Subjects algebra

Floor Absolute Sets

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1. **Problem 1:** Solve the equation $[x] + x + [x] = 1$, where $[x]$ denotes the floor function (greatest integer less than or equal to $x$). 2. **Step 1:** Let $n = [x]$, then the equation becomes $n + x + n = 1$ or $2n + x = 1$. 3. **Step 2:** Since $n = [x]$, we have $n \leq x < n+1$. 4. **Step 3:** Substitute $x = 1 - 2n$ from the equation into the inequality: $$n \leq 1 - 2n < n + 1$$ 5. **Step 4:** Solve the inequalities: - Left: $n \leq 1 - 2n \Rightarrow 3n \leq 1 \Rightarrow n \leq \frac{1}{3}$ - Right: $1 - 2n < n + 1 \Rightarrow -2n < n \Rightarrow -3n < 0 \Rightarrow n > 0$ 6. **Step 5:** Since $n$ is an integer, $0 < n \leq \frac{1}{3}$ implies $n = 0$. 7. **Step 6:** For $n=0$, $x = 1 - 2(0) = 1$ and check $[1] + 1 + [1] = 1 + 1 + 1 = 3 \neq 1$, so no solution here. 8. **Step 7:** Check $n=0$ carefully: $n=0$ means $0 \leq x < 1$, but $x=1$ is not in $[0,1)$, so no solution. 9. **Step 8:** Try $n=0$ with $x$ in $[0,1)$, then $x = 1 - 2(0) = 1$ is outside the interval, so no solution. 10. **Step 9:** Try $n=1$: - Check inequalities: $1 \leq 1 - 2(1) = -1 < 2$ is false. 11. **Step 10:** Try $n=-1$: - $-1 \leq 1 - 2(-1) = 3 < 0$ is false. 12. **Step 11:** No integer $n$ satisfies the inequalities, so no solution for problem 1. --- 13. **Problem 2:** Solve $|x-1| + |x-1| = 3$. 14. **Step 1:** Simplify: $2|x-1| = 3 \Rightarrow |x-1| = \frac{3}{2}$. 15. **Step 2:** Solve absolute value: $$x - 1 = \frac{3}{2} \quad \text{or} \quad x - 1 = -\frac{3}{2}$$ 16. **Step 3:** Solutions: $$x = 1 + \frac{3}{2} = \frac{5}{2}$$ $$x = 1 - \frac{3}{2} = -\frac{1}{2}$$ --- 17. **Problem 3:** Given sets $C = \{1, 2\}$, $B = \{2, 3\}$, $A = \{1, 2, 3\}$, and $M = \{1, ..., 1\}$ (likely a typo or singleton set $\{1\}$), find: - $A \times B$ (Cartesian product) - $(A - B) \cap C$ 18. **Step 1:** Compute $A \times B$: $$A \times B = \{(1,2), (1,3), (2,2), (2,3), (3,2), (3,3)\}$$ 19. **Step 2:** Compute $A - B$ (elements in $A$ not in $B$): $$A - B = \{1\}$$ 20. **Step 3:** Compute $(A - B) \cap C$: $$\{1\} \cap \{1, 2\} = \{1\}$$ --- **Final answers:** - Problem 1: No solution. - Problem 2: $x = \frac{5}{2}$ or $x = -\frac{1}{2}$. - Problem 3: - $A \times B = \{(1,2), (1,3), (2,2), (2,3), (3,2), (3,3)\}$ - $(A - B) \cap C = \{1\}$