Question

Verify that |a+b| <_ |a| + |b| .Take a=1/3 , b = -2/9
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Answer 1

Step-by-step explanation:

a + b = $\frac{1}{3} + \frac{-2}{9}$ = $\frac{(3*3) - 2}{9}$ = $\frac{9-2}{9}$ = $\frac{7}{9}$

LHS = |a+b| = | $\frac{7}{9}$| =  $\frac{7}{9}$

|a| = |$\frac{1}{3}$ | =  $\frac{1}{3}$    and  |b| =  | $\frac{-2}{9}$| =  $\frac{2}{9}$

RHS = |a| + |b| =  $\frac{1}{3}$ + $\frac{2}{9}$ =  $\frac{(3*3) + 2}{9}$ =  $\frac{9+2}{9}$ = $\frac{11}{9}$

$\frac{7}{9}$ <  $\frac{11}{9}$

|a+b| < |a| + |b|

Hence proved.