Question

For \(x \in \mathbb{R}\), define \[ \langle x \rangle = x-[x] \] where \([x]\) denotes the greatest integer less than or equal to \(x\). Then for any arbitrary pair of real numbers \(x,y \in \mathbb{R}\), which of the following is always true?

• A. \(\langle x+y \rangle \le \langle x \rangle+\langle y \rangle\)
• B. \(\langle x+y \rangle \ge \langle x \rangle+\langle y \rangle\)
• C. \(\langle x+y \rangle = \langle x \rangle+\langle y \rangle\)
• D. \(\langle x+y \rangle \ne \langle x \rangle+\langle y \rangle\)

Hint:

For fractional part functions, \[ \langle x+y \rangle= \begin{cases} \langle x \rangle+\langle y \rangle, & \text{if } \langle x \rangle+\langle y \rangle<1,\\ \langle x \rangle+\langle y \rangle-1, & \text{if } \langle x \rangle+\langle y \rangle\ge1. \end{cases} \] This identity is extremely useful in problems involving greatest integer and fractional part functions.

Answer 1

The Correct Option is A

Concept: The quantity \[ \langle x \rangle = x-[x] \] represents the

fractional part of the real number \(x\). Since \([x]\) is the greatest integer less than or equal to \(x\), the fractional part always satisfies \[ 0 \le \langle x \rangle < 1. \] Similarly, \[ 0 \le \langle y \rangle < 1. \] To determine the correct relationship between \(\langle x+y \rangle\) and \(\langle x \rangle+\langle y \rangle\), we express \(x\) and \(y\) in terms of their integer and fractional parts and then analyze the resulting expression carefully.

Step 1: Express \(x\) and \(y\) in terms of their integer and fractional parts. Using the definition of fractional part, \[ x=[x]+\langle x \rangle \] and \[ y=[y]+\langle y \rangle. \] Adding these two equations, we obtain \[ x+y=[x]+[y]+\langle x \rangle+\langle y \rangle. \] This decomposition separates the integer part from the fractional contribution.

Step 2: Analyze the sum of the fractional parts. Since \[ 0\le \langle x \rangle<1 \] and \[ 0\le \langle y \rangle<1, \] their sum satisfies \[ 0\le \langle x \rangle+\langle y \rangle<2. \] Therefore, there are only two possible cases: \[ 0\le \langle x \rangle+\langle y \rangle<1 \] or \[ 1\le \langle x \rangle+\langle y \rangle<2. \] We examine each case separately.

Step 3: Case 1 --- When \(\langle x \rangle+\langle y \rangle<1\). In this case, the sum of the fractional parts is itself a fractional number. Therefore, \[ \langle x+y \rangle = \langle x \rangle+\langle y \rangle. \] Thus, equality holds.

Step 4: Case 2 --- When \(\langle x \rangle+\langle y \rangle\ge1\). When the sum of the fractional parts reaches or exceeds \(1\), one extra integer is generated. Hence, \[ \langle x+y \rangle = \langle x \rangle+\langle y \rangle-1. \] Since \(1>0\), \[ \langle x+y \rangle < \langle x \rangle+\langle y \rangle. \] Therefore, in this case the fractional part of the sum is strictly less than the sum of the fractional parts.

Step 5: Combine both cases. From Case 1, \[ \langle x+y \rangle = \langle x \rangle+\langle y \rangle. \] From Case 2, \[ \langle x+y \rangle < \langle x \rangle+\langle y \rangle. \] Thus, in every possible situation, \[ \boxed{\langle x+y \rangle \le \langle x \rangle+\langle y \rangle}. \] Hence the correct option is \[ \boxed{(A)\ \langle x+y \rangle \le \langle x \rangle+\langle y \rangle}. \]