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M4 defines 1 <> 0 as a part of an axiom. However, 1.18 (d) proves 1 > 0 which implies 1 can't be < 0 and can't be = 0 (by definition of ordering relation)

Iaroslav Baranov
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  • Please use LaTeX to typeset your equations, or if you absolutely have to use ASCII, typeset the $\neq$ ($\neq$) sign as "=/=". Writing it as "<>" is unclear. – Jam Oct 15 '22 at 09:07
  • Without looking into Rudin & going by the attached Context : M4 says that there is some non-zero element & hence 1.18D can talk about non-zero x. Hence M4 is necessary. – Prem Oct 15 '22 at 09:08
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    See MSE Q1169389 for why we specifically set $1\neq 0$ in the field axioms. In short, we are avoiding an inconvenient trivial case of the field with one element. – Jam Oct 15 '22 at 09:12
  • Also, note that Rudin's definitions of the less than relation $<$ (Def. 1.17) do not define it to be irreflexive (i.e. for all $x$, that $x \not< x$). That is a derived property. We infer that if $x>y$ or $x<y$, then $x\neq y$. So, even if we had neglected to define $1\neq 0$ and somehow had $1=0$, we would not directly have $1>0$ or $1<0$ and would still ought to prove that. – Jam Oct 15 '22 at 09:36

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As it is already mentioned in the comments:

If you do not require that $1\neq 0$, Then the set F could consist of only one element, the null element 0. However, at least two elements, the null element $0$ and the identity $1$, are required for proposition 1.18.

Josef
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