Page 56 - DMTH401_REAL ANALYSIS

Real Analysis

Notes
4 4
= .
5 5
All this lead us to the following properties:

Property 2: For any real number x
2
2
|x| = x = |–x| 2
Proof: We know that |x| = x for x 2'0.
2
Thus |x| = |x||x| = x, x = x, for x  0
Again for x < 0, we know that |x| = –x. Therefore
|x | = |x| |x| = –x. –x = x 2
2
Therefore, it follows that
2
2
|x| = x for any xR.
Now you should try the other part as an exercise.
Property 3: For any two real numbers x and y, prove that |x.y| = |x|.|y|.
Proof: Since x and y are any two real numbers, therefore, either both are positive or one is
positive and the other is negative or both are negative i.e. either x  0, y  0 or x  0, y  0 or
x  0, y  0 or x  0, y  0. We discuss the proof for all the four possible cases separately.

Case (i): When x  0, y  0.
Since x  0, therefore, we have, by definition,
|x| = x, |y| = y

Also x  0, y  0 simply that xy 0 and hence
|xy| = xy = |x||y|

which proves the property.
Case (ii): When x  0, y  0. Then obviously xy  0. Consequently by definition, it, follows that
|x| = x, |y| = –Y, |xy| = –xy

Hence
|xy| = –xy = x(–y) = |x||y|

which proves the property.
Case (iii): When x  = 0, y 0.
Interchange x and y in (ii).

Case (iv): When x  0, y  0, then xy = 20. Accordingly, we have
|x| = –x, |y| = –y, |xy| = xy.
Hence

|xy| = xy = (–x)(–y) = |x||y|
using the field properties stated. This concludes the proof of the property.

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