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How do you prove $|y|<x \Leftrightarrow -x<y<x$

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

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

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OK so let's assume $|y|<x$

$y$ is either positive or negative. If it is positive then the assumption just gives us $y<x$. 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, $-x<y$. 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 $-x<y<x$. $x$ 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 :)

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

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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

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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.

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