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)
2026-04-03 04:54:01.1775192041
Why it is specified that 1 <> 0 as a part of an axiom of multiplication in Rudin? We can derive it
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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.