The Rudin Reader: Section 1.18

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"If

then

, so that

"Since

, we have

"Hence

This is true by the second rule for ordered fields.

"And therefore

$\begin{align} xz&=0+xz &&\text{(A4)}\notag\\ &=xz+0 &&\text{(A2)}\notag\\ &=xz+(xy+(-(xy))) &&\text{(A5)}\notag\\ &=xz+(xy+x(-y)) &&\text{(1.16(\textit{c}))}\notag\\ &=xz+(x(-y)+xy) &&\text{(A2)}\notag\\ &=(xz+x(-y))+xy &&\text{(A3)}\notag\\ &=(x(z+(-y)))+xy &&\text{(D)}\notag\\ &=x(z-y)+xy &&\text{(common notation)}\notag\\ &>0+xy &&\text{(ordered rule i)}\notag\\ &=xy &&\text{(A4)}\notag \end{align}$

Thus we have

"By (a), (b), and Proposition 1.16(c),

I find that using (b) makes this proof unnecessarily complicated. Instead, here is another proof.

First, since

, by (a) we have

. Furthermore, since

, we have

(see above for proof). By 1.16(c),

. Since both elements of the right-hand side of the equation

exceed zero, the second rule for ordered fields mandates that the product of these elements also exceed zero. Thus,

"...so that

Because

, we have

"Hence

therefore, by the first rule for ordered fields,

. By (A4),

, and also,

Thus,

"If

, then

This is because of (a).

"Hence

This is because of the second rule for ordered fields.

"If

and

, then

, we have

$\begin{align} yv&=vy &&\text{(M2)} \notag\\ &=0y &&\text{(substitution)}\notag\\ &=0. &&\text{(1.16(\textit{a}))} \notag \end{align}$

, consider (c). Take

as the

used in (c),

as the

used in (c), and our current

as (c)'s

. Then, by (c),

. We have

and (M2) gives

. Thus, if

, we have

Therefore, if

and

, we have

"But $y \cdot (1/ y)=1>0$ ."

$\begin{align} y \cdot \frac{1}{y}&=1 &&\text{(M5)} \notag\\ &>0 &&\text{(\textit{d})}\notag \end{align}$

"Hence $(1/y)>0$ ."

We know $y>0$ , and $y \cdot \frac{1}{y}>0$ . Now assume that $\frac{1}{y}<0$ . Then, taking our $y$ as the $x$ in (b), $\frac{1}{y}$ as (b)'s $y$ , and $0$ as (b)'s $z$ , we must have

$\begin{align} y \cdot \frac{1}{y}&<y \cdot 0 &&\text{(substitution)}\notag\\ &=0 \cdot y &&\text{(M2)} \notag\\ &= 0. &&\text{(1.16(\textit{a}))} \notag \end{align}$

This, however, is exactly contrary to what was just proved above. Therefore our assumption must be false, and $\frac{1}{y}>0$ .

"If we multiply both sides of the inequality $x<y$ by the positive quantity $(1/x)(1/y)$ ..."

As we have just proved, both $\frac{1}{y}$ and $\frac{1}{x}$ exceed zero. Therefore, by the second rule for ordered fields, their product also exceeds zero.

"...we obtain $1/y<1/x$ ."

Take $\left(\frac{1}{x}\right)\cdot \left(\frac{1}{y}\right)$ as $x$ in (b), our $x$ as (b)'s $y$ , and our $y$ as (b)'s $z$ . Then $\left(\frac{1}{x}\right) \left( \frac{1}{y}\right) \cdot x < \left(\frac{1}{x}\right) \left(\frac{1}{y}\right) \cdot y$ . Now consider:

$\begin{align} \left(\frac{1}{x}\right) \left(\frac{1}{y}\right) \cdot x &=\left( \left(\frac{1}{x}\right) \left(\frac{1}{y}\right) \right)\cdot x &&\text{(common notation)}\notag\\ &=\left( \left(\frac{1}{y}\right) \left(\frac{1}{x}\right) \right)\cdot x &&\text{(M2)}\notag\\ &=\left(\frac{1}{y}\right) \left(\left(\frac{1}{x}\right) \cdot x \right) &&\text{(M3)}\notag\\ &=\left(\frac{1}{y}\right) \left(x \cdot \left(\frac{1}{x}\right) \right) &&\text{(M2)}\notag\\ &=\left(\frac{1}{y}\right) \cdot 1 &&\text{(M5)}\notag\\ &=1\cdot \left(\frac{1}{y}\right) &&\text{(M2)}\notag\\ &=\left(\frac{1}{y}\right). &&\text{(M4)}\notag \end{align}$