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I was checking out the consequences of property 9 of chapter 1 in Spivak's calculus book but the way he proves

(-x)(-y) = xy

using the aforementioned properties (particularly p2 and p3), its never made clear as to why the distributive law would itself hold for any negative numbers. We can argue that the distributive law for positive number holds easily by taking some samples, but we cannot do so for negative numbers because (-x)(-y) is not yet defined. So how is it that he can use distributive law?

Also what law would be violated if xy = xy, (-x)y=xy, x(-y)=xy and (-x)(-y)=-xy?

Edit: I guess the question is: if there any reason to define (-x)(-y)=xy other than to ensure that distributive law holds?

  • How does he define negative numbers, their addition and multiplication? it is possible to construct negative numbers as ordered pairs of natural numbers. – Somos May 22 '20 at 11:37
  • $(-x)y=xy$ and $x(-y)=-(xy)$ would undermine commutativity as you would have $(-x)x\not = x(-x)$ unless $xx=-(xx)$ – Henry May 22 '20 at 11:38
  • @Henry sorry i have edited the question now –  May 22 '20 at 11:40
  • The sign multiplication is about the behaviour of the multiplication and the additive inverse, so unformally in order to define the sign law unequivocally you need an axiom relating both the addition part and the multiplication part of your ring, and the only such axiom is the distributive law. In the example you give you also violate the fact that 1 is a unit, but this can be fixed by changing the sign of the result in the two identities with mixed signs – Louis Hainaut May 22 '20 at 11:53
  • @LouisHainaut can you list some resource where i can learn about the sign law and its construction in more detail? –  May 22 '20 at 11:55
  • With your edited version $(-x)y=xy$ and $x(-y)=xy$ and $(-x)(-y)=-(xy)$ and supposing $y=-z$ you get $-xz=(-x)(-z)=(-x)y=xy=x(-y)=xz$. The only way to avoid this – Henry May 22 '20 at 11:56
  • For further discussion of this and related topics see here and here (and see also here on structure preservatiuon in extensions, axiomatization, Hankel / Peacock permanence principle) – Bill Dubuque May 22 '20 at 15:25
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    Concerning your edit question at the bottom. Whether there are other reasons or not (there are), ensuring that the distributive property holds is a very, very important reason for the definition. The distributive property permeates all mathematical activity involving real or complex numbers in ways we rarely even recognize because we are just so used to seeing it that we overlook its presence. I think you would be genuinely surprised as how much harder things would be without the distributive law working for negative numbers. – Paul Sinclair May 22 '20 at 20:43
  • It might make things feel more satisfying to note that $\mathbb{Z}$ (with the usual arithmetic) in fact shares all the "universal" properties of $\mathbb{N}$: any universal equality between terms which holds in $\mathbb{N}$ (e.g. "For all $x$, we have $x\cdot 1=x$" or "For all $x,y$, we have $x\cdot (y+y)=(x\cdot y)+(x\cdot y)$") is also true in $\mathbb{Z}$. This is basically because of how we can "build" $\mathbb{Z}$ from $\mathbb{N}$, by thinking of an integer $z$ as the equivalence class of ordered pairs of natural numbers with respect to the relation $$(a,b)\sim (c,d)\iff a+d=b+c.$$ – Noah Schweber May 23 '20 at 05:23
  • (Or, using more jargon: $\mathbb{Z}$ is a quotient of a power of $\mathbb{N}$, and equations are preserved by powers and quotients of structures.) So, it's more than just singling out a couple algebraic properties as especially important. $${}{}{}{}{}{}{}{}{}{}{}{}{}$$ The general study of what sorts of constructions preserve what sorts of algebraic properties shows up throughout mathematics, but in particular universal algebra is especially relevant. Of course this is going way advanced, but it may be comforting to know that this sort of question is seriously studied in the first place. – Noah Schweber May 23 '20 at 05:25

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If $(1)(2)=(1)(-2)$ then $1$ is not a multiplicative identity, unless $2=-2.$

With conventional multiplication, we can define $f(x,y)=|x|\cdot |y|\cdot sgn(x).$ In the book A Budget Of Trisectors, by the mathenatician Underwood Dudley, he relates a letter he received that insisted that $f(x,y)$ is the "real" multiplication.

DanielWainfleet
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