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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 $0$.


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.

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It's basically just to rule out the trivial case so that in all proofs, theorems and such we don't have to say "except in this trivial case". –  Ragib Zaman Jun 30 '12 at 5:27
If all properties $P_i$ $\ (1\leq i\leq 6)$ hold in a system containing only one element $0$ then it's impossible to prove $1\ne0$ from the $P_i$ alone. –  Christian Blatter Jun 30 '12 at 8:43
I had a go at the tags, but I'm probably missing some, I don't know that many. –  Ben Millwood Jun 30 '12 at 12:43

4 Answers 4

up vote 25 down vote accepted

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.

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So based on your definition of addition and multiplication with the one element set, if we have ${0}$ then all 6 properties will be satisfied where in (P6) we replace $1$ with $0$? But then for ${1}$ how do we check (P3)? We haven't defined what $(-1)+1$ or $1+(-1)$ is? –  tcmtan Jun 30 '12 at 7:16
@user22678: There is only one element in the set, namely $\cdot$. "$1$", "$0$", "$-0$", "$-1$", "$--0$" and so on are all just different ways of referring to $\cdot$. –  Chris Eagle Jun 30 '12 at 8:20
@user22678: "0" and "1" are names; absent a prohibition (which is the new axiom $1\neq 0$) they can be two different names of the same thing. In this particular object, every single name that you see in the axioms refers to the same object, $\bullet$. So "$1$" refers to $\bullet$, and "$-1$" also refers to $\bullet$, and "$0$" refers to $\bullet$. So "$(-1)+1$" has been defined: it's $\bullet + \bullet$, which we defined to be $\bullet$. Same for "$1+(-1)$". –  Arturo Magidin Jun 30 '12 at 17:35
@user22678: yes; except that the set is not defined by the operations; the operations are defined on the set, which "pre-exists" the operations. –  Arturo Magidin Jul 1 '12 at 1:01
@user22678: It's important to keep clear what is what. An operation is a function on a set; you cannot define an operation before you have a set (just like you cannot define a function without first having a domain and a codomain). You don't get the set "out of" the operation. You start with a set, and the operation is defined on the set. –  Arturo Magidin Jul 1 '12 at 1:19

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


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 $0 = 1$.

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I understand that if you have just the set ${0}$ then yes, $0$ satisfies the "$1$" in (P6). But what if you have ${1}$ and replace all the $0$'s in the six properties into $1$'s, namely (P2) and (P3)? For $(-0)+0=0+(-0)$ it seems quite comfortable to verify this, but when $(-1)+1=1+(-1)=1$ it seems like something isn't right? –  tcmtan Jun 30 '12 at 9:58
In this context, "1" is just a label for the multiplicative identity (i.e. the number that satisfies $a\cdot 1 = 1 \cdot a = a$), and so it has different meanings in different rings, and you can't transfer the meaning from the integers to the meaning in the zero ring. –  dbaupp Jun 30 '12 at 13:31

(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).

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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 $0 = 1$.

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