1. ## Inequality problem

I can´t grasp what the inequality is asking for?
For what positive number (s) is the following statement true?
$\displaystyle Ix-3I<s=>Ix-4I<2$ (those strange I:s represent abs: )

I solved $\displaystyle Ix-4I<2$
That gives me $\displaystyle (-2)<x-4<2$ / add +4
$\displaystyle 2<x<6$
And now I don´t understand what I am suppose to do.

2. Dear Henryt999,

We have to find $\displaystyle S\in{Z^+}$ such that, $\displaystyle \left|x-3\right|< S\Rightarrow\left|x-4\right|<2$

As you have mentioned, $\displaystyle \left|x-4\right|<2\Rightarrow2<x<6$-------------A

Similarly, $\displaystyle \left|x-3\right|< S\Rightarrow-S<x-3<S$

Therefore, $\displaystyle 3-S<x<3+S$--------------B

From A and B, $\displaystyle 3-S\leq{2}$ and $\displaystyle S+3\geq{6}$

Therefore, $\displaystyle S\geq{1}$ and $\displaystyle S\geq{3}$

So we would get, $\displaystyle S\geq{3}$ as the answer.

Hope this helps.

3. Hello, Henryt999!

For what positive number $\displaystyle s$ is the following statements true?

. . $\displaystyle \begin{array}{cccc}|x-3| &<& s & [1]\\ \\[-3mm] |x-4| &<& 2 & [2] \end{array}$
Note that: .$\displaystyle s \,\geq\,0$

From [1], we have: .$\displaystyle -s \:<\:x-3 \:<\:s \quad\Rightarrow\quad -s+3 \:<\:x\:<\:s + 3$ .[3]

From [2], we have: .$\displaystyle -2 \:<\:x-4\:<\:2 \quad\Rightarrow\quad 2 \:<\:x\:<\:6$ .[4]

[3] tells us that $\displaystyle x$ is between $\displaystyle 3-s$ and $\displaystyle 3+s$
Code:

- - - o = = = = = = o - - -
3-s           3+s

[4] tells us that $\displaystyle x$ is between 2 and 6.
Code:

- - o = = = = = = = = o - -
2                 6

For both statements to be true,
. . the first interval must be "inside" the second interval.
Code:

- - - o = = = = = = o - - -
3-s           3+s

- - o = = = = = = = = o - -
2                 6

So we have: .$\displaystyle \begin{array}{ccccccc}3-s \:\geq\:2 & \Rightarrow & -s \:\geq \:-1 & \Rightarrow & s \:\leq\:1 \\ 3+s \:\leq\:6 & \Rightarrow & s \:\leq\:3 \end{array}$

. . which are both true for: .$\displaystyle s \,\leq\,1$

Therefore, the inequalities are true for: .$\displaystyle 0 \,\leq s\,\leq\,1$

4. Dear Sorban,

You had considered the situation where the 3-s<x<3+x is inside 2<x<6. But I have considered the situation where 2<x<6 is inside 3-s<x<3+x. I think both situations must be considered since in both situations, the two inequalities hold.

Therefore the answer could be either, $\displaystyle 0 \,\leq S\,\leq\,1$ or $\displaystyle S\geq{3}\Rightarrow{S}\in[0,1]\cup[3,\infty)$

5. ## Thanks too both of you guys but..

Therefore, the inequalities are true for: .$\displaystyle 0 \,\leq s\,\leq\,1$

Thanks for the help, one part I dont understand.
We get that $\displaystyle s<1$
how did this get too $\displaystyle 0<s<1$?

6. [QUOTE=Henryt999;445369]For what positive number (s) is the following statement true? $\displaystyle |x-3|<s=>|x-4|<2$ [/QUOTE

Originally Posted by Sudharaka
Therefore the answer could be either, $\displaystyle 0 \,\leq S\,\leq\,1$ or $\displaystyle S\geq{3}\Rightarrow{S}\in[0,1]\cup[3,\infty)$
Originally Posted by Henryt999
Therefore, the inequalities are true for: .$\displaystyle 0 \,\leq s\,\leq\,1$
Sudharaka, the solution Sorban gave is correct and is the only solution.
I think that you have misread the question or you failed to see the implication.
$\displaystyle |x-3|<s$ is the set of numbers $\displaystyle (3-s,3+s)$.
$\displaystyle |x-4|<2$ is the set of numbers $\displaystyle (2,6)$.

The implication in the question, $\displaystyle |x-3|<s~{\color{blue}\Rightarrow} |x-4|<2$, means that $\displaystyle x \in \left( {3 - s,3 + s} \right) \Rightarrow ~ x \in \left( {2,6} \right)$.
We know that $\displaystyle 7\in [3,\infty)$ but we cannot have $\displaystyle s=7$.
For then we would have $\displaystyle 0\in (-4,10)$ but $\displaystyle 0 \notin (2,6)$.

You want to find all values of $\displaystyle s$ for which $\displaystyle \left( {3 - s,3 + s} \right) \subseteq \left( {2,6} \right)$ is true.

7. Dear Plato,

What you have said is correct. Now I see the mistake I had made and thank you so much for telling me that.