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Some true/false questions

by RoRi

For each statement, prove that it is true or show that it is false.

  1. 2^{\log 5} = 5^{\log 2}.
  2. \displaystyle{\log_2 5 = \frac{\log_3 5}{\log_2 3}}.
  3. \displaystyle{ \sum_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}} for every n \geq 1.
  4. 1 + \sinh x \leq \cosh x for all x.
  1. True.
    Proof. We can compute using the definition of the exponential

        \[ 2^{\log 5} = e^{\log \left( 2^{\log 5} \right)} = e^{\log 5 \cdot \log 2} = e^{\log 2 \cdot \log 5} = e^{\log \left(5^{\log 2} \right)} = 5^{\log 2}. \qquad \blacksquare\]

  2. False.
    On the left we have

        \[ \log_2 5 = \frac{\log 5}{\log 2}. \]

    While on the right we have,

        \[ \frac{\log_3 5}{\log_2 3} = \frac{\frac{\log 5}{\log 3}}{\frac{\log 3}{\log 2}}} = \frac{\log 5 \log 2}{(\log 3)^2} = \left( \frac{\log 5}{\log 2} \right) \left( \frac{\log 2}{\log 3} \right)^2.\]

    But since \frac{\log 2}{\log 3} \neq 1, these two quantities cannot be equal.

  3. True.
    Proof. The proof is by induction. For the case n = 1 on the left we have

        \[ \sum_{k=1}^1 \frac{1}{\sqrt{k}} = 1. \]

    While on the right we have

        \[ 2 \sqrt{1} = 2. \]

    Therefore, indeed for the case n = 1.

        \[ \sum_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}. \]

    Assume then that the statement is true for some positive integer m. Then,

        \begin{align*} \sum_{k=1}^{m+1} \frac{1}{\sqrt{k}} &= \frac{1}{\sqrt{m+1}} + \sum_{k=1}^m \frac{1}{\sqrt{k}} \\[9pt] &< \frac{1}{\sqrt{m+1}} + 2 \sqrt{m} &(\text{Ind. hyp.}) \\[9pt] &= \frac{2\sqrt{m}\sqrt{m+1} + 1}{\sqrt{m+1}} \\[9pt] &= 2 \left( \frac{\sqrt{m^2 + m} + \frac{1}{2}}{\sqrt{m+1}} \right) \\[9pt] &< 2 \left( \frac{\sqrt{m^2+m+\frac{1}{4}} + \frac{1}{2}}{\sqrt{m+1}} \right) \\[9pt] &= 2 \left( \frac{\sqrt{(m+\frac{1}{2})^2} + \frac{1}{2}}{\sqrt{m+1}} \right)\\[9pt] &= 2 \left( \frac{m+\frac{1}{2} +\frac{1}{2}}{\sqrt{m+1}} \right) \\[9pt] &= 2 \sqrt{m+1}. \end{align*}

    Thus, the inequality holds for the case m+1; hence, it holds for all positive integers n. \qquad \blacksquare

  4. False.
    From the definitions of \sinh and \cosh we have

        \[ \sinh x =\frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}. \]

    Using these definitions, the inequality states

        \begin{align*} 1 + \sinh x \leq \cosh x && \implies && 1 + \frac{e^x - e^{-x}}{2} &\leq \frac{e^x + e^{-x}}{2} \\[9pt] && \implies && 2 + e^x - e^{-x} &\leq e^x + e^{-x} \\[9pt] && \implies && 2 \leq 2e^{-x}. \end{align*}

    However, this is false if x < 0 since e^{-x} > 1 for x < 0.

One comment

  1. M says:

    The inequality in d) seems incorrect for positive x, and correct for negative ones. Therefore, the answer here and in the book seems to have a typo.

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