1. [SOLVED] Absolute distance

"Write it down as an inequality: number $\displaystyle x$ is (on a number line, I guess) less distant from $\displaystyle 1$ (one), than it is from $\displaystyle x^2$."

$\displaystyle |x-1|<|x-x^2|=|x(x-1)|$

For $\displaystyle x\geq1$: $\displaystyle x-1<x-x^2 \Rightarrow x^2<1$ FALSE ($\displaystyle x$ is greater than $\displaystyle 1$)
For $\displaystyle 0\leq x<1$: $\displaystyle -(x-1)<-(x-x^2) \Rightarrow x^2>1$ FALSE ($\displaystyle x$ is less than $\displaystyle 1$ and greater than $\displaystyle 0$)
For $\displaystyle x<0$: $\displaystyle -(x-1)<x-x^2 \rightarrow x^2-2x+1<0 \Rightarrow (x-1)^2<0$ FALSE (always positive)

$\displaystyle \Rightarrow$ inequality has no solutions

Is this absolutely(!) correct?

2. Originally Posted by courteous
"Write it down as an inequality: number $\displaystyle x$ is (on a number line, I guess) less distant from $\displaystyle 1$ (one), than it is from $\displaystyle x^2$."

$\displaystyle |x-1|<|x-x^2|=|x(x-1)|$

For $\displaystyle x\geq1$: $\displaystyle x-1<x-x^2 \rightarrow x^2<1$ FALSE ($\displaystyle x$ is greater than $\displaystyle 1$)
If x > 1 then $\displaystyle x-x^2 < 0$ for all x. Therefore you have to solve the inequality:

$\displaystyle x\geq1$: $\displaystyle x-1<x^2-x \rightarrow 0 < x^2 -2x +1 ~\implies~ 0< (x-1)^2$

This is true for all x > 1

For $\displaystyle 0\leq x<1$: $\displaystyle -(x-1)<-(x-x^2) \rightarrow x^2>1$ FALSE ($\displaystyle x$ is less than $\displaystyle 1$ and greater than $\displaystyle 0$)
For $\displaystyle x<0$: $\displaystyle -(x-1)<x-x^2 \implies x^2-2x+1<0 \rightarrow (x-1)^2<0$ FALSE (always positive)

$\displaystyle \Rightarrow$ inequality has no solutions

Is this absolutely(!) correct?
I'm not quite sure, but in my opinion the last line of your solution isn't correct either. (Unfortunately I haven't spotted the error yet)

By first inspection and playing around with some values I assume that the solution could be:

$\displaystyle x\in \left((-\infty,-1) \cup (1,\infty)\right)$

3. Originally Posted by courteous
"Write it down as an inequality: number $\displaystyle x$ is less distant from $\displaystyle 1$ (one), than it is from $\displaystyle x^2$."
If the problem is to find real numbers x such that the distance from x to 1 is less than the distance from x to $\displaystyle x^2$, then the correct expression is $\displaystyle \left| {x - 1} \right| < \left| {x^2 - x} \right|$.
It is worth noting that $\displaystyle \left| {x - 1} \right| = \left| {1 - x} \right|\;\& \,\left| {x^2 - x} \right| = \left| {x - x^2} \right|$, that is distance is symmetric.
As pointed out above the solution set is $\displaystyle \left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)$.
To see this note that $\displaystyle \left| {x^2 - x} \right| = \left| x \right|\left| {x - 1} \right|$ and if $\displaystyle x \not= 1$ then
$\displaystyle \left| {x - 1} \right| < \left| x \right|\left| {x - 1} \right|\; \Rightarrow \;1 < \left| x \right|$

4. Originally Posted by Plato
If the problem is to find real numbers x such that the distance from x to 1 is less than the distance from x to $\displaystyle x^2$, then the correct expression is $\displaystyle \left| {x - 1} \right| < \left| {x^2 - x} \right|$.[/tex]
So, even as distance is symmetric, then the correct initial expression is only $\displaystyle \left| {x - 1} \right| < \left| {x^2 - x} \right|$. Why (it makes sense, besides, the other way around yields "no solution" result)? What if you've had some hard-line initial conditions, not $\displaystyle x^2$?

5. Originally Posted by courteous
Why (it makes sense, besides, the other way around
Originally Posted by courteous
yields "no solution" result)? What if you've had some hard-line initial conditions, not $\displaystyle x^2$?
$\displaystyle \begin{array}{l} \left| {x - 1} \right| < \left| {x - x^2 } \right| \\ \left| {1 - x} \right| < \left| {x - x^2 } \right| \\ \left| {x - 1} \right| < \left| {x^2 - x} \right| \\ \left| {1 - x} \right| < \left| {x^2 - x } \right| \\ \end{array}$
Each of the above has the solution $\displaystyle ( - \infty , - 1) \cup (1,\infty )$.
The order makes no difference.

6. Originally Posted by courteous
$\displaystyle |x-1|<|x-x^2|=|x(x-1)|$
Indeed! The mischievous $\displaystyle |x-x^2|=|x(x-1)|$ has mislead me. All clear (now).