Page 57 - DMTH401_REAL ANALYSIS

Unit 3: Matric Spaces

Alternatively, the proof can be given by using property 2 in following way: Notes
2
2
|xy| = (xy) = x y = |x| .|y| 2
2
2
2
= (|x|.|y|) 2
Therefore
|xy| =  (|x||y|)
Since |xy|, |x| and |y| are non-negative, therefore we take the positive sign only and we have
|xy| = |x||y|
which proves the property.

You can use any of the two methods to try the following exercise.
The next property is related to the modulus of the sum of two real members. This is one of the
most important properties and is known as Triangular Inequality:

Property 4: Triangular Inequality
For any two real numbers x and y, prove that
|x + y|  |x| + |y|.
Proof: For any two real numbers x and y the number x + y  0 or x + y < 0.

If x + y  0, then by definition
|x + y| = x + y. . . .(1)
Also, we know that

|x| x " xR
|x| x " yR
Therefore
|x| + |y|  x + y

x + y  |x| + |y|. . . .(2)
From (1) and (2), it follows that
|x + y| |x| + |y|

Now, if x + y < 0, then again by definition, we have.
|x + y|= – (x + y)
or |x + y|= (–x) + (–y) . . .(3)
Also we know that (see property 1)
–x  |x| and –y  |y|.

Consequently, we get
(–x) + (–y)  |x|+|y|
or (–x) + (–y)  |x|+|y| . . .(4)

From (3) and (4), we get
|x + y| |x|+|y|

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