Question

sqrt(x^2)=|x| how can i proof it by real analysis

Answer 1

Conditions

sqrt(x^2) = |x| how can i proof it by real analysis

Solution

To prove this we must consider the definitions of square root and modulus.

In mathematics, a square root of a number $a$ is a number $y$ such that $y^2 = a$, or, in other words, a number $y$ whose square (the result of multiplying the number by itself, or $y \times y$) is $a$. For example, 4 is a square root of 16 because $4^2 = 16$.

In mathematics, the absolute value (or modulus) $|a|$ of a real number $a$ is the non-negative value of $a$ without regard to its sign. Namely, $|a| = a$ for a positive $a$, $|a| = -a$ for a negative $a$, and $|0| = 0$. For example, the absolute value of 3 is 3, and the absolute value of $-3$ is also 3. The absolute value of a number may be thought of as its distance from zero.

Let's fix $x, x \in \mathbb{R}$

1) If $x = 0$, then by property of a square root, $\sqrt{0} = 0 = |0|$

2) If $x \geq 0$, then by property of a square root, since the square-root notation without sign represents the positive square root, it follows that $\sqrt{x^2} = x = |x|$.

3) If $x \leq 0$, then by property of a square root, since the square-root notation without sign represents the positive square root, it follows that $\sqrt{x^2} = -x = |x|$.

As we see, for all $x \in \mathbb{R}$ the domains and ranges of these functions are identical. This means, that the proposition is proved.