Exercise 54

The set of points on the real number line whose distances to point 3 are smaller than their distances to point 5 is:

  1. \( -\infty , \, 3 \)

  2. \( (4 , +\infty) \)

  3. \( (4 , \, 5) \)

  4. \( (-\infty , \, 4) \cap (5, +\infty) \)

  5. \( (-\infty , \, 4) \)

Try to solve it before checking the answer.

  1. \( (-\infty, \, 4) \)

Let \(x\) be a point of such set. According the statement of the problem, we have:

\[ \boldsymbol{(1)} \hspace{3em} |x – 3| < |x - 5| \]

Points 3 and 5 divide the line into three intervals that will allow us to solve equation \( \boldsymbol{(1)} \) considering three cases.

Los intervalos son: \( (-\infty, \, 3) \),   \( [3, \, 5) \)   y   \( [ 5, +\infty ) \)

Case 1. \(\boldsymbol{x}\) in the interval \(\boldsymbol{ (-\infty, \, 3) }\); that is \(\boldsymbol{ x < 3 }\):

\[ \begin{aligned} x < 3 &\Rightarrow x < 3 \quad \text{ and } \quad x < 5 \\[2em] &\Rightarrow x - 3 < 0 \quad \text{ and } \quad x - 5 < 0 \end{aligned} \]

Entonces:

\(|x – 3| = -(x – 3) \)

and

\(|x – 5| = -(x – 5)\)

\[ \begin{aligned} x < 3 &\Rightarrow x < 3 \quad \text{ and } \quad x < 5 \\[2em] &\Rightarrow x - 3 < 0 \quad \text{ and } \quad x - 5 < 0 \\[2em] &\Rightarrow |x - 3| = -(x - 3) \quad \text{ and } \quad |x - 5| = -(x - 5) \end{aligned} \]

Replacing in \((1)\):

\[ \begin{aligned} – (x – 3) < - (x - 5) &\Leftrightarrow -x + 3 < -x + 5 \\[2em] &\Leftrightarrow 3 < 5 \end{aligned} \]

Since \(3 < 5\) is true, regardless of the value of \(x\), we have that \( x \in (-\infty, +\infty) \).

But we are dealing with case 1, in which \( x \in (-\infty, \, 3) \), so the solution for this case is the intersection of both intervals; that is:

\[ (-\infty, +\infty) \cap (-\infty, \, 3) = \boldsymbol{(-\infty, \, 3)} \]

Case 2. \(\boldsymbol{x}\) in the interval \( \boldsymbol{[3,\,5)} \); that is \(\boldsymbol{3 \leq x < 5 } \):

\[ \begin{aligned} 3 \leq x < 5 &\Rightarrow 3 \leq x \quad \text{ and } \quad x < 5 \\[2em] &\Rightarrow x - 3 \geq 0 \quad \text{ and } \quad x - 5 < 0 \\[2em] &\Rightarrow |x - 3| = x - 3 \quad \text{ and } \quad |x - 5| = - (x - 5) \end{aligned} \]
\[ \begin{aligned} 3 \leq x < 5 &\Rightarrow 3 \leq x \quad \text{ and } \quad x < 5 \\[2em] &\Rightarrow x - 3 \geq 0 \quad \text{ and } \quad x - 5 < 0 \end{aligned} \]

Entonces:

\( |x – 3| = x – 3 \)

and

\( |x – 5| = – (x – 5) \)

Replacing in \((1)\):

\[ \begin{aligned} x – 3 < -(x - 5) &\Leftrightarrow x - 3 < -x + 5 \\[2em] &\Leftrightarrow 2x < 8 \\[2em] &\Leftrightarrow x < 4 \Leftrightarrow x \in (-\infty, \, 4) \end{aligned} \]

Case 3. \(\boldsymbol{x}\) in the interval \( \boldsymbol{ [5, +\infty) } \); that is \(\boldsymbol{ 5 \leq x }\):

\[ \begin{aligned} 5 \leq x &\Rightarrow 3 < x \quad \text{ and } \quad x \geq 5 \\[2em] &\Rightarrow x - 3 > 0 \quad \text{ and } \quad x – 5 \geq 0 \\[2em] &\Rightarrow |x – 3| = x – 3 \quad \text{ and } \quad |x – 5| = x – 5 \end{aligned} \]
\[ \begin{aligned} 5 \leq x &\Rightarrow 3 < x \quad \text{ and } \quad x \geq 5 \\[2em] &\Rightarrow x - 3 > 0 \quad \text{ and } \quad x – 5 \geq 0 \\[2em] &\Rightarrow |x – 3| = x – 3 \quad \text{ and } \quad |x – 5| = x – 5 \end{aligned} \]

Entonces:

\( |x – 3| = x – 3 \)

and

\( |x – 5| = x – 5 \)

Replacing in \((1)\):

\[ x – 3 < x - 5 \Leftrightarrow -3 < -5 \]

Since \( -3 < -5 \) is not true, regardless the value of \(x\), we have that \( x \in \emptyset \), the empty set.

But we are dealing with case 3, in which \( x \in [ 5, +\infty ) \), so the solution for this case is the intersection of both intervals; that is:

\[ [5, +\infty) \cap \emptyset = \emptyset \]

Finally, the overall solution is the union of the solutions to the three cases. That is:

\[ (-\infty, \, 3) \cup [3, \, 4) \cup \emptyset = \boldsymbol{(-\infty, \, 4 )} \]

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