Explanation for why $1\neq 0$ is explicitly mentioned in Chapter 1 of Spivak's Calculus for properties of numbers.

Solution 1:

To show that $1\neq 0$ cannot be proven from the other six properties, consider the set that contains only one element, $\{\bullet\}$. Define $+$ by $\bullet+\bullet = \bullet$ and $\cdot$ by $\bullet\cdot\bullet = \bullet$. Then letting $0=\bullet$, $-\bullet =\bullet$, and $1=\bullet$, all six axioms are satisfied, but $1=0$. Thus, $1\neq 0$ cannot be proven from the first six axioms, since you have a model in which the first six axioms are true, but $1\neq 0$ is not.

Yes: the reason we need to specify it is so that we don't just have the one-element "field". Basicallly, the condition that $1\neq 0$ is formally undecidable from the first six properties, so it needs to be specified.

Solution 2:

To prove that $1 \neq 0$ can't be proven from those properties, one can just construct an example where $1 = 0$ and those properties hold, since this means that you have structures that satisfy the axioms where some have $1 = 0$ and others have $1 \neq 0$, so this property is independent of the axioms.

Specifically, the zero ring (what you get if you have a single number) satisfies all of them. If you want, you can easily check manually since the only choice for any variable is $0$:

P1) $0 + (0 + 0) = 0 = (0 + 0) + 0$

P2) $0 + 0 = 0 = 0 + 0$

etc.

P6 is the one we need to look at specifically, it can be written as "There exists a number $x$ such that $a \cdot x = x \cdot a = a$, and we call this $x$ the $1$ of the ring". In the zero ring, choose $x = 0$, so

$0 \cdot x = 0 \cdot 0 = 0 = x \cdot 0$

Thus $0$ satisfies the rule for the "$1$" in the ring, and so $0 = 1$.

Solution 3:

The important thing to realise here is that $1$, $0$, $+$, $\cdot$, $-$ need not mean the things that you're used to them meaning. We can come up for any rule for combining any things, and if it behaves a bit like $+$ (i.e. follows rules (1)-(4)) we may call it $+$, and if it behaves a bit like $\cdot$ we may call it $\cdot$. Then if an object follows rule (2) we might call it $0$ and if an object follows rule (6) we might call it $1$. Given just these six rules, we can't even be sure that $+$ and $\cdot$ are different operations¹, so we can't be sure that the object we called $0$ and the object we called $1$ are different objects.

¹ Conspicuous by its omission from your list is the distributive law that defines how multiplication and addition interact: $a\cdot(b + c)=a\cdot b+a\cdot c$. If that were included, we'd have a sort of way of telling the difference between addition and multiplication, but there'd still be the completely uninteresting case (only one object, and all operations just give you that object back again) where it worked but still $0 = 1$.

Solution 4:

(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).