Why is the absolute value of $| e - π |$ $|e-pi |$ = $π - e$ $pi-e$ ?

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Why is the absolute value of $| e - π |$ $|e-pi |$ = $π - e$ $pi-e$ ?

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Answer 1 · 2018-02-27T23:47:43

It can be either $e - π$ $e-pi$ or $π - e$ $pi-e$ . See below.

Explanation:

Recall that

$| x | = \pm x$ $|x|=+-x$ , because $| x | = x$ $|x|=x$ and $| - x | = x$ $|-x|=x$ (the absolute value of a negative number becomes that same number, but positive).

Therefore, $| e - π | = \pm (e - π)$ $|e-pi|=+-(e-pi)$

$\pm (e - π)$ $+-(e-pi)$ can be either:

$+ (e - π) = e - π$ $+(e-pi)=e-pi$

or

$- (e - π) = - e + π = π - e$ $-(e-pi)=-e+pi=pi-e$

Answer 2 · 2018-02-28T01:01:51

Because $e < π$ $e < pi$

Explanation:

The absolute value of a number is essentially its non-negative distance from $0$ $0$ .

So:

$abs(x) = { (x " if " x >= 0), (-x " if " x < 0) :}$

With $e ~~ 2.71828$ and $pi ~~ 3.14159$ we have:

$e < pi$

and hence:

$e - pi < 0$

So:

$abs(e-pi) = -(e-pi) = pi - e$

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Why is the absolute value of $| e - π |$ $|e-pi |$ = $π - e$ $pi-e$ ?