If $a I am guessing that I can say y
$y-a Also we can say that $a-x<0$ So if we add these, we get: $y-x WLOG, $x-y Does this look ok or am I missing some steps?
Thank you for your help!
Analysis Question regarding absolute value
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real-analysis
3 Answers
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Without loss of generality, suppose that $y \leq x$. Then
$x - y \leq b - a$
since $x < b$ and $a < y$.
If $x \leq y$, then by the same argument
$y - x = -(x - y) \leq b - a$.
Since $a < b$, $|b - a| = b- a$. Hence in both cases
$|x - y| = |-(x - y)| \leq |b - a| = b - a$
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$$a
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The absolute value of the difference of two numbers has a geometric interpretation: that's the Euclidean distance between two points in one dimension. Imagine a line with four points on it: $a, b$ and $x, y$ between them. Then the distance from $x$ to $y$ $(|x - y|)$ is shorter than from $a$ to $b$ $(|a-b| = b - a$, as $b > a$ in this case$)$.