18 SPECIAL TOPICS IN THEORETICAL ARITHMETIC

   

i ‘ Since b is positiv, |b| = b.
k| Case 1. a is positiv or zero.
k| \” Then |a| = a
s ‘- Hence we have |a| = a < b = |b]
ki Therfor la| < |b]
] Case 2. a is negativ.
Then |a| = — a
Since — b <a, —a <b
Therfor la| < |b]

If b < a < — b, we have a similar proof.

For the converse, since |a| < |b| and |a| is positiv or zero

(§ 45), it follows that |b| must be positiv and that therfor b
i must be either positiv or negativ.
b Then by taking two main cases according as b is positiv or
‘ negativ and in each main case taking two subcases, first with
a positiv or zero, second with a negativ, our theorem is easily
proved.

For example,

—4<2<4 12| < |4] 2<4
—4<0<4 0] < |4] 0<4
—4< —-2<4 | — 2] < |4] 2<4
=g e 3= (=5  |—=3| <|=-5] 3<5

47. Th. If a and b are both positiv or zero, or both megativ
or zero, |la + b| = |a| + |b].

For, if @ and b are both positiv or zero, @ + b is positiv or
Zero.

Hence |a +b| =a + b = |a| + |D]

If @ and b are both negativ or zero, @ + b is negativ or
zero.

Then ¢ +b| = —(@+b) = —a+ — b= |a| + [D]
For example, |3 + 5| = |3]| + |5

|4+ 0| = |4] + |0

[ =2 e U e g -