Yes.

- Sep 23rd 2012, 12:07 PMemakarovRe: Solving inequalities with absolute values within absolute values
Yes.

- Sep 23rd 2012, 12:08 PMBobRossRe: Solving inequalities with absolute values within absolute values
Okay I get it now, thanks a lot!!

- Sep 23rd 2012, 05:32 PMHallsofIvyRe: Solving inequalities with absolute values within absolute values
The absolute value function is

**continuous**and so is any combination of absolute value functions. And since such a function is continuous, the only place it can change from ">" to "<" is where it is "=". That is, a very good way to solve inequalities involving absolute values (or any complicated but continuous functions) is to solve the associated**equation**to separate the real line into intervals. You can then determine which of the intervals give ">" and which "<" by evaluating at one point in each interval.

For example, the inequality |||x - 1| - 1| - 1| >= 4 has the associated equation |||x-1|- 1|- 1|= 4. Removing the "outer" absolute value, we must have ||x-1|- 1|- 1= 4 or ||x-1|-1|- 1= -4.

a) ||x-1|- 1|- 1= 4 is the same as ||x-1|- 1|= 5. Removing the next absolute value, either |x-1|- 1= 5 or |x-1|- 1= -5.

a1) |x- 1|- 1= 5 is the same as |x-1|= 6. Removing the last absolute value, either x- 1= 6 or x- 1= -6.

a1I) x- 1=6 gives x= 7.

a1II) x- 1= -6 gives x= -5.

a2) |x- 1|- 1= -5 is the same as |x- 1|= -4. That's impossible, of course.

b) ||x-1|-1|- 1= -4 is the same as ||x-1|- 1|= -3. That's impossible.

So the only solutions to the**equation**are x= 7 and x= -5. Those two points divide the real number line into three intervals:

i) x< -5. x=-6 is in that set. |||-6-1|- 1|- 1|= |||-7|- 1|- 1|= ||7-1|- 1|= |6- 1|= 5 which is larger than 4 so the inequality is true for all x less than or equal to -5.

ii) -5< x< 7. x= 0 is in that set. |||0- 1|- 1|- 1|= |||-1|- 1|- 1|= ||1- 1|- 1|= |0 - 1|= -1 which is less than 4 so the inequality is false for all x between -5 and 7.

iii) x> 7. x= 8 is in that set. |||8- 1|- 1|- 1|= ||7- 1|- 1|= |6- 1|= 5 which is larger than 4. The inequality is true for all x larger than or equal to 7.

||x-1|-1|- 1= -4. - Sep 23rd 2012, 07:21 PMjohnsomeoneRe: Solving inequalities with absolute values within absolute values
This thread has algebraic derivations - and I showed before a solution using distances.

Here's yet another way to do it:

Think of the set of x satisfying |||x-1|-1|-1|>=4 as the complement of the set of x satisfying |||x-1|-1|-1|<4. (Every real x will satisfy exactly one of those two inequalities.)

The point of this is that it's algebraically easier to work with |U| < V than it is to work with |U| > V.

That's because of this (you should know this by heart!): |Z - W| < K if and only if -K < Z - W < K.

Since it will be used twice in what follows, also convince yourself of this trivially true but perhaps odd looking statement:

"Given a and b positive reals, then for any real Y: -a < |Y| < b if and only if |Y| < b."

Now, verify that each line is true if and only if the line above it is true:

|||x-1|-1|-1|<4

-4 < ||x-1|-1|-1 < 4

-3 < ||x-1|-1| < 5

||x-1|-1| < 5

-5 < |x-1|-1 < 5

-4 < |x-1| < 6

|x-1| < 6

-6 < x-1 < 6

-5 < x < 7

Therefore the set of x such that |||x-1|-1|-1|<4 is the set: -5 < x < 7, or, in interval notation (-5, 7).

Therefore the set of x such that |||x-1|-1|-1|>=4 is the complement of that set, which is: x<=-5 or 7<= x. In interval notation (-infinity, -5] Union [7, infinity).