Proof of The Maximum and Minimum of Two Elements of an Ordered Field: Sum, Difference and Absolute Value

Each result cited below is universally quantified over the data in its own statement and is applied to the data named where it is cited. By hypothesis, the order $\le$ of $F$ is a total order on the set $F$, so the clauses of The Maximum and Minimum of Two Elements of a Total Order apply to it with $X=F$; in particular $a\le b$ or $b\le a$. By Fields §field and Commutative Rings §ring, addition in $F$ is commutative, and $x-y=x+(-y)$ for $x,y\in F$ by Negatives, Differences, Reciprocals and Quotients §negative. The absolute value is that of Absolute Value in an Ordered Field §absolute-value, and the rules of Rules of Arithmetic and Order in an Ordered Field are in force by Commutative Rings, Fields and Ordered Fields: Standard Notation §ordered-fields; the clauses used are cited below.

Sum and difference. Suppose $a\le b$. By The Maximum and Minimum of Two Elements of a Total Order §values, $\max\{a,b\}=b$ and $\min\{a,b\}=a$, so $\max\{a,b\}+\min\{a,b\}=b+a=a+b$. By Rules of Arithmetic and Order in an Ordered Field §order-sum, $0\le b-a$, so $|b-a|=b-a$ by Absolute Value in an Ordered Field §absolute-value. By Rules of Arithmetic and Order in an Ordered Field §signs and commutativity of addition,

$$-(a-b)=-(a+(-b))=(-a)+(-(-b))=(-a)+b=b+(-a)=b-a,$$

and by Rules of Arithmetic and Order in an Ordered Field §absolute-value, applied to $a-b$, $|a-b|=|-(a-b)|=|b-a|=b-a=\max\{a,b\}-\min\{a,b\}$.

Suppose $b\le a$. By The Maximum and Minimum of Two Elements of a Total Order §values, $\max\{a,b\}=a$ and $\min\{a,b\}=b$, so $\max\{a,b\}+\min\{a,b\}=a+b$. By Rules of Arithmetic and Order in an Ordered Field §order-sum, $0\le a-b$, so $|a-b|=a-b=\max\{a,b\}-\min\{a,b\}$ by Absolute Value in an Ordered Field §absolute-value.

As one of the two cases holds by totality, this proves the clauses sum and difference.

Absolute. By Rules of Arithmetic and Order in an Ordered Field §absolute-value, $-|a|\le a\le|a|$. If $0\le a$, then $|a|=a$ by Absolute Value in an Ordered Field §absolute-value, so $-a=-|a|\le a$, and The Maximum and Minimum of Two Elements of a Total Order §values, applied to the elements $a$ and $-a$ of $F$ in the case $-a\le a$, gives $\max\{a,-a\}=a=|a|$. Otherwise $|a|=-a$ by Absolute Value in an Ordered Field §absolute-value, so $a\le|a|=-a$, and The Maximum and Minimum of Two Elements of a Total Order §values, applied to $a$ and $-a$ in the case $a\le-a$, gives $\max\{a,-a\}=-a=|a|$.