6: The Induction Principle
- Page ID
- 204824
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The classical mathematical induction is a concept that helps to prove mathematical results and theorems for all natural numbers. The principle of the classical mathematical induction is a specific technique that is used to prove certain statements in algebra which are formulated in terms of \(n\), where \(n\) is a natural number. Any mathematical statement, expression is proved based on the premise that it is true for \(n=1, n=k\), and then it is proved for \(n=k+1\).
Let \(t_0 \in \mathbb{T}\) and assume that
\[\left\{S(t): t \in\left[t_0, \infty\right)\right\}\notag \]
is a family of statements satisfying
(i) \(S\left(t_0\right)\) is true.
(ii) If \(t \in\left[t_0, \infty\right)\) is right-scattered and \(S(t)\) is true, then \(S(\sigma(t))\) is true.
(iii) If \(t \in\left[t_0, \infty\right)\) is right-dense and \(S(t)\) is true, then there exists a neighbourhood \(U\) of \(t\) such that \(S(s)\) is true for all \(s \in U \cap(t, \infty)\).
(iv) If \(t \in\left(t_0, \infty\right)\) is left-dense and \(S(s)\) is true for \(s \in\left[t_0, t\right)\), then \(S(t)\) is true.
Then \(S(t)\) is true for all \(t \in\left[t_0, \infty\right)\).
Proof.
Let
\[S^*=\left\{t \in\left[t_0, \infty\right): S(t) \quad \text { is not true } \quad\right\} \notag \]
We assume that \(S^* \neq \emptyset\). Let \(\inf S^*=t^*\). Because \(\mathbb{T}\) is closed, we have \(t^* \in \mathbb{T}\).
1. If \(t^*=t_0\), then \(S\left(t^*\right)\) is true.
2. If \(t^* \neq t_0\) and \(t^*=\rho\left(t^*\right)\), then, using (iv), we get that \(S\left(t^*\right)\) is true.
3. If \(t^* \neq t_0\) and \(\rho\left(t^*\right)<t^*\), then \(\rho\left(t^*\right)\) is right-scattered. Since \(S\left(\rho\left(t^*\right)\right)\) is true, we get that \(S\left(t^*\right)\) is true.
Consequently, \(t^* \notin S^*\). If we suppose that \(t^*\) is right-scattered, then, using that \(S\left(t^*\right)\) is true and (ii), we conclude that \(S\left(\sigma\left(t^*\right)\right)\) is true, which is a contradiction. From the definition of \(t^*\), it follows that \(t^* \neq \sup \mathbb{T}\). Since \(t^*\) is not right-scattered and \(t^* \neq \sup \mathbb{T}\), we obtain that \(t^*\) is right-dense. Because \(S\left(t^*\right)\) is true, using (iii), there exists a neighbourhood \(U\) of \(t^*\) such that \(S(s)\) is true for all \(s \in U \cap\left(t^*, \infty\right)\), which is a contradiction. Consequently, \(S^*=\emptyset\).
Let \(t_0 \in \mathbb{T}\) and assume that
\[\left\{S(t): t \in\left(-\infty, t_0\right]\right\} \notag\]
is a family of statements satisfying
(i) \(S\left(t_0\right)\) is true.
(ii) If \(t \in\left(-\infty, t_0\right]\) is left-scattered and \(S(t)\) is true, then \(S(\rho(t))\) is true.
(iii) If \(t \in\left(-\infty, t_0\right]\) is left-dense and \(S(t)\) is true, then there exists a neighbourhood \(U\) of \(t\) such that \(S(s)\) is true for all \(s \in U \cap(-\infty, t)\).
(iv) If \(t \in\left(-\infty, t_0\right)\) is right-dense and \(S(s)\) is true for \(s \in\left(t, t_0\right)\), then \(S(t)\) is true.
Then \(S(t)\) is true for all \(t \in\left(-\infty, t_0\right]\).
Proof.
Let
\[S^*=\left\{t \in\left(-\infty, t_0\right]: S(t) \quad \text { is not true } \quad\right\} \notag \]
Assume that \(S^* \neq \emptyset\). Note that \(t_0 \notin S^*\). Let
\[t^*=\sup S^*\]
1. If \(t^*=t_0\), then \(S\left(t^*\right)\) is true.
2. If \(t^* \neq t_0\) and \(t^*=\sigma\left(t^*\right)\), then, using (iv), we obtain that \(S\left(t^*\right)\) is true.
3. If \(t^* \neq t_0\) and \(t^*<\sigma\left(t^*\right)\), then \(\sigma\left(t^*\right)\) is left-scattered, \(\rho\left(\sigma\left(t^*\right)\right)=t^*\), and \(S\left(\sigma\left(t^*\right)\right)\) is true. Therefore, we conclude that \(S\left(t^*\right)\) is true.
Therefore, \(t^* \notin S^*\). If \(t^*\) is left-scattered, then, using (ii), we get that \(S\left(\rho\left(t^*\right)\right)\) is true, which is a contradiction. If \(t^*\) is left-dense, then, using (iii), there exists a neighbourhood \(U\) of \(t^*\) such that \(S(s)\) is true for all \(s \in U \cap\left(-\infty, t^*\right)\), which is a contradiction. Consequently, \(S^*=\emptyset\).
Let \(\mathbb{T}=3^{\mathbb{N}}\). We will prove the inequality
\[\frac{1}{4}\left(s^2-r^2\right) \leq \frac{1}{13}\left(s^3-r^3\right) \quad \text { for any } \quad r \leq s, \quad r, s \in \mathbb{T} .\]
Solution
Fix \(r, s \in \mathbb{T}, r \leq s\). Let
\[S(t): \frac{1}{4}\left(t^2-r^2\right) \leq \frac{1}{13}\left(t^3-r^3\right), \quad t \in[r, s]_{\mathbb{T}}\notag\]
1. The statement \(S(r)\) is true.
2. Observe that any \(t \in \mathbb{T}\) is right-scattered. Assume that \(S(t)\) is true. Then
\[\sigma(t)=3 t\notag\]
and
\[t^2 \leq t^3\notag\]
Hence, using (), we find
\[\frac{1}{4}\left(t^2-r^2\right)+2 t^2 \leq \frac{1}{13}\left(t^3-r^3\right)+2 t^3\notag\]
or
\[\frac{1}{4}\left(9 t^2-r^2\right) \leq \frac{1}{13}\left(27 t^3-r^3\right)\notag\]
Thus,
\[\frac{1}{4}\left((\sigma(t))^2-r^2\right) \leq \frac{1}{3}\left((\sigma(t))^3-r^3\right)\notag\]
Consequently \(S(\sigma(t))\) is true.
Applying the induction principle, we conclude that ( 0.2 ) is true for any \(t \in[r, s]_{\mathbb{T}}\) and then ( 0.1 ) is true for any \(r, s \in \mathbb{T}, r \leq s\).
Example 0.97. Let \(\mathbb{T}=[-2,-1) \cup\left\{-\frac{1}{n}\right\}_{n \in \mathbb{N}} \cup\{0\} \cup\left\{\frac{1}{n}\right\}_{n \in \mathbb{N}} \cup\left(1, \frac{3}{2}\right] \cup\left[\frac{7}{4}, \frac{11}{6}\right]\). We will prove
\[S(t): \sqrt{t+2}>t \quad \text { for any } \quad t \in \mathbb{T} .\notag\]
Solution
We have the following cases.
1. For \(t=-2\), we have
\[0>-2\notag\]
i.e., the given inequality is true.
2. Let \(t \in \mathbb{T}\) be right0scattered. Then, we have the following cases.
a. Let \(t=-\frac{1}{n}\) for some \(n \in \mathbb{N}\). Then
\[\begin{aligned}
\sqrt{-\frac{1}{n}+2} & >0 \\
& >-\frac{1}{n}
\end{aligned}\notag\]
i.e., \(S(t)\) is true. Note that
\[\sigma(t)=-\frac{1}{n+1}\notag\]
and
\[\begin{aligned}
\sqrt{-\frac{1}{n+1}+2} & >0 \\
& >-\frac{1}{n+1}
\end{aligned}\notag\]
i.e., \(S(\sigma(t))\) is true.
b. Let \(t=\frac{1}{n}\) for some \(n \in \mathbb{N}, n \geq 2\). Then
\[\frac{1}{n}>\frac{1}{n^2}\notag\]
and
\[\frac{1}{n}+2>\frac{1}{n^2},\notag\]
whereupon
\[\sqrt{\frac{1}{n}+2}>\frac{1}{n},\notag\]
i.e., \(S(t)\) is true. Note that
\[\sigma(t)=\frac{1}{n-1} .\notag\]
Then
\[\frac{1}{n-1}>\frac{1}{(n-1)^2}\notag\]
and
\[\frac{1}{n-1}+2>\frac{1}{(n-1)^2},\notag\]
whereupon
\[\sqrt{\frac{1}{n-1}+2}>\frac{1}{n-1} .\notag\]
Thus, \(S(\sigma(t))\) is true.
c. Let \(t=\frac{3}{2}\). Then
\[\begin{aligned}
\sqrt{\frac{3}{2}+2} & =\sqrt{\frac{7}{2}} \\
& >\sqrt{\frac{9}{4}} \\
& =\frac{3}{2}
\end{aligned}\notag\]
i.e., \(S(t)\) is true. Note that
\[\sigma(t)=\frac{7}{4} \notag\]
and
\[\begin{aligned}
\sqrt{\dfrac{7}{4}+2} & =\sqrt{\dfrac{15}{4}} \\
& >\sqrt{\dfrac{49}{16}} \\
& =\dfrac{7}{4}
\end{aligned}\notag \]
i.e., \(S(\sigma(t))\) is true.
3. Let \(t\) is right-dense. Then \(t \in[-2,-1) \cup\{0\} \cup\left(1, \dfrac{3}{2}\right] \cup\left[\dfrac{7}{4}, \dfrac{11}{6}\right]\). We have the following cases.
a. Let \(t \in[-2,-1) \cup\left(1, \dfrac{3}{2}\right] \cup\left[\dfrac{7}{4}, \dfrac{11}{6}\right]\). Then
\[t^2-t-2<0, \quad t \in\left(1, \dfrac{3}{2}\right] \cup\left[\dfrac{7}{4}, \dfrac{11}{6}\right]\notag \]
and hence,
\[t^2<t+2, \quad t \in\left(1, \dfrac{3}{2}\right] \cup\left[\dfrac{7}{4}, \dfrac{11}{6}\right],\notag \]
whereupon
\[\sqrt{t+2}>t, \quad t \in\left(1, \dfrac{3}{2}\right] \cup\left[\dfrac{7}{4}, \dfrac{11}{6}\right] .\notag \]
For \(t \in[-2,-1)\) the inequality is true.
b. Let \(t=0\). Then
\[\sqrt{2}>0\notag \]
and by point 2 it follows that there is a neighbourhood \(U\) of 0 so that \(S(s)\) is true for any \(s \in U \cap \mathbb{T}\).
c. Let \(t\) is left0dense. Then \(t \in[-2,-1) \cup\{0\} \cup\left(1, \dfrac{3}{2}\right] \cup\left(\dfrac{7}{4}, \dfrac{11}{6}\right]\). By points 2 and 3 , we get that if \(s \in\left[t_0, t\right)_{\mathrm{T}}\) for some \(t_0 \in \mathbb{T}\) and \(S(s)\) is true, then \(S(t)\) is true.
By the induction principle, it follows that \(S(t)\) is true for any \(t \in \mathbb{T}\).
Let \(\mathbb{T}=(-\infty,-8) \cup\left\{-7-\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\{-7\} \cup\left\{-7+\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\{8\} \cup\left\{8+\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup 10^{\mathbb{N}}\). We will prove
\[S(t): \dfrac{2 t-15}{t+5} \geq 0 \quad \text { for any } \quad t \in \mathbb{T} .\notag \]
1. We will prove that ( 0.3 ) holds for any \(t \in(-\infty,-6]_{\mathrm{T}}\) using the dual induction principle.
a. We have
\[\begin{aligned}
\dfrac{2 \cdot(-6)-15}{-6+5} & =\dfrac{-12-15}{-1} \\
& =27 \\
& >0
\end{aligned}\notag \]
i.e., \(S(-6)\) is true.
b. Let \(t \in(-\infty,-6]_{\mathrm{T}}\) is left-scattered. Then we have the following cases.
i. Let \(t=-7-\dfrac{1}{n}\) for some \(n \in \mathbb{N}, n \geq 2\). Then
\[\rho(t)=-7-\dfrac{1}{n-1} .\notag \]
We have
\[\begin{aligned}
\dfrac{2\left(-7-\dfrac{1}{n}\right)-15}{-7-\dfrac{1}{n}+5} & =\dfrac{-14-\dfrac{2}{n}-15}{-2-\dfrac{1}{n}} \\
& =\dfrac{29+\dfrac{2}{n}}{2+\dfrac{1}{n}} \\
& >0
\end{aligned}\notag \]
i.e., \(S(t)\) is true. Moreover, we have
\[\begin{aligned}
\dfrac{2\left(-7-\dfrac{1}{n-1}\right)-15}{-7-\dfrac{1}{n-1}+5} & =\dfrac{-14-\dfrac{2}{n-1}-15}{-2-\dfrac{1}{n-1}} \\
& =\dfrac{29+\dfrac{2}{n-1}}{2+\dfrac{1}{n-1}} \\
& >0
\end{aligned}\notag \]
Thus, \(S(\rho(t))\) is true.
ii. Let \(t=-8\). Then \(t\) is left-dense and ( 0.3 ) holds for any \(t \in(-\infty,-8]_{\mathrm{T}}\). Thus, there is a neighbourhood \(U\) of -8 such that ( 0.3 ) holds for any \(t \in U \cap(-\infty,-6]\) T.
iii. Let \(t=-7+\dfrac{1}{n}\) for some \(n \in \mathbb{N}\). Then
\[\rho(t)=-7+\dfrac{1}{n+1} .\notag \]
\[
\begin{aligned}
\dfrac{2\left(-7+\dfrac{1}{n}\right)-15}{-7+\dfrac{1}{n}+5} & =\dfrac{-14+\dfrac{2}{n}-15}{-2+\dfrac{1}{n}} \\
& =\dfrac{29-\dfrac{2}{n}}{2-\dfrac{1}{n}} \\
& >0
\end{aligned}
\notag \]
because \(29-\dfrac{2}{n}>0\) and \(2-\dfrac{1}{n}>0\). Thus, \(S(t)\) is true. Next,
\[\begin{aligned}
\dfrac{2\left(-7+\dfrac{1}{n+1}\right)-15}{-7+\dfrac{1}{n+1}+5} & =\dfrac{-14+\dfrac{2}{n+1}-15}{-2+\dfrac{1}{n+1}} \\
& =\dfrac{29-\dfrac{2}{n+1}}{2-\dfrac{1}{n+1}} \\
& >0
\end{aligned}\notag \]
\(29-\dfrac{2}{n+1}>0\) and \(2-\dfrac{1}{n+1}>0\). Therefore \(S(\rho(t))\) is true.
c. Let \(t \in(-\infty,-6]_{\mathrm{T}}\) is left-dense. Then we have the following cases.
i. Let \(t \in(-\infty,-8)_{\mathrm{T}}\). Then
\[\dfrac{2 t-15}{t+5}>0\notag \]
and
\[\dfrac{2 s-15}{s+5}>0 \quad \text { for any } \quad s \in(-\infty, t)_{\mathrm{T}} .\notag \]
ii. Let \(t=-7\). By 1.2.1, it follows that there is a neighbourhood \(U\) of -7 such gthat \(S(s)\) holds for any \(s \in U \cap(-\infty,-7)_{\mathbb{T}}\).
d. Let \(t \in(-\infty,-6]_{\mathrm{T}}\) is right-dense. Then we have the following cases.
i. Let \(t \in(-\infty,-8)_{\mathrm{T}}\). By the previous cases, we get that \(S(s)\) is true for any \(s \in(t,-6)_{\mathrm{T}}\) and \(S(t)\) is true.
ii. Let \(t=-7\). By 1.2 .1 , it follows that there is a neighbourhood \(U\) of -7 such gthat \(S(s)\) holds for any \(s \in U \cap(-\infty,-7)_{\mathrm{T}}\).
Now, applying the dual induction principle, we conclude that \(S(t)\) is true for any \(t \in(-\infty,-6]_{\mathrm{T}}\).
2. Now, we will prove that \((0.3)\) holds for any \(t \in[8, \infty)_{\mathrm{T}}\) using the induction principle.
a. We have
\[\begin{aligned}
\dfrac{2 \cdot 8-15}{8+5} & =\dfrac{16-15}{13} \\
& =\dfrac{1}{13} \\
& >0
\end{aligned}\notag \]
Thus, \(S(8)\) is true.
b. Let \(t \in[8, \infty)_{\mathrm{T}}\) is right-scattered and \(S(t)\) is true. Then we have the following cases.
i. Let \(t=8+\dfrac{1}{n}\) for some \(n \in \mathbb{N}, n \geq 2\). Then
\[\sigma(t)=8+\dfrac{1}{n-1}\notag \]
We have
\[\begin{aligned}
\dfrac{2\left(8+\dfrac{1}{n}\right)-15}{8+\dfrac{1}{n}+5} & =\dfrac{16+\dfrac{2}{n}-15}{13+\dfrac{1}{n}} \\
& =\dfrac{1+\dfrac{2}{n}}{13+\dfrac{1}{n}} \\
& >0
\end{aligned}\notag \]
and
\[\begin{aligned}
\dfrac{2\left(8+\dfrac{1}{n-1}\right)-15}{8+\dfrac{1}{n-1}+5} & =\dfrac{16+\dfrac{2}{n-1}-15}{13+\dfrac{1}{n-1}} \\
& =\dfrac{1+\dfrac{2}{n-1}}{13+\dfrac{1}{n-1}} \\
& >0
\end{aligned}\notag \]
i.e., \(S(t)\) and \(S(\sigma(t))\) is true.
ii. Let \(t=9\). Then
\[\sigma(9)=10\notag \]
We have
\[\begin{aligned}
\dfrac{2 \cdot 9-15}{9+5} & =\dfrac{18-15}{14} \\
& =\dfrac{3}{14} \\
& >0
\end{aligned}\notag \]
and
\[\begin{aligned}
f r a c 2 \cdot 10-1510+5 & =\dfrac{20-15}{15} \\
& =\dfrac{5}{15}
\end{aligned}\notag \]
\[\begin{array}{l}
=\dfrac{1}{3} \\
>0 .
\end{array}\notag \]
So, \(S(9)\) and \(S(10)\) are true.
iii. Let \(t=10^n\) for some \(n \in \mathbb{N}\). Then
\[\sigma(t)=10^{n+1}\notag \]
We have
\[\dfrac{2 \cdot 10^n-15}{10^n+5}>0\notag \]
and
\[\dfrac{2 \cdot 10^{n+1}-15}{10^{n+1}+5}>0 .\notag \]
Thus, \(S(t)\) and \(S(\sigma(t))\) are true.
c. Let \(t=8\). We have
\[\begin{aligned}
\dfrac{2 \cdot 8-15}{8+5} & =\dfrac{16-15}{13} \\
& =\dfrac{1}{13} \\
& >0
\end{aligned}\notag \]
i.e., \(S(8)\) is true. By 2.2.1, it follows that there is a neighbourhood \(U\) of 8 such that \(S(t)\) is true for any \(t \in U \cap[8, \infty)\) T.
d. Note that there is no left-dense points in \((8, \infty)_{\mathrm{T}}\).
Now, applying the induction principle, we conclude that \(S(t)\) is true for any \(t \in[8, \infty)_{\mathbb{T}}\).
Let \(\mathbb{T}=\{-1\} \cup\left\{-1-\dfrac{1}{n^2}\right\}_{n \in \mathbb{N}} \cup\left\{\dfrac{1}{n^3}\right\}_{n \in \mathbb{N}} \cup(1,2]\). We will prove that
\[S(t): t^2-t-2 \leq 2 \text { for any } t \in \mathbb{T} .\notag \]
1. We have
\[\begin{aligned}
(-1)^2-(1-1)-2 & =1+1-2 \\
& =0
\end{aligned}\notag \]
i.e., \(S(-1)\) is true.
2. Let \(t \in \mathbb{T}\) be right-scattered. We have the following cases.
a. Let \(t=-1+\dfrac{1}{n^2}\) for some \(n \in \mathbb{N}, n \geq 2\). Then
\[\sigma(t)=-1+\dfrac{1}{(n-1)^2}\notag \]
We have
\[\begin{aligned}
\left(-1+\dfrac{1}{n^2}\right)^2-\left(-1+\dfrac{1}{n^2}\right)-2 & =1-\dfrac{2}{n^2}+\dfrac{1}{n^4}+1-\dfrac{1}{n^2}-2 \\
& =\dfrac{1}{n^4}-\dfrac{3}{n^2} \\
& =\dfrac{1-3 n^2}{n^4} \\
& \leq 0
\end{aligned}\notag \]
i.e., \(S(t)\) is true. Next,
\[\begin{aligned}
\left(-1+\dfrac{1}{(n-1)^2}\right)^2-\left(-1+\dfrac{1}{(n-1)^2}\right)-2 & =1-\dfrac{2}{(n-1)^2}+\dfrac{1}{(n-1)^4}+1-\dfrac{1}{(n-1)^2}-2 \\
& =\dfrac{1}{(n-1)^4}-\dfrac{3}{(n-1)^2} \\
& =\dfrac{1-3(n-1)^2}{(n-1)^4} \\
& \leq 0
\end{aligned}
\notag \]
Thus, \(S(\sigma(t))\) is true.
b. Let \(t=\dfrac{1}{n^3}\) for some \(n \in \mathbb{N}, \geq 2\). Then
\[\sigma(t)=\dfrac{1}{(n-1)^3}\notag \]
We have
\[\begin{aligned}
\left(-1+\dfrac{1}{n^3}\right)^2-\left(-1+\dfrac{1}{n^3}\right)-2 & =1-\dfrac{2}{n^3}+\dfrac{1}{n^6}+1-\dfrac{1}{n^3}-2 \\
& =\dfrac{1}{n^6}-\dfrac{3}{n^3} \\
& =\dfrac{1-3 n^3}{n^6}
\end{aligned}\notag \]
\[
\leq 0
\notag \]
i.e., \(S(t)\) is true. Moreover,
\[\begin{aligned}
\left(-1+\dfrac{1}{(n-1)^3}\right)^2-\left(-1+\dfrac{1}{(n-1)^3}\right)-2 & =1-\dfrac{2}{(n-1)^3}+\dfrac{1}{(n-1)^6}+1-\dfrac{1}{(n-1)^3}-2 \\
& =\dfrac{1}{(n-1)^6}-\dfrac{3}{(n-1)^3} \\
& =\dfrac{1-3(n-1)^3}{(n-1)^6} \\
& \leq 0
\end{aligned}\notag \]
and then \(S(\sigma(t))\) is true.
3. Let \(t \in \mathbb{T}\) be right-dense and \(S(t)\) be true. We have the following cases.
a. Let \(t=-1\). Then, by 2.1 , it follows that there is a neighbourhood \(U\) of -1 such that \(S(s)\) is true for any \(s \in U \cap \mathbb{T}\).
b. Let \(t=0\). Then, by 2.2 , it follows that there is a neighbourhood \(U\) of 0 such that \(S(s)\) is true for any \(s \in U \cap \mathbb{T}\).
c. Let \(t \in(1,2]\). Then
\[t^2-t-2 \leq 0\notag \]
and there is a neighbourhood \(U\) of \(t\) such that \(S(s)\) is true for any \(s \in U \cap \mathbb{T}\).
d. Let \(t \in \mathbb{T}\) be left0dense and \(S(s)\) is true for any \(s \in[-1, t)\). We have that \(t \in(1,2]\). By the previous cases it follows that \(S(t)\) is true.
Now, applying the induction principle, we conclude that \(S(t)\) is true for any \(t \in \mathbb{T}\).
Let \(\mathrm{T}=\mathbb{N}\). We will prove that
\[S(n): 1^4+2^4+\cdots+n^4=\dfrac{1}{30} n(n+1)(2 n+1)\left(3 n^2+3 n-1\right) \quad \text { for any } \quad n \in \mathbb{T} .\notag \]
Solution
1. We have
\[\begin{aligned}
\dfrac{1}{30} \cdot 1 \cdot(1+1) \cdot(2 \cdot 1+1)\left(3 \cdot 1^2+3 \cdot 1-1\right) & \\
& =\dfrac{1}{30} \cdot 2 \cdot 3 \cdot 5 \\
& =1 \\
& =1^4
\end{aligned}\notag \]
Thus, \(S(1)\) is true.
2. Firstly, note that any point of \(\mathbb{T}\) is right-scattered. Assume that \(S(n)\) is true for some \(n \in \mathbb{T}\). We have
\[\sigma(n)=n+1 .\notag \]
Then
\[\begin{aligned}
& 1^4+2^4+\cdots+(n+1)^4=1^4+2^4+\cdots+n^4+(n+1)^4 \\
= & \dfrac{1}{30} n(n+1)(2 n+1)\left(3 n^2+3 n-1\right)+(n+1)^4 \\
= & (n+1)\left(\dfrac{1}{30} n(2 n+1)\left(3 n^2+3 n-1\right)+(n+1)^3\right) \\
= & \dfrac{1}{30}(n+1)\left(\left(2 n^2+n\right)\left(3 n^2+3 n-1\right)+30(n+1)^3\right) \\
= & \dfrac{1}{30}(n+1)\left(6 n^4+6 n^3-2 n^2+3 n^3+3 n^2-n+30 n^3+90 n^2+90 n+30\right) \\
= & \dfrac{1}{30}(n+1)\left(6 n^4+39 n^3+91 n^2+89 n+30\right) \\
= & \dfrac{1}{30}(n+1)(n+2)\left(6 n^3+27 n^2+37 n+15\right) \\
= & \dfrac{1}{30}(n+1)(n+2)(2 n+3)\left(3 n^2+9 n+5\right) \\
= & \dfrac{1}{30}(n+1)(n+2)(2 n+3)\left(3 n^2+6 n+3+3 n+3-1\right) \\
= & \dfrac{1}{30}(n+1)(n+2)(2 n+3)\left(3(n+1)^2+3(n+1)-1\right)
\end{aligned}\notag \]
Thus, \(S(\sigma(n))\) is true.
Applying the induction principle, we conclude that \(S(n)\) is true for any \(n \in \mathbb{N}\).
Let
\[\mathbb{T}=-15^{\mathbb{N}} \cup\left\{-14-\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\left\{-14+\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup(-12,-10] \cup\{-9\} \cup\left\{-9+\dfrac{1}{n}\right\}_{n \in \mathbb{N}}\notag \]
Prove that
\[\dfrac{2 t-1}{3}<\dfrac{t-3}{2} \text { for any } t \in \mathbb{T}\notag \]
- Answer
-
Add texts here. Do not delete this text first.
Let
\[\mathbb{T}=\{2\} \cup\left\{2+\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\left\{3-\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\left\{3+\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\{3\} \cup 4^{\mathbb{N}}\notag \]
Prove that
\[\dfrac{2-t}{t+1}+\dfrac{t+2}{t-1}>\dfrac{t-3}{2} \text { for any } t \in \mathbb{T}\notag \]
- Answer
-
Add texts here. Do not delete this text first.
Let
\[\mathbb{T}=(-\infty, 0] \cup\left\{\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\{2\} \cup\left\{2-\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\{5\} \cup\left\{5-\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\left\{5+\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup 7^{\mathbb{N}}\notag \]
Prove that
\[5(t-3)(t-4)>0 \quad \text { for any } \quad t \in \mathbb{T}\notag \]
- Answer
-
Add texts here. Do not delete this text first.
\[\begin{aligned}
\mathbb{T}= & (-\infty,-10] \cup\{-9\} \cup\left\{-9-\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\left\{-9+\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup(-8,-1] \cup\{4\} \cup\left\{4+\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \\
& \cup\{6\} \cup\left\{6-\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup\left\{6+\dfrac{1}{n}\right\}_{n \in \mathbb{N}} \cup 9^{\mathbb{N}} .
\end{aligned}\notag \]
Prove that
\[5 t^2-2(t+3)(t-1)>4 t+9 \quad \text { for any } \quad t \in \mathbb{T}\notag \]
- Answer
-
Add texts here. Do not delete this text first.
Let \(\mathbb{T}=\mathbb{N}\). Prove that
\[1^5+2^5+\cdots+n^5=\dfrac{1}{12} n^2(n+1)^2\left(2 n^2+2 n-1\right) \quad \text { for any } \quad n \in \mathbb{T}\notag \]
- Answer
-
Add texts here. Do not delete this text first.