How do you simplify $abs(-14)$ ?

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Answer 1 · 2015-12-22T20:58:37

$abs(-14) = 14$

If $x$ is a Real number then $abs(x)$ can be defined as follows:

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

In our case $-14 < 0$ so $abs(-14) = -(-14) = 14$

If $z$ is a Complex number then $abs(z)$ can be defined as follows:

$abs(z) = sqrt(z bar(z))$

where $bar(z)$ is the Complex conjugate of $z$ .

For example:

$abs(3+4i) = sqrt((3+4i)(3-4i)) = sqrt(3^2+4^2) = sqrt(9+16)$

$= sqrt(25) = 5$

This gives the same value as the above definition for Real numbers, if $z$ happens to be Real.

How do you simplify abs(-14)abs(-14)?