First slide

Introduction to sets

Question

If $[\frac{x}{2}] + [\frac{x}{3}] = \frac{5 x}{6}$ , then $x$ is any term of
the following

Moderate

A

$3, 6, 9, 12, \dots$
B

$9, 18, 27, 36, \dots$
C

$6, 12, 18, 24, \dots$
D

$\frac{6}{5}, \frac{12}{5}, \frac{18}{5} . . .$

Solution

Since $[\frac{x}{2}], [\frac{x}{3}] \in I$ , so $[\frac{x}{2}] + [\frac{x}{3}]$ $\in I$
Thus $\frac{5 x}{6} \in I \Rightarrow x = \frac{6}{5} n, n \in I$
Substituting this value in $[\frac{x}{2}] + [\frac{x}{3}]$ $= \frac{5 x}{6}$ , we have
$\begin{array}{l} [\frac{3}{5} n] + [\frac{2}{5} n] = n \\ \Rightarrow \frac{3}{5} n - \{\frac{3}{5} n\} + \frac{2}{5} n - \{\frac{2}{5} n\} = n \\ \Rightarrow \{\frac{3}{5} n\} + \{\frac{2}{5} n\} = 0 \\ \Rightarrow \{\frac{3}{5} n\} = 0 = \{\frac{2}{5} n\} (Since 0 \leq {x} < 1) \end{array}$
Thus $3 n = 5 m_{1}$ , $2 n = 5 m_{2}$ Therefore $x n = \frac{2 n . 3 n}{5}$
$\begin{array}{l} = \frac{5 m_{1} \times 5 m_{2}}{5} = 5 m_{1} m_{2} \Rightarrow \frac{x \cdot 5 m_{1}}{3} = 5 m_{1} m_{2} \\ \Rightarrow x = 3 m_{2} \end{array}$
Similarly $\frac{x . 5 m_{2}}{2} = 5 m_{1} m_{2} \Rightarrow x = 2 m_{1}$
Hence $x$ is multiple of $2$ and $3$ so of $6$ and $x \in I$

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