# Proof: $|y|<x \Leftrightarrow -x<y<x$ [duplicate]

How do you prove $$|y|

I tried using the fact that $$|a|=\sqrt{a^{2}}$$ but it got messy.how would I go about proving this

• use the definition of the absolute value function. Jul 22, 2020 at 13:38
• Rather than using $|a|=\sqrt{a^2}$ it is far cleaner to use $|a| = \begin{cases} a&\text{if }a\geq 0\\ -a&\text{if }a<0\end{cases}$ Jul 22, 2020 at 13:45
• Does this answer your question? prove that |a| < b if and only if -b < a < b Jul 22, 2020 at 13:55

OK so let's assume $$|y|

$$y$$ is either positive or negative. If it is positive then the assumption just gives us $$y. Now $$x$$ must also be positive (or actually non-negative) because it is greater than $$y$$ and so $$-x$$ is negative and so, for sure, $$-x. Now what if $$y$$ is negative? Again $$x$$ must be non-negative because it is greater than $$|y|$$ so certainly $$y < x$$. Now we need to show that $$-x < y$$ well we have $$|y| < x$$. If we multiply both sides by $$-1$$ this flips the inequality and we get $$-x < y$$ (because $$y$$ is negative, $$-|y|=y$$).

Now we assume $$-x. $$x$$ must be positive (as we know $$-x < x$$). This tells us that $$|y| < x$$ as if $$y$$ is positive this is just $$y < x$$ (which we already know) and if $$y$$ is negative then we multiple $$-x < y$$ by $$-1$$ to give $$-y < x$$ and, of course, $$-y=|y|$$ as $$y$$ is negative.

It is often helpful to use proof by cases with $$|y|$$ as when $$y$$ is positive $$|y|=y$$ and when $$y$$ is negative $$|y|=-y$$. These two cases cover all of the possibilities. If $$y$$ is $$0$$ then both hold :)

Suppose $$|y|, note that $$y\le|y| and $$-y\le|y|, thus $$-y. Or you can also check by definition of absolute value, but I think the facts($$y\le|y|$$ and $$y\le|y|\forall y\in \Bbb R)$$ here I used is worth knowing.

Suppose$$-x, if $$0\le y$$, then $$|y|=t. if $$y<0$$, then $$|y|=-y

Definition of absolute value: $$|x|=\left\{\begin{array}{ll} x & \text { if } x \geq 0 \\ -x & \text { if } x<0 \end{array}\right.$$

First note that $$\left| a \right| \ge 0$$ for any $$a \in \mathbb{R}$$ by the definition. So if we know $$\left| y \right| < x$$ then $$x > 0$$

$$\Rightarrow$$ Assume $$\left| y \right| < x$$

• If $$y \ge 0$$ then we have $$\left| y \right| = y$$ and $$y < x$$, and since $$x > 0$$ then $$-x < 0 \le y$$ and we have $$-x < y < x$$
• If $$y < 0$$ then we have $$\left| y \right| = -y$$ and $$-y < x$$, and so $$y > -x$$, we know that $$y$$ is negative and so it's surely less than $$x$$ which is 0 or positive. so we can conclude that $$y < x$$, thus we have. $$-x < y < x$$

$$\Leftarrow$$ Assume that $$-x < y < x$$

• If $$y \ge 0$$ then we have $$\left| y \right| = y$$ and so $$-x < \left| y \right| < x$$
• If $$y < 0$$ then we have $$\left| y \right| = -y$$ and so $$-x < y < x$$ and so $$x > -y > -x$$ thus we have $$x > \left| y \right| > -x$$

In either case we have $$x > \left| y \right|$$

Thus we've proven the statement

You know $$-|y| \leq y \leq |y|$$ and similarly $$-|y| \leq -y \leq |y|$$ it follows that $$x \geq |y| \geq y$$ and similarly $$x \geq |y| \geq -y$$. The second inequality can be rewritten as $$-x \leq y$$ which combined with the first inequality gives $$-x \leq y \leq x$$ as desired.