Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 1, Problem 3
Problem
Prove Proposition 1.15.
The axioms for multiplication imply the following statements.
(a) If $x \neq 0$ and $xy = xz$ then $y = z$.
(b) If $x \neq 0$ and $xy = x$ then $y = 1$.
(c) If $x \neq 0$ and $xy = 1$ then $y = 1/x$.
(d) If $x \neq 0$ then $1/(1/x) = x$.
Answer
(a) $xy = xz$ $\overset{(M5)}{\Longrightarrow}$ $(1/x)(xy) = (1/x)(xz)$ $\overset{(M3)}{\Longrightarrow}$ $((1/x)x)y = ((1/x)x)z$ $\overset{(M2)}{\Longrightarrow}$ $(x(1/x))y = (x(1/x))z$ $\overset{(M5)}{\Longrightarrow}$ $1y = 1z$ $\overset{(M4)}{\Longrightarrow}$ $y = z$.
(b) $xy = x$ $\overset{(M4)}{\Longrightarrow}$ $xy = 1x$ $\overset{(M2)}{\Longrightarrow}$ $xy = x1$ $\overset{(a)}{\Longrightarrow}$ $y = 1$.
(c) $xy = 1$ $\overset{(M5)}{\Longrightarrow}$ $xy = x(1/x)$ $\overset{(a)}{\Longrightarrow}$ $y=1/x$.
(d) $x(1/x)=1$ $\overset{(M2)}{\Longrightarrow}$ $(1/x)x=1$ $\overset{(c)}{\Longrightarrow}$ $x = 1/(1/x)$. In fact, we need a condition $1/x \neq 0$ to use (c), but it is not guaranteed. To prove $1/x \neq 0$, the distributive law is necessary. Hence, (d) cannot be derived only by the axioms for multiplication.
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