My question is:

Solve: $|x-4|< a$, where $a$ belongs to the real numbers. Solve this by considering various cases depending upon whether $a$ is negative, positive or zero.

What I have tried so far: If $a>0$ then: $x < a+4$ and $x>4-a$, if $a=0$ then there is no solution.

My doubt is: Should I consider the case $a<0$ as again $|x-4|<a$ which is not possible as absolute value cannot be negative.

  • $\begingroup$ Note that if $a \le 0$, there is no $x$ such that $|x-4|\lt a$, because $|w|$ is always $\ge 0$. That will com out if you use the general machinery described by robjohn, but you might as well handle it separately. $\endgroup$ – André Nicolas Jun 2 '12 at 19:07
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    $\begingroup$ Probably the hint should be: depending on whether $x-4$ is negative, positive, or zero. $\endgroup$ – GEdgar Jun 3 '12 at 23:29

Hint: $|x-4|<a$ means that $x$ is closer than $a$ units to $4$.

Another Hint: $|x-4|<a$ means that $(x-4)<a$ and $-(x-4)<a$.

  • $\begingroup$ I really dont understand these modulus based questions. plz can you try to show it on a number line? $\endgroup$ – mgh Jun 2 '12 at 14:27
  • $\begingroup$ There is no modulus here; there is an absolute value. Try this: draw a number line from $0$ to $10$, then color the part of the number line that is closer than $2$ units to $4$. Try the same thing for the part closer than $3$ units to $4$. $\endgroup$ – robjohn Jun 2 '12 at 14:33
  • $\begingroup$ @ robjohn ya I did that $\endgroup$ – mgh Jun 2 '12 at 14:43
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    $\begingroup$ "modulus" is another word for "absolute value", robjohn. Yes, the same word is used in connection with congruences, but there are only so many words to go around, some of them have to work overtime. $\endgroup$ – Gerry Myerson Jun 4 '12 at 0:52
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    $\begingroup$ @Gerry: I see that in the link I provided, it says "In mathematics, the absolute value (or modulus)". I apologize for being so narrow-minded. $\endgroup$ – robjohn Jun 4 '12 at 8:01

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