First time posting on here. Started a class with proof based calculus using Spivak's Calculus.
This problem is from the first chapter on proving absolute values.
Below is my attempt at a solution, there is a hint under the question which says to look at the exponent but I couldn't figure it out doing that.
Can anyone guide me here, I'm doing a second degree in applied mathematics after being away from school for almost a decade 😥
$$\require{cancel} \begin{aligned}|x| &=\sqrt{x^{2}} \text { for all } x \in R \\\left|\frac{1}{x}\right| &=\frac{1}{|x|} \\ \cancel{\sqrt{x^2} }\left(\frac{\sqrt{(1)^{2}}}{\cancel{\sqrt{x^{2}}}}\right) &=\left(\frac{1}{\cancel{\sqrt{x^{2}}}}\right)(\cancel{\sqrt{x^2}}) \\ \sqrt{(1)^{2}} &=1 \\ 1 &=1 \end{aligned}$$
Best Answer
Case 1: $x=0$
$\frac{1}{x}$ is undefined.
LS = $|\frac{1}{x}|$. This is undefined since $\frac{1}{x}$ is undefined.
RS = $\frac{1}{|x|} = \frac{1}{|0|} = \frac{1}{0}$. This is also undefined.
So both sides are undefined.
Case 2: $x>0$
$\frac{1}{x} > 0$. This is because a positive number, divided by a positive number, is always positive. (not sure if this requires some proof also).
LS = $|\frac{1}{x}| = \frac{1}{x}$ since $\frac{1}{x}$ is positive.
RS = $\frac{1}{|x|} = \frac{1}{x}$ since $x$ is positive.
LS = RS.
Case 3: x<0
$\frac{1}{x} < 0$. This is because a positive number divided by a negative number is negative.
LS = $|\frac{1}{x}| = -\frac{1}{x}$, since $\frac{1}{x}<0$
RS = $\frac{1}{|x|} = \frac{1}{-x}$ since $x<0$
Then we can bring the negative outside the fraction
RS = $-\frac{1}{x}$
LS = RS.
So in all 3 cases the left and right sides are equal.