Exercise 1.2.12

Question

Let $y_{1}=6$, and for each $n \in \mathbf{N}$ define $y_{n+1}=\left(2 y_{n}-6\right) / 3$
  \begin{enumerate}[label=(\alph*)] 
  \item Use induction to prove that the sequence satisfies $y_{n}>-6$ for all $n \in \mathbf{N}$.
  \item Use another induction argument to show the sequence $\left(y_{1}, y_{2}, y_{3}, \ldots\right)$ is decreasing.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Suppose for induction that $y_n > -6$, our base case clearly satisfies $y_1 > -6$. then $$ \begin{aligned} y_{n+1} = (2y_n - 6)/3 &\implies y_n = (3y_{n+1} + 6)/2 > -6 \ &\implies y_{n+1} > (2\cdot (-6) - 6)/3 = -6 \end{aligned} $$ Thus $y_{n+1} > -6$ \item Suppose $y_{n+1} < y_{n}$, the base case $2 < 6$ works. Now $$ \begin{aligned} y_{n+1} < y_{n} &\implies 2y_{n+1} < 2y_n \ &\implies 2y_{n+1} - 6 < 2y_n - 6 \ &\implies (2y_{n+1} - 6)/3 < (2y_n - 6)/3 \ &\implies y_{n+2} < y_{n+1} \end{aligned} $$ Thus $(y_n)$ is decreasing. \end{enumerate}

Exercise 1.2.12

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Exercise 1.2.12

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