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.