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(P6) If $a$ is any number, then $a⋅1=1⋅a=a$

http://i.imgur.com/vFJQLQI.png

Cannot understand the importance of this. Do we know what 0 is but don't yet know what 1 is?

And what does it mean to prove $1 \neq 0$?

Max
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  • We need more context. Is he constructing the real numbers or talking about any arbitrary ring? – Eigenfield Apr 23 '17 at 11:45
  • Constructing reals I suppose. Here's some more context: https://math.stackexchange.com/questions/164811/explanation-for-why-1-neq-0-is-explicitly-mentioned-in-chapter-1-of-spivaks-c The link may contain the answer, but if so then I don't understand it. – Max Apr 23 '17 at 11:55
  • If the properties being listed are intended to be all that is assumed about the real numbers, then, as Spivak says, the assumption $0\neq1$ is all that prevents us from believing that the only real number is 0 (and that therefore $1=0$). That belief would completely mess up calculus (and most of the rest of mathematics). – Andreas Blass Apr 23 '17 at 11:58
  • I've also found in the notes "...and that there are distinguished numbers called 0 and 1." here http://math.uga.edu/~pete/math2400_lecture_1.pdf I guess it's important to note that multiplicative (1) is not the same as additive identity (0), because we don't yet know what 1 and 0 means precisely? – Max Apr 23 '17 at 12:01
  • Andreas Blass, thank you. – Max Apr 23 '17 at 12:01
  • That link contains the answer. He defines the real numbers from those axioms and those axioms can not be used to show that $0≠1$, so he must include that as an additional axiom. – Eigenfield Apr 23 '17 at 12:07
  • Okay, I think I get it now. – Max Apr 23 '17 at 12:10
  • It is for Uniqueness – smokeypeat Jul 04 '17 at 07:28

1 Answers1

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The axioms can be used to show that if $z$ is an element with the property that $z+x=x+z=x$ for every $x$, then $z=0$. Indeed, $$ z+0=0 $$ from the stated property of $z$ (with $x=0$) and also $$ z+0=z $$ from the property of $0$ stated as an axiom. Hence $z=0$.

The proof that if $ux=xu=x$ for all $x$ then $x=1$ is just the same (substitute $u$ for $z$, $1$ for $0$ and multiplication for addition).

On the other hand, the set $\{?\}$ ($?$ means any object you like) with the operations $$ {?}+{?}={?},\qquad {?}{?}={?} $$ satisfies all axioms regarding addition and multiplication (and also order, actually) stated for the real numbers. Here $0={?}=1$, because the unique element satisfies the requirements for $0$ and $1$ stated by the axioms.

This means that we cannot prove $0\ne1$ only with the stated axioms and so we need to add $0\ne1$ as a further axiom, if we want the axioms to reflect what we expect from the real numbers.

egreg
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  • Isn't 1+1=2? :) – Max Apr 23 '17 at 13:27
  • @Max If $0=1$, then $1+1=0$. – egreg Apr 23 '17 at 13:30
  • Thanks. Really hard to know what I'm "supposed to know" and what I'm not supposed to know. He later shows how $1 > 0$ because $1=1^2$ and $1^2 > 0$, I guess we're truly not supposed to know much about the symbols "1" and "0" at this point. One is additive identity and the the other is multiplicative identity, and maybe they're the same thing, we don't really yet know. Using other symbols rather than "1" and "0" would help disassociate from preconceived notions about the two identities. – Max Apr 24 '17 at 05:29