TITLE: Explanation for why $1\neq 0$ is explicitly mentioned in Chapter 1 of Spivak's Calculus for properties of numbers.
QUESTION [11 upvotes]: During the first few pages of Spivak's Calculus (Third edition) in chapter 1 it mentions six properties about numbers.

(P1) If $a,b,c$ are any numbers, then $a+(b+c)=(a+b)+c$
(P2) If $a$ is any number then $a+0=0+a=a$
(P3) For every number $a$, there is a number $-a$ such that $a+(-a)=(-a)+a=0$
(P4) If $a$ and $b$ are any numbers, then $a+b=b+a$
(P5) If $a,b$ and $c$ are any numbers, then $a\cdot(b\cdot c)=(a\cdot b)\cdot c$
(P6) If $a$ is any number, then $a\cdot 1=1\cdot a=a$

Then it further states that $1\neq 0$. In the book it says that it was an important fact to list because there is no way that it could be proven on the basis of the $6$ properties listed above - these properties would all hold if there were only one number, namely $0$. 
Questions:
1) How does one rigorously prove that $1\neq0$ cannot be proven from the $6$ properties listed? 
2) It says that "these properties would all hold if there were only one number, namely $0$." Is a reason as to why this is explicitly mentioned is to avoid this trivial case where we only have the number $0$? Is there another deeper reason as to why this sentence was mentioned in relation to $1\neq 0$?
NB: Can someone please check if the tags are appropriate and edit if necessary? Thanks.

REPLY [5 votes]: (1) How does one rigorously prove that 1≠0  cannot be proven from the 6  properties listed?
(2) It says that "these properties would all hold if there were only one number, namely 0 ." Is a reason as to why this is explicitly mentioned is to avoid this trivial case where we only have the number 0 ? Is there another deeper reason as to why this sentence was mentioned in relation to 1≠0 ?

Any equational algebraic theory whose axioms are all universal, i.e. that assert equalities of terms composed of operations, variables, and constants, for all values of the variables, necessarily has a one element model. Indeed, defining all of the constants to be the one element (say $0)$ and defining all the operations to have value $0$ makes all axioms true, since they  evaluate to $\,0 = 0.$
Hence $\,1\ne 0\,$ is not deducible from your axioms since it is not true in a one element model.
The reason that $\rm\:1\ne 0\:$ is adjoined as an axiom for fields (and domains) is simply a matter of convenience. For example, it proves a very convenient target for proofs by contradiction, which often conclude by deducing $\rm\:1 = 0.\:$ Also, it avoids the inconvenience of needing to explicitly exclude in proofs motley degenerate cases that occur in one element rings, e.g. that $\rm\:0\:$ is invertible, since $\rm\:0\cdot 0 = 1\, (= 0).\:$ Much more so than proofs by contradiction, this confuses many students (and even some experienced mathematicians) as witnessed here in the past, e.g. see the long comment threads here and here (see esp. my comments in Hendrik's answer).