Chapter 3: Problem 16
Given any rational number \(r,-r\) is also a rational number.
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Chapter 3: Problem 16
Given any rational number \(r,-r\) is also a rational number.
These are the key concepts you need to understand to accurately answer the question.
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For any odd integer \(n\), $$ \left\lfloor\frac{n^{2}}{4}\right\rfloor=\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right) . $$
A calculator display shows that \(\sqrt{2}=1.414213562\), and \(1.414213562=\frac{1414213562}{1000000000}\). This suggests that \(\sqrt{2}\) is a rational number, which contradicts Theorem 3.7.1. Explain the discrepancy.
Suppose \(n\) and \(d\) are integers and \(d \neq 0\). Prove each of the following. a. If \(d \mid n\), then \(n=\lfloor n / d\rfloor \cdot d\). b. If \(n=\lfloor n / d\rfloor \cdot d\), then \(d \mid n\). c. Use the floor notation to state a necessary and sufficient condition for an integer \(n\) to be divisible by an integer \(d\).
If \(k\) is an integer, what is \(\lceil k\rceil ?\) Why?
a. Prove that if \(a, d, q\), and \(r\) are integers such that \(a=\) \(d q+r\) and \(0
\leq r
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