Why is the absolute value of 16 not equal to 4?

[basty]

If $\sqrt{x^2} = |x|$, why $\sqrt{16} ≠ |4|$ instead of 4 (please see below image)?

 

[HomogenousCow]

The absolute value of 4 is indeed 4. The absolute value is only important when x is negative.

 

[basty]

I mean if $\sqrt{16} = 4$ why $\sqrt{x^2}$ is not x?

 

[HomogenousCow]

Because when x is negative, $\sqrt{x^2} = -x$

 

[Mark44]

I mean if $\sqrt{16} = 4$ why $\sqrt{x^2}$ is not x?

But is $\sqrt{(-4)^2} = -4$?

 

[basty]

But is $\sqrt{(-4)^2} = -4$?

Yes indeed.

Because $\sqrt{(-4)^2} = (-4)^{\frac{2}{2}} = (-4)^1 = -4$

 

[Mark44]

But is $\sqrt{(-4)^2} = -4$?

Yes indeed.

Absolutely not! $\sqrt{(-4)^2} = \sqrt{16} = 4 = |-4|$

 

[basty]

Absolutely not! $\sqrt{(-4)^2} = \sqrt{16} = 4 = |-4|$

What about this?

$\sqrt{(-4)^2} = (-4)^{\frac{2}{2}} = (-4)^1 = -4$

Isn't from the above shows that $\sqrt{(-4)^2} = -4$?

 

[Mark44]

What about this?

$\sqrt{(-4)^2} = (-4)^{\frac{2}{2}} = (-4)^1 = -4$

Isn't from the above shows that $\sqrt{(-4)^2} = -4$?

No, it doesn't.

The exponent properties you are using apply only to numbers that are nonnegative.

$\sqrt{(-4)^2} = [(-4)^2]^{1/2} = 16^{1/2} = + 4$
The rule that you are misusing says that $(a^m)^n = a^{mn}$, provided that $a \ge 0$.