Subjects algebra

Floor Inequalities

Step-by-step solutions with LaTeX - clean, fast, and student-friendly.

Search Solutions

Floor Inequalities


1. **Stating the problem:** We are given inequalities involving the function $E$ applied to variables $m$, $n$, $y$, and expressions involving sums like $m + y$. The goal is to understand the chain of inequalities and the conclusions about $E(m + y)$ in relation to $E(m)$ and $E(y)$. 2. **Understanding the given inequalities:** - $E(n) \leq n \leq E(m) + 1$ - $E(y) \leq y \leq E(y) + 1$ From these, the combined inequality is: $$E(m) - E(y) \leq m - y \leq E(m) + E(y) + 2$$ 3. **Interpreting the integer property:** Since $E(m) - E(y)$ is an integer, it implies that the floor function $E$ satisfies $E(a) = E(b)$ for some expressions $a$ and $b$ related to $x + y$. 4. **Inequality involving sums:** We have: $$E(m) + E(y) \leq E(m + y)$$ and also: $$E(m + y) \leq x + y < E(m) + E(y) + 2 \leq 2$$ This leads to the conclusion: $$E(m + y) \leq E(m) + E(y) + 1$$ 5. **Final conclusion:** Combining the inequalities, we get: $$E(a) + E(b) \leq E(m + y) \leq E(m) + E(y) + 1$$ **Explanation:** This shows that the floor of the sum $m + y$ is at least the sum of the floors $E(m) + E(y)$ and at most one more than that sum. This is consistent with the property of the floor function where: $$E(m) + E(y) \leq E(m + y) \leq E(m) + E(y) + 1$$ This inequality accounts for the possible carryover when adding fractional parts of $m$ and $y$. Thus, the text explains the bounds on the floor of a sum in terms of the floors of the individual terms.