# Why does the second term require $|x|\leq|y|$ not $|x|\leq y$?

The book says that the relation $\sqrt{x^2-y^2} + \arccos\dfrac{x}{y} = 0$ where $y \neq 0$ implies $\left|x\right| \geq \left|y\right|$ and $|x| \leq |y|$.

I understand that $x^2 - y^2 \geq 0$ from where $x^2 \geq y^2 \implies |x| \geq |y|$. But I cannot understand how to get the second constraint. What I can get is that $-1 \leq \dfrac{x}{y} \leq 1 \implies |x| \leq y$.

• What I can get is ... Only if $y \gt 0$, otherwise you get $|x| \le -y$. Now combine the two. – dxiv Jun 25 '18 at 18:20
• $y^2$ is never negative, regardless of whether or not $y$ is positive or negative. $|x|$ is never negative regardless of whether or not $y$ is positive or negative or if $x$ is positive or negative. Tell me... is it possible for $|x|<y$ in the case that $y$ is negative? – JMoravitz Jun 25 '18 at 18:21
• $-1\le \frac xy\le 1\Rightarrow \left|\frac xy\right|\le 1$ – AnotherJohnDoe Jun 25 '18 at 18:23
• In an attempt to rephrase what other users have said: $\left(-1\leq \frac{x}{y}\leq 1\right)\implies \begin{cases} -y\leq x\leq y&\text{in the case that}~y>0\\y\leq x\leq -y&\text{in the case that}~y<0\end{cases}$. It is not always true that $-1\leq \frac{x}{y}\leq 1\implies -y\leq x\leq y$ in every case as evidenced by pairs such as $(3,-6)$ or $(-2,-6)$ etc... If we say an implication like this, we must either specify in what cases it actually works, or we must make the statement actually true for all cases. – JMoravitz Jun 25 '18 at 18:43
• It is eligible for a separate individual inequality, but only with the caveat that it reminds the reader that it only works in certain restrictive scenarios, what those restrictive scenarios are, and that there are feasible scenarios where it doesn't work. – JMoravitz Jun 25 '18 at 18:48

Given that you accept $-1 \le \frac xy \le 1$ from the $\arccos$ function you need $|x| \le |y|$ to satisfy it. Your version of $|x| \le y$ would fail for $x=-1,y=-2$ but $\arccos \frac {(-1)}{(-2)}$ is perfectly well defined.
$-1 \le \frac xy \le 1$ then if $y > 0$ then $-y \le x \le y \implies |x| \le y=|y|$.
And if $y < 0$ then $-y \ge x \ge y \implies |x| \le -y = |y|$.