Can x be equal to the square root of its own absolute value?

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The discussion centers on proving that the absolute value of x, |x|, is equal to the square root of x squared, sqrt(x^2). Two proofs are presented: the first uses the property of square roots, while the second considers two cases based on the sign of x. The importance of establishing that both x^2 and |x| are non-negative is emphasized, as this is crucial for applying the definition of square roots. The conclusion follows from confirming that a = |x| and b = x^2, satisfying the conditions needed for the proof. Overall, the proofs aim to validate the equality |x| = sqrt(x^2).
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Homework Statement


Prove that |x| = sqrt(x^2)

The Attempt at a Solution


I've written two proofs but I don't know if they can be justified as real proofs or whether they are valid or not.
Proof 1:
\surd x^{2} = \surd \vert x \vert ^{2} = \vert x \vert

Proof 2:
First Case ) Suppose x \geq 0 then \surd x^{2} = x = \vert x \vert
Second Case ) Suppose x < 0 then \surd x^{2} = -x where -x > 0 therefore -x = \vert x \vert
 
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basil32 said:
Prove that |x| = sqrt(x^2)

Let's look at the definition of the square root:

If a2 = b and a ≥ 0, then a = √b. Now look at your problem. What two things do we have to prove?
 
gb7nash said:
Let's look at the definition of the square root:

If a2 = b and a ≥ 0, then a = √b. Now look at your problem. What two things do we have to prove?

That x^{2} \geq 0 and \vert x \vert \geq 0 ?
 
gb7nash said:
If a2 = b and a ≥ 0, then a = √b. Now look at your problem. What two things do we have to prove?

The two bolded things are what you want to prove. Once you have those, then the conclusion follows. Before you do anything, what is a in your problem? What is b? Once you have a and b, what is the first thing you need to prove?
 
gb7nash said:
The two bolded things are what you want to prove. Once you have those, then the conclusion follows. Before you do anything, what is a in your problem? What is b? Once you have a and b, what is the first thing you need to prove?

a = \vert x \vert and b = x^{2}

a^{2} = \vert x \vert ^{2} = x ^ {2} = b
a = \vert x \vert which is nonnegative. correct?
 
Correct.
 
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