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

I am self-studying the wonderful book, Elementary Geometry from an Advanced Standpoint.

In chapter 1, problem 18 it says: Postulate M-6 (which says 1 not equal to 0) may seem superfluous. Is it? Can it be proved, on the basis of the other postulates, that there is any number at all other than 0?

The "other postulates" include the commutativity, associativity, identity, and distributive laws for addition and multiplication.

I don't see how to show that without the postulate 1 not equal to 0 it cannot be shown that there is any number other than 0. Would you help please?

Community
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Roger Costello
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3 Answers3

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Show that the set $\{0\}$ with the usual addition and multiplication satisfies all of your postulates except M-6. This gives a model of the other postulates in which M-6 is false, which shows that it’s independent of the others.

Brian M. Scott
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Assuming what I expect is there, the set consisting of just a number $0$ satisfies all axioms, with operations $0+0=0$ and $0 \cdot 0 = 0.$ The usual role of the symbol $1$ is that $1 \cdot x = x,$ but that is taken care of here, with $1=0,$ by $0 \cdot 0 = 0.$

Will Jagy
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The axioms are the Field axioms. As a hint, consider the analogous problem of showing that the Field axioms do not imply that $\:1 + 1 \ne 0.\:$ If the implication held true then $1 + 1 \ne 0\:$ would be true in every field. To refute the implication it suffices to find a counterexample: a field with $1 + 1= 0.$ But everyone knows such a field, namely Parity Arithmetic, i.e. the integers modulo $2$, which one easily checks satisfies all the Field axioms, using the operation tables below, e.g. the commutative laws hold because the operation tables are symmetric.

$$\begin{array}{rcl} \rm Parity\ Arithmetic & &\rm\!\!\! modulo\ 2 \\ \begin{array}{|c|c|c|}\hline + & \bf\text{even} &\rm\bf odd\\\hline \bf\text{even} & \text{even} & \text{odd}\\\hline \rm\bf odd & \text{odd} &\rm even\\\hline \end{array} \!\!\!\!\!\!\!& \iff &\!\!\!\!\!\!\! \begin{array}{|c|c|c|}\hline + & \bf 0 & \bf 1\\\hline \bf 0 & 0 & 1\\\hline \bf 1 & 1 & 0\\\hline \end{array} \\ \begin{array}{|c|c|c|}\hline \times & \bf\text{even} &\rm odd\\\hline \bf\text{even} & \text{even} & \text{even}\\\hline \rm\bf odd & \text{even} &\rm odd\\\hline \end{array} \!\!\!\!\!\!\!& \iff &\!\!\!\!\!\!\! \begin{array}{|c|c|c|}\hline \times & \bf 0 & \bf 1 \\\hline \bf 0 & 0 & 0\\\hline \bf 1 & 0 & 1\\\hline \end{array} \end{array} $$

Your problem is simpler since a counterexample has $\:1 = 0,\:$ hence $\rm\: x = 1\cdot x = 0\cdot x = 0,\:$ i.e. every element $= 0,\:$ so the structure has only the single element $0.\:$ The axiom for inverses is vacuously satisfied, since there are no nonzero elements to check for inverses. All other axioms are universal, i.e. identities such as $\rm\:\forall\, x,y\!:\ x+y = y + x,\:$ and they hold true for all elements simply because there is only a single value $\:0\:$ for the operations to take, so it can only evaluate to $\:0 = 0\:$ (vs. possibly $\rm\:1 = 0\:$ when verifying the two-element field above), see the operation tables below.

$$\rm\begin{array}{} Null\ Arithmetic\ (mod\ 1)\\ \begin{array}{|c|c|}\hline + & \bf 0 \\\hline \bf 0 & 0 \\\hline \end{array}\quad \begin{array}{|c|c|}\hline * & \bf 0 \\\hline \bf 0 & 0 \\\hline \end{array} \end{array} $$

Bill Dubuque
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