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Ordered Ring Integer Ring Positive Contradictory Statement Characteristic Contrapostive

1	Positive
1	Integer
1	Ring
1	Ordered Ring
2	Contradictory Statement
3	Characteristic
3	Contrapostive

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Basic Facts About Ordered Rings

Throughout this entry, $(R, \le)$ is an ordered ring.

Lemma 1 If $a,b,c \in R$ with , then .

Proof. The contrapositive will be proven.

Let $a,b,c \in R$ with $a+c \ge b+c$ . Note that $-c \in R$ . Thus,

$\begin{displaymath}\begin{array}{rl} b & =b+0 \\ & =b+c+(-c) \\ & \le a+c+(-c) \\ & =a+0 \\ & =a. \qedhere \end{array}\end{displaymath}$

$\qedsymbol$

Lemma 2 If $\vert R\vert \neq 1$ and has a characteristic, then it must be 0.

Proof. Suppose not. Let

be a positive integer such that $\operatorname{char}~R=n$ . Since $\vert R\vert \neq 1$ , it must be the case that

Let $r \in R$ with . By the previous lemma, $\displaystyle 0<r \le \ldots \le \sum_{j=1}^{n-1} r \le \sum_{j=1}^n r=0$ , a contradiction. $\qedsymbol$

Lemma 3 If $a,b \in R$ with $a \le b$ and $c \in R$ with , then $ac \ge bc$ .

Proof. Note that $-c \in R$ and

. Since $a \le b$ , $-(ac)=a(-c) \le b(-c)=-(bc)$ . Thus,

$\begin{displaymath}\begin{array}{rl} bc & =bc+0 \\ & =bc+(ac+(-(ac))) \\ & =(b... ...\ & =(-(bc)+bc)+ac \\ & =0+ac \\ & =ac. \qedhere \end{array}\end{displaymath}$

$\qedsymbol$

Lemma 4 Suppose further that is a ring with multiplicative identity $1 \neq 0$ . Then .

Proof. Suppose that $0 \not< 1$ . Since

is an ordered ring, it must be the case that

. By the previous lemma, $1 \cdot 1 \ge 0 \cdot 1$ . Thus, $1 \ge 0$ , a contradiction. $\qedsymbol$