What is the Modulus of the Square Root of a Number?

[sachin123]

Consider this function:
(x^2)^0.5

My textbook considers this to be |x| (modulus of x).
Is it correct?

x^2 gives a positive quantity.On taking square root of that,we'll get either -x or +x.
So why modulus?

Also,
(x^0.5)^2
This is modulus isn't it?

Thank You

 

[Mentallic]

For (x²)^0.5 consider that if you plug in a value of x=-1, squaring that gives 1 and then taking the square root (since we define the square root of a number to be its positive square root) this will be 1, which is |-1|. So if we plug in any positive number for x, we get back the positive but if we plug in a negative, we get back the negative of that, or its modulus. So the answer is (x²)^0.5=|x|

For the other, yes the answer is just x, but remember that you can't take the square root of a negative number so it's only defined for $x\geq 0$

 

[HallsofIvy]

Consider this function:
(x^2)^0.5

My textbook considers this to be |x| (modulus of x).
Is it correct?

x^2 gives a positive quantity.On taking square root of that,we'll get either -x or +x.

This is wrong. Every operation (or function) on one or more numbers gives a single value. $\sqrt{a}$ is defined as the positive number whose square is a. Think about it: What are the solutions to $x^2= a$? answer: $x= \pm\sqrt{a}$. If $\sqrt{a}$ itself were both plus and minus, you would NOT need to write the "$\pm$".

So why modulus?

Also,
(x^0.5)^2
This is modulus isn't it?

Thank You

No, it is not. If x is already negative then either its square is not defined (it you are talking about real numbers only) and so you cannot do that calculation, or it is an imaginary number whose square is negative.

 

[Mark44]

So why modulus?
Also,
(x^0.5)^2
This is modulus isn't it?

No, it is not. If x is already negative then either its square is not defined (it you are talking about real numbers only) and so you cannot do that calculation, or it is an imaginary number whose square is negative.

I'm sure you meant "... then either its square root is not defined ..."

 

[HallsofIvy]

If, as I said, we are talking about real numbers only, and x= -4, then $(-4)^{1/2}$ "is not defined"- there is no such real number and we simply cannot do that calculation.

If we are talking about complex numbers, then $(-4)^{1/2}= 2i$ and then $((-4)^{1/2})^2= (2i)^2= -4$.