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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