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MATH 447: Real Variables - Homework #1

Jerich Lee

December 16, 2024

Problem.
[1.8 a] The principle of mathematical induction can be extended as follows. A list $P_{m}, P_{m+1}, \dots{}$ of propositions is true provided (i) $P_{m}$ is true, (ii) $P_{n+1}$ is true whenever $P_{n}$ is true and $n\geq{} m$. Prove $n^2>n+1$ for all integers $n\geq2{}$.
Solution.
Proof. We will prove the base case of $n=2$ first: $$\begin{align} (2)^2>(2)+1 \end{align}$$ $$\begin{align} 4>3 \end{align}$$ The above verifies the base case. The inductive hypothesis is as follows: $$\begin{align} n^2>n+1, \ n\geq 2 \end{align}$$ We want to prove the case such that $P_{n+1}$ is true whenever $P_{n}$ is true and $n\geq{} m$. The inductive step is as follows: $$\begin{align} (n+1)^2 &> (n+1)+1 \\ n^2+2n+1 &> n+2 \\ n^2+n &> 1 \\ n(n+1) &> 1 \end{align}$$ By the inductive hypothesis, $n^2>n+1, n\geq{} 2$, $$\begin{align} n(n^2) &> n(n+1) > 1 \\ n^3 &> 1 \end{align}$$ The last line is true for all $n\geq{} 2$.
∎

Problem.
[2.8] Find all rational solutions of the equation $x^8-4x^5+13x^3-7x+1=0$.
Solution. To find all rational solutions of the equation $x^8-4x^5+13x^3-7x+1=0$, we can use the Rational Zeros Theorem. Corollary—Rational Zeros Theorem: If a polynomial equation $$\begin{align} a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0 \end{align}$$ with integer coefficients has a rational solution $x = \frac{p}{q}$ —where $p$ and $q$ are integers with no common factors and $q \neq{} 0$—, then:
  1. $p$ must be a factor of the constant term $a_0$
  2. $q$ must be a factor of the leading coefficient $a_n$
First, let's identify the coefficients:
  1. $c_8 = 1$
  2. $c_5 = -4$
  3. $c_3 = 13$
  4. $c_1 = -7$
  5. $c_0 = 1$
According to the theorem, if $\frac{c}{d}$ is a rational solution (where $c$ and $d$ are integers with no common factors and $d \neq{} 0$), then:
  1. $c$ must divide $c_0 = 1$
  2. $d$ must divide $c_8 = 1$
The only integers that divide 1 are 1 and -1. Therefore, the only possible rational solutions are:
  1. $\frac{1}{1} = 1$
  2. $\frac{-1}{1} = -1$
Now, we need to check if these candidates actually satisfy the equation: For $x = 1$: $$\begin{align} 1^8 - 4(1^5) + 13(1^3) - 7(1) + 1 = 1 - 4 + 13 - 7 + 1 = 4 \neq 0 \end{align}$$ For $x = -1$: $$\begin{align} (-1)^8 - 4((-1)^5) + 13((-1)^3) - 7(-1) + 1 = 1 + 4 - 13 + 7 + 1 = 0 \end{align}$$ Therefore, the only rational solution to the equation $x^8-4x^5+13x^3-7x+1=0$ is $x = -1$.
Problem.
[3.5]
  1. Show $|b|\leq a$ if and only if $-a\leq b\leq a$.
  2. Prove $||a|-|b||\leq|a-b|$ for all $a,b \in \mathbb{R}$.
Solution.
Proof.
  1. If $|b| \leq a$, then by definition, $-a \leq b \leq a$. This is because $|b| \leq a$ implies $b$ is within the interval $[-a, a]$.
  2. If $-a \leq b \leq a$, then $b$ is within the interval $[-a, a]$. This directly implies $|b| \leq a$ since the maximum deviation of $b$ from zero is $a$.
Thus, $|b| \leq{} a$ if and only if $-a \leq{} b \leq{} a$.
∎

Proof. We will prove this inequality using the triangle inequality and considering both possible cases. Using the Triangle Inequality:
  1. The triangle inequality states $|a| = |(a - b) + b| \leq |a - b| + |b|$.
  2. Rearranging gives $|a| - |b| \leq |a - b|$.
Consider the Reverse Situation:
  1. Similarly, $|b| = |(b - a) + a| \leq |b - a| + |a| = |a - b| + |a|$.
  2. Rearranging gives $|b| - |a| \leq |a - b|$.
Combine the Results:
  1. From the two inequalities, we have: $$\begin{align} |a| - |b| \leq |a - b| \quad \text{and} \quad |b| - |a| \leq |a - b| \end{align}$$
Conclusion:
  1. The absolute value $||a| - |b||$ is defined as: $$\begin{align} ||a| - |b|| = \max(|a| - |b|, |b| - |a|) \end{align}$$
  2. Therefore, $||a| - |b|| \leq |a - b|$.
Thus, we have proven that $||a|-|b||\leq{}|a-b|$ for all $a,b \in{} \mathbb{R}$.
∎

Problem.
[3.8] Let $a,b \in{} \mathbb{R}$. Show if $a\leq{} b_{1}$ for every $b_{1} > b$, then $a\leq{} b$.
Solution.
Proof. We will prove this statement using a proof by contradiction. Assume the hypothesis: For every $b_1 > b$, we have $a \leq{} b_1$. Suppose, for the sake of contradiction, that $a > b$. Consider $b_1 = \frac{a + b}{2}$. Note that:
  1. $b_1 > b$ (because $a > b$)
  2. $b_1 < a$ (because it's the midpoint between $a$ and $b$)
By our initial assumption, since $b_1 > b$, we must have $a \leq{} b_1$. However, we also showed that $b_1 < a$. This is a contradiction: we can't have both $a \leq{} b_1$ and $b_1 < a$. Therefore, our supposition that $a > b$ must be false. We conclude that $a \leq{} b$. Thus, we have shown that if $a \leq{} b_1$ for every $b_1 > b$, then $a \leq{} b$.
∎

Problem.
[4.1 r] For each set below that is bounded above, list three upper bounds for the set. Otherwise write "NOT BOUNDED ABOVE" or "NBA". $\cap_{n=1}^\infty{}( 1-\frac{1}{n}, 1+\frac{1}{n} )$
Solution. It is observed that the intersection of the sequences above converges to 1 as $n$ approaches infinity. Therefore, 2, 3, and 4 are all upper bounds for the set.
Problem.
[4.8] Let $S$ and $T$ be nonempty subsets of $\mathbb{R}$ with the following property: $s\leq{} t$ for all $s \in{} S$ and $t \in{} T$.
Solution. Observe $S$ is bounded above and $T$ is bounded below. Prove $\text{sup}\ S \leq{} \text{inf}\ T$ Give an example of such sets $S$ and $T$ where $S \ \cap{} \ T$ is nonempty. Give an example of sets $S$ and $T$ where $\text{sup}\ S = \text{inf}\ T$ and $S \ \cap{} \ T$ is the empty set.
  1. Let $M=t,\ t \in T$. Then $S\leq M$ for all $s\in S$. By Def 4.2, $M$ is an upper bound of $S$, and $S$ is bounded above.
  2. Let $m=s$, such that $s\in S$. Then, $m\leq t$ for all $t \in T$. Then, $m$ is a lower bound of $T$, and $T$ is bounded below.
To prove $\sup{} S \leq{} \inf{} T$:
  1. By the given property, we know that $s \leq t$ for all $s \in S$ and all $t \in T$.
  2. Let $M = \sup S$. By definition of supremum, $s \leq M$ for all $s \in S$.
  3. For any $t \in T$, we have $s \leq t$ for all $s \in S$.
  4. Therefore, $M = \sup S \leq t$ for all $t \in T$.
  5. Since $M \leq t$ for all $t \in T$, $M$ is a lower bound for $T$.
  6. By definition of infimum, $\inf T$ is the greatest lower bound of $T$.
  7. Thus, $M \leq \inf T$.
Therefore, we have proven that $\sup{} S \leq{} \inf{} T$. An example of such sets $S$ and $T$ where $S \ \cap{} \ T$ is nonempty: $S=[0,1], T=[1,2]$ An example of sets $S$ and $T$ where $\text{sup}\ S = \text{inf}\ T$ and $S \ \cap{} \ T$ is the empty set: $S=[0,1), T=(1,2]$

Problem 7 (8.5)

Problem.
[8.5]
  1. Consider three sequences $(a_{n}), (b_n)$ and $(s_{n})$ such that $a_{n}\leq s_{n}\leq b_{n}$ for all $n$ and $\lim_{ n \to \infty }a_{n}=\lim_{ n \to \infty }b_{n}=s$. Prove $\lim_{ n \to \infty }s_{n}=s$. This is called the "squeeze lemma".
Solution.
  1. Given: For all $\varepsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$ such that for $n > N_1$ and $n > N_2$:
    1. $|a_n - s| < \varepsilon$, which implies $s - \varepsilon < a_n < s + \varepsilon$
    2. $|b_n - s| < \varepsilon$, which implies $s - \varepsilon < b_n < s + \varepsilon$
  2. We also know that for all $n \in \mathbb{N}$, $a_n \leq s_n \leq b_n$
  3. Combining these facts, we can conclude that for $n > \max(N_1, N_2)$: $$\begin{align} s - \varepsilon < a_n \leq s_n \leq b_n < s + \varepsilon \end{align}$$
  4. This implies: $$\begin{align} s - \varepsilon < s_n < s + \varepsilon \end{align}$$
  5. Therefore: $$\begin{align} |s_n - s| < \varepsilon \end{align}$$
  6. By the definition of a limit of a sequence, this proves that $\lim_{n \to \infty} s_n = s$
Thus, we have proven the squeeze lemma.
  1. Suppose $(s_{n})$ and $(t_{n})$ are sequences such that $|s_{n}|\leq t_{n}$ for all $n$ and $\lim_{ n \to \infty }t_{n}=0$. Prove $\lim_{ n \to \infty }s_{n}=0$.
  1. Given: $\lim_{n \to \infty} t_n = 0$, so for any $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $n > N$:
  2. $|t_n - 0| < \varepsilon$
  3. This implies: $|t_n| < \varepsilon$
  4. We know that $|s_n| \leq t_n$ for all $n$, so: $|s_n| \leq |t_n| < \varepsilon$
  5. This means: $-\varepsilon < s_n < \varepsilon$
  6. Therefore: $|s_n - 0| < \varepsilon$
By the definition of a limit, this proves that $\lim_{n \to \infty} s_n = 0$.
Problem.
[8.6] Let $(s_{n})$ be a sequence in $\mathbb{R}$.
  1. Prove $\lim_{ n \to \infty }s_{n}=0$ if and only if $\lim_{ n \to \infty }|s_{n}|=0$
Solution.
Proof. We will prove this statement in two parts:
  1. If $\lim_{n \to \infty} s_n = 0$, then $\lim_{n \to \infty} |s_n| = 0$:
    1. Given $\lim_{n \to \infty} s_n = 0$, for every $\epsilon > 0$, there exists an $N$ such that for all $n \geq N$, $|s_n - 0| < \epsilon$.
    2. This simplifies to $|s_n| < \epsilon$.
    3. Therefore, $\lim_{n \to \infty} |s_n| = 0$.
  2. If $\lim_{n \to \infty} |s_n| = 0$, then $\lim_{n \to \infty} s_n = 0$:
    1. Given $\lim_{n \to \infty} |s_n| = 0$, for every $\epsilon > 0$, there exists an $N$ such that for all $n \geq N$, $|s_n| < \epsilon$.
    2. This directly implies that $|s_n - 0| < \epsilon$, so $\lim_{n \to \infty} s_n = 0$.
Thus, we have proven that $\lim_{n \to \infty} s_n = 0$ if and only if $\lim_{n \to \infty} |s_n| = 0$.
  1. Observe that if $s_{n}=(-1)^n$, then $\lim_{ n \to \infty }|s_{n}|$ exists, but $\lim_{ n \to \infty }s_{n}$ does not exist.
Observation: For the sequence $s_n = (-1)^n$, we can see that $s_n$ alternates between -1 and 1. To prove that $\lim_{n \to \infty} s_n$ does not exist, we can establish two subsequences:
  1. $s_{n_1}$: the subsequence of even terms, where $s_{n_1} = 1$ for all $n$
  2. $s_{n_2}$: the subsequence of odd terms, where $s_{n_2} = -1$ for all $n$
Clearly, $\lim_{n \to \infty} s_{n_1} = 1$ and $\lim_{n \to \infty} s_{n_2} = -1$ Since these two subsequences converge to different values, we can conclude that $\lim_{n \to \infty} s_n$ does not exist, demonstrating that the sequence is divergent. However, $\lim_{n \to \infty} |s_n| = 1$ does exist, as $|s_n| = 1$ for all $n$.
∎

Problem.
[Bonus] Use the completeness of $\mathbb{R}$ to show the existence of $x>0$ with $x^2=2$. Specifically, consider $S=\{t\in{} \mathbb{R} : t>0, t^2<2\}$. Clearly, $S$ is nonempty $(1\in{} S)$. Further, $2$ is an upper bound for $S$. Indeed, suppose $t\in{} S$, then $(2-t)(2+t)=4-x^2>4-2>0$. Clearly, $2+t>0$, hence $2-t>0$. Let $x=\text{sup} \ S$. Prove that $x^2=2$, by establishing that (i) $x^2\leq{} 2$, and (ii) $x^2\geq{} 2$. Once these inequalities are established, we can conclude that $x^2=2$.
Solution.
Proof.
  1. Let $\alpha = \sup S$. We will prove that $\alpha^2 = 2$ by showing that $\alpha^2 \leq 2$ and $\alpha^2 \geq 2$.
  2. First, let's prove $\alpha^2 \leq 2$:
    1. Assume $\alpha^2 > 2$.
    2. For $n \in \mathbb{N}$, consider $(\alpha - \frac{1}{n})^2$: $$\begin{align} \left( \alpha-\frac{1}{n} \right)^2=\alpha^2-\frac{2\alpha}{n}+\frac{1}{n^2}>\alpha^2-\frac{2\alpha}{n} \end{align}$$
    3. Since $\alpha^2 > 2$, we have: $$\begin{align} 2<\alpha^2-\frac{2\alpha}{n} \\ 2-\alpha^2<-\frac{2\alpha}{n} \\ \alpha^2-2> \frac{2\alpha}{n} \\ \frac{\alpha^2-2}{2\alpha}> \frac{1}{n} \end{align}$$
    4. Choose $n_0 \in \mathbb{N}$ such that $\frac{1}{n_0} < \frac{\alpha^2-2}{2\alpha}$.
    5. Then: $$\begin{align} \left( \alpha-\frac{1}{n_0} \right)^2>\alpha^2-(\alpha^2-2)=2 \end{align}$$
    6. This contradicts the fact that $\alpha$ is an upper bound for $S$.
    7. Therefore, our assumption must be false, and $\alpha^2 \leq 2$.
  3. Now, let's prove $\alpha^2 \geq 2$:
    1. Assume $\alpha^2 < 2$.
    2. For $n \in \mathbb{N}$, consider $(\alpha + \frac{1}{n})^2$: $$\begin{align} \left( \alpha + \frac{1}{n} \right)^2 = \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n^2} < \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n} = \alpha^2 + \frac{2\alpha + 1}{n}. \end{align}$$
    3. Choose $n_0 \in \mathbb{N}$ such that $\frac{1}{n_0} < \frac{2 - \alpha^2}{2\alpha + 1}$.
    4. Then: $$\begin{align} \left( \alpha + \frac{1}{n_0} \right)^2 < \alpha^2 + (2 - \alpha^2) = 2. \end{align}$$
    5. This means $\alpha + \frac{1}{n_0} \in S$, contradicting $\alpha$ as an upper bound for $S$.
    6. Therefore, our assumption must be false, and $\alpha^2 \geq 2$.
  4. Since we have shown $\alpha^2 \leq 2$ and $\alpha^2 \geq 2$, we can conclude that $\alpha^2 = 2$.
Thus, we have proven the existence of a real number $\alpha{} > 0$ such that $\alpha{}^2 = 2$.
∎