Question

The solution set for $[x]\{x\} = 1$ [where $\{x\}$ and $[x]$ are respectively fractional part and greatest integer function.] is:

• A. $R^+ - (0,1)$
• B. $R^+ - \{1\}$
• C. $\{m + \frac{1}{m} | m \in I - \{0\}\}$
• D. $\{m + \frac{1}{m} | m \in N - \{1\}\}$

Answer 1

Answer: (D)

$\{m + \frac{1}{m} | m \in N - \{1\}\}$

Answer 2

To solve the equation $[x]\{x\} = 1$, we use the definitions of the greatest integer function $[x]$ and the fractional part function $\{x\}$. We know that any real number $x$ can be uniquely written as $x = [x] + \{x\}$, where $[x]$ is an integer and $0 \le \{x\} < 1$.

Let $[x] = n$ and $\{x\} = f$. The given equation becomes $n \cdot f = 1$.

From the properties of $n$ and $f$:

1. $n$ must be an integer.
2. $0 \le f < 1$.

From $n \cdot f = 1$, we can express $f$ in terms of $n$: $f = \frac{1}{n}$.

Now, substitute this expression for $f$ into the inequality $0 \le f < 1$: $0 \le \frac{1}{n} < 1$.

We need to consider different cases for the integer $n$:

Case 1: $n = 0$. If $n = 0$, the equation $n \cdot f = 1$ becomes $0 \cdot f = 1$, which simplifies to $0 = 1$. This is a contradiction, so $n$ cannot be $0$.

Case 2: $n < 0$ (i.e., $n$ is a negative integer). If $n$ is a negative integer, then $\frac{1}{n}$ will be a negative number. However, the fractional part $f$ must satisfy $f \ge 0$. Since $\frac{1}{n} < 0$ for $n < 0$, there are no solutions in this case.

Case 3: $n > 0$ (i.e., $n$ is a positive integer). If $n$ is a positive integer, then $\frac{1}{n}$ will be positive, so $0 \le \frac{1}{n}$ is always satisfied. Now we need to satisfy the second part of the inequality: $\frac{1}{n} < 1$. Since $n$ is positive, we can multiply both sides by $n$ without changing the direction of the inequality: $1 < n$.

So, $n$ must be an integer greater than 1. This means $n \in \{2, 3, 4, \dots\}$.

For these values of $n$, the corresponding fractional part is $f = \frac{1}{n}$. The value of $x$ is $x = [x] + \{x\} = n + f = n + \frac{1}{n}$.

Therefore, the solution set for $x$ is $\{n + \frac{1}{n} \mid n \in \{2, 3, 4, \dots\}\}$.

Thus, the solution set is $\{m + \frac{1}{m} | m \in N - \{1\}\}$.