-1
$\begingroup$

In the document accessible at the following link [1]: https://i.stack.imgur.com/P2AvU.jpg, I've tried to explain through purely logical means ( mainly DeMorgan's law) why a $<$ absolute value inequality must be translated by an $\text{AND}$ statement :

$\endgroup$
  • 2
    $\begingroup$ Have you considered a specific example? For instance, if $|x|<3$, what is the possible range of valid $x$ values? $\endgroup$ – Arthur Mar 17 at 20:40
  • 1
    $\begingroup$ The question is simple but a little bit Tricky: IMO it is not possible ti use only distributivity (see your comment below). We have a conditional definition of abs that - when used with the inequality - gives : $( x \ge 0 \land x < N) \lor (x < 0 \land -x < N)$. $\endgroup$ – Mauro ALLEGRANZA Apr 10 at 15:12
  • 1
    $\begingroup$ Consider 1st disjunct : from $x \ge 0$ we derive (using properties of numbers) $x > -N$. Thus we have : $(x < N) \text { and } (x > -N)$. $\endgroup$ – Mauro ALLEGRANZA Apr 10 at 15:14
  • 1
    $\begingroup$ 2nd disjunct : from $-x < N$ we derive (using properties of numbers) : $x > -N$. And from $x < 0$ we have (again by arithmetic) : $x < N$. Thus we have : $(x < N) \text { and } (x > -N)$. $\endgroup$ – Mauro ALLEGRANZA Apr 10 at 15:16
  • 1
    $\begingroup$ Now we may conclude by logic alone, having derived the same result under both disjuncts, using disjunction elimination. $\endgroup$ – Mauro ALLEGRANZA Apr 10 at 15:18
0
$\begingroup$

We have $|x|<N$. Whenever you have an absolute value you can do a case by case investigation.

Case 1: Assume $x>0$ then $|x|=x \implies x<N$. Combining $x>0$ and $x<N$ yields $0<x<N$.

Case 2: Assume $x\leq 0$ then $|x|=-x \implies -x<N \implies -N < x$. Combining $x\leq 0 $ and $-N<x$ results in $-N<x<0$.

If we combine both cases we get $|x|<N$ is equivalent to $-N<x<N$, which is equivalent to $-N<x \text{ AND } x<N$.

$\endgroup$
  • $\begingroup$ $-N<x \text{ OR } x<N$ would imply that $x = 2N$ (if we look at $-N<x$) is also a feasible solution. Which is not the case. Both conditions need to be satisfied for any feasible solution. $\endgroup$ – MachineLearner Mar 17 at 21:10
  • $\begingroup$ @RayLittlerock You are welcome :). $\endgroup$ – MachineLearner Mar 17 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.