now. Aug 17, 2020 at 16:36

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.)