w3

True or false? Use the axioms of $\mathbb{R}$ to determine the truth or falsehood of the following statements.

(a) $-\left( -\left( \sqrt{2} + \sqrt{2^{-1}} - 1 \right) + \frac{1 - \sqrt{2}}{\sqrt{2}} \right) \in \mathbb{Q}$.

$= - \left( - \sqrt{2} - \frac{\sqrt{2}}{3} + 1 + \frac{\sqrt{2} - 2}{3} \right)$

$= \sqrt{2} + \frac{\sqrt{2}}{3} - 1 - \frac{\sqrt{2}}{3} + 1$

$= \sqrt{2} \notin \mathbb{Q}$

Since $\sqrt{2} \notin \mathbb{Q}$, then the statement is false.

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(b) If $a, b \in \mathbb{R}$ satisfy $a^2 = b^2$, then $a = b$.

Since $a^2 = b^2$, then $a^2 - b^2 = 0$.

$(a + b)(a - b) = 0$.

$a = b \, \text{or} \, a = -b$.

Therefore, the statement is false.

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(c) If $a, b \in \mathbb{R}$ satisfy $a^2 = b^2$, then $a^3 = b^3$.

Since $a^2 = b^2$, then $a = \pm b$.

Thus, $a^3 = (\pm b)^3$, then $a^3 = \pm b^3$.

Therefore, the statement is false.

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(d) There is no real number $T$ such that $T^2 - 1$ is negative and $2T > 1$.

Since $T^2 - 1 < 0$, then $-1 < T < 1$.

Since $T > \frac{1}{2}$, then $\frac{1}{2} < T < 1$.

However, $\exists T \in \left( \frac{1}{2}, 1 \right)$ such that $T^2 - 1 < 0$ and $2T > 1$.

Then $T$ can be a real number.

Hence, the statement is false.

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Find the fractions whose decimal representations are:

(a) $0.\overline{23}$

Let $0.\overline{23} = a$

$100a = 23.\overline{23}$

$10000a = 2323.\overline{23}$

$10000a - 100a = 2323.\overline{23} - 23.\overline{23}$

$9900a = 2300$

$a = \frac{2300}{9900} = \frac{23}{99}$

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(b) $2.2\overline{83}$

Let $2.2\overline{83} = a$

$1000a = 2283.\overline{83}$

$10a = 22.\overline{83}$

$1000a - 10a = 2283.\overline{83} - 22.\overline{83}$

$990a = 2261$

$a = \frac{2261}{990}$
Compute the sum $\begin{aligned}\sum_{n \in \mathbb{N}} \frac{3^n + 5^n}{7^{n+1}}\end{aligned}$

\[ \begin{aligned} &= \sum_{n \in \mathbb{N}} \frac{3^n}{7^{n+1}} + \sum_{n \in \mathbb{N}} \frac{5^n}{7^{n+1}} \\ &= \left( \frac{3^1}{7^2} + \frac{3^2}{7^3} + \dots + \frac{3^n}{7^{n+1}} \right) + \left( \frac{5^1}{7^2} + \frac{5^2}{7^3} + \dots + \frac{5^n}{7^{n+1}} \right) \\ &= \frac{\frac{3^1}{7^2} \left( 1 - (\frac{3}{7})^n \right)}{1 - \frac{3}{7}} + \frac{\frac{5^1}{7^2} \left( 1 - (\frac{5}{7})^n \right)}{1 - \frac{5}{7}} \\ &= \frac{\frac{3}{7} - \frac{3^{n+1}}{7^{n+1}}}{4} + \frac{\frac{5}{7} - \frac{5^{n+1}}{7^{n+1}}}{2} \\ &= \frac{3}{4 \times 7} - \frac{3^{n+1}}{4 \times 7^{n+1}} + \frac{5}{2 \times 7} - \frac{5^{n+1}}{2 \times 7^{n+1}} \\ &= \frac{26}{2 \times 4 \times 7} - \frac{2\times3^{n+1} + 4 \times 5^{n+1}}{2 \times 4 \times 7^{n+1}} \\ &= \frac{13}{28} - \frac{3^{n+1}}{4 \times 7^{n+1}} - \frac{5^{n+1}}{2 \times 7^{n+1}} \\ &= \frac{13}{28} - \frac{1}{4 \times \left( \frac{7}{3} \right)^{n+1}} - \frac{1}{2 \times \left( \frac{7}{5} \right)^{n+1}} \end{aligned} \]

When $n \to \infty$, $\left( \frac{7}{3} \right)^{n+1}$ and $\left( \frac{7}{5} \right)^{n+1} \to \infty$

Then $\frac{1}{4 \times \left( \frac{7}{3} \right)^{n+1}}$ and $\frac{1}{2 \times \left( \frac{7}{5} \right)^{n+1}} \to 0$

Thus the sum is $\frac{13}{28}$
Use the arithmetic mean-geometric mean inequality$^1$ to prove that, with the $n + 1$ positive numbers
$1, 1 + \frac{1}{n}, 1 + \frac{1}{n}, \dots, 1 + \frac{1}{n}$, we have that, for any positive integer $n$, $\left( 1 + \frac{1}{n} \right)^n \leq \left( 1 + \frac{1}{n+1} \right)^{n+1}.$

Using AM-GM inequality

$\frac{1+1+\frac{1}{n}+\cdots+1+\frac{1}{n}}{n+1}>\sqrt[n+1]{\left(1+\frac{1}{n})\cdot\left(1+\frac{1}{n}\right)\cdots\left(1+\frac{1}{n}\right)\right)}$

$\frac{n+1 + 1}{n+1} > \sqrt[n+1]{\left( 1 + \frac{1}{n} \right)^n}$

$\left( 1 + \frac{1}{n+1} \right)^{n+1} \geq \left( 1 + \frac{1}{n} \right)^n$

Thus $\left( 1 + \frac{1}{n} \right)^n \leq \left( 1 + \frac{1}{n+1} \right)^{n+1}$

But $1 \neq 1 + \frac{1}{n}, (n > 0)$, then $\left( 1 + \frac{1}{n} \right)^n < \left( 1 + \frac{1}{n+1} \right)^{n+1}$

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Prove that $\left( \left( 1 + \frac{1}{n} \right)^n \right)_{n \in \mathbb{N}}$ is a bounded sequence. Hint: you may use the binomial theorem.

$$ \begin{aligned} &\left(1 + \frac{1}{n}\right)^n ,n\rightarrow\infty\ &= C_n^0 \left( \frac{1}{n} \right)^0 + C_n^1 \left( \frac{1}{n} \right)^1 + C_n^2 \left( \frac{1}{n} \right)^2 + \cdots + C_n^n \left( \frac{1}{n} \right)^n + \cdots \

&= 1 + \frac{n!}{(n-1)! 1!} \cdot \frac{1}{n} + \frac{n!}{2!(n-2)!} \cdot \frac{1}{n^2} + \cdots + \frac{n!}{n!} \left( \frac{1}{n} \right)^n + \cdots \

&= 1 + 1 + \frac{n(n-1)}{2!} \cdot \frac{1}{n^2} + \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{n^3} + \cdots + \frac{1}{n^n} + \cdots \

&< 1 + 1 + \frac{n^2}{2!} \cdot \frac{1}{n^2} + \frac{n^3}{3!} \cdot \frac{1}{n^3} + \cdots + \frac{n^n}{n!} \cdot \frac{1}{n^n} + \cdots \

&= 2 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} + \cdots \

&= 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} + \cdots \

&< 1 + 1 + \frac{1}{2^1} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}} + \cdots \

&= 2 + \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 3 \end{aligned} $$

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