Verify that |x+y|<|x|+|y| by taking x= 2/3, y= -3/5.
Given: x= $\frac{2}{3}$, y= $\frac{-3}{5}$
To do: Verify that |x+y|<|x|+|y|
Solution:
|x+y|:
x + y = $\frac{2}{3} + \frac{-3}{5} = \frac{1}{15}$
|x + y| = $\frac{1}{15}$
|x| + |y|:
|x| = |$\frac{2}{3}$| = $\frac{2}{3}$
|y| = |$\frac{-3}{5}$| = $\frac{3}{5}$
|x| + |y| = $\frac{2}{3} + \frac{3}{5} = \frac{19}{15}$
So, we can observe that |x + y| < |x| + |y|
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