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\title{Math 104 HW 1}
\author{schel337 }
\date{8/27/2020}

\begin{document}

\maketitle

\section*{Ross 1.10}
\begin{align*}
    (2n + 1) + (2n+3) + \dots + (4n-1) &= 
    2n * n + (1 + 3 + \dots + 2n-1) \\
    &= 2n^2 + n + 2*(1 + 2 + \dots + n-1) \\
    &= 2n^2 + n + 2 \frac{n(n-1)}{2} \\
    &= 2n^2 + n + n(n-1) \\
    &= 3n^2
\end{align*}

\section*{Ross 1.12}
\begin{enumerate}[(a)]
    \item 
    \begin{align*}
        (a + b)^1 &= a + b \\
        (a+b)^2 &= a^2 + 2 ab + b^2 \\
        (a+b)^3 &= a^3 + 3 a^2 b + 3 a b^2 + b^3
    \end{align*}
    \item 
    \begin{align*}
        {n \choose k} + {n \choose k-1} &= \frac{n!}{(n-k)! k!} + \frac{n!}{(n-k+1)!(k-1)!} \\
        &= n!\frac{(n-k)! k! + (n-k+1)!(k-1)!}{(n-k)! k!(n-k+1)!(k-1)!} \frac{{n+1 \choose k}}{{n+1 \choose k}}\\
        &= {n+1 \choose k} \frac{n! k! (n+1-k)!}{(n+1)!} \frac{(n-k)! k! + (n-k+1)!(k-1)!}{(n-k)! k!(n-k+1)!(k-1)!}  \\
        &= {n+1 \choose k} \frac{1}{n+1} \frac{(n-k)! k! + (n-k+1)!(k-1)!}{(n-k)! (k-1)!} \\
        &= {n+1 \choose k} \frac{1}{n+1} \left( \frac{k!}{(k-1)!} + \frac{(n-k+1)!}{(n-k)!}\right) \\
        &= {n+1 \choose k} \frac{k + (n-k+1)}{n+1} \\
        &= {n + 1 \choose k}
    \end{align*}
    \item We showed the base cases in part a). Then suppose the binomial theorem holds for all $k \leq n$. Then
    \begin{align*}
        (a+b)^{n+1} &= (a+b)^{n} (a+b) \\
        &= \left( \sum_{k=0}^{n} {n \choose k} a^{k} b^{n-k} \right) (a+b) \\
        &= \sum_{k=0}^{n} {n \choose k} a^{k+1} b^{n-k} + \sum_{k=0}^{n} {n \choose k} a^{k} b^{n-k + 1} \\
        &= \sum_{k=0}^{n+1} \left({n \choose k} + {n \choose k - 1}  \right) a^{k} b^{n-k + 1} \\
        &=\sum_{k=0}^{n+1} {n + 1 \choose k} a^{k} b^{n+1-k}
    \end{align*}
    So the inductive argument shows the binomial theorem holds for all $n$
    
\end{enumerate}
\section*{Ross 2.1}
For the polynomial $x^2 - p$ for prime $p$, the only possible rational roots are $\pm 1$ and $\pm p$, which will not be satisfied for anything other than $p=1$ so $\sqrt{3},\sqrt{5},\sqrt{7}, \sqrt{31}$ are all irrational. For $\sqrt{24}$, the possible rational roots of $x^2 - 24$ would be $1,2,3,4,6,8,12,24$, none of which are roots so it is irrational.

\section*{Ross 2.2}
Once again for roots of prime numbers $p$ the possible rational roots are $\pm 1$ and $\pm p$. None of the options will satisfy the required polynomials.

\section*{Ross 2.7}
\begin{align*}
    x + \sqrt{3} - \sqrt{4 + 2 \sqrt{3}}&= 0 \\
    x^2 + 3- 2( \sqrt{3} \sqrt{4 + 2 \sqrt{3}}) + 4 + 2 \sqrt{3} &= 0 \\
    x^2 + 3 - 2 \sqrt{12 + 6 \sqrt{3}} + 4 + 2 \sqrt{3} &= 0 \\
    x^2 + 7 - 2 \sqrt{12 + 6 \sqrt{3}} + 2 \sqrt{3} &= 0 
\end{align*}
This is quite gnarly to further simplify so I omit further latex to evaluate a good polynomial:
\begin{align*}
    (x + \sqrt{3} - \sqrt{4 + 2 \sqrt{3}})^4 &= x^4 - 4 x^3 + 6 x^2 - 4 x + 1
\end{align*}
Which has a nice root at $x=1$, so $\sqrt{4 + 2 \sqrt{3}} - \sqrt{3} = 1$

\section*{Ross 3.6}
\begin{enumerate}[(a)]
    \item 
\begin{align*}
    |a+b+c| = |(a+b)| + c| \leq |a+b| + |c| \leq |a| + |b| + |c|
\end{align*}
    \item The case $n=1$ is automatically true. The case $n=2$ is true by the triangle inequality. So assuming the inequality holds for $n \geq 1$, then
    \begin{align*}
        |a_1 + a_2 + \dots + a_n + a_{n+1}| &\leq |a_1 + a_2 + \dots + a_n| + |a_{n+1}| \\
        &\leq |a_1| + \dots + |a_n| + |a_{n+1}|
    \end{align*}
\end{enumerate}

\section*{4.11}
Let $a < b \in \mathbb{R}$. Then there exists $q_1 \in \mathbb{Q}$ such that $a < q_1 < b$. Then $a < q_1 \in \mathbb{R}$ so there exists $q_2 \in \mathbb{Q}$ such that $a < q_2 < q_1 < b$ and so on for any number of $q_n$ which shows there are countably infinite rationals between $a$ and $b$.

\section*{4.14}
\begin{enumerate}[(a)]
    \item First, to show that $\sup (A+B)$ exists, note that $A+B$ will also be bounded as $A$ and $B$ are (simply add their upper bounds and their lower bounds).\\
Second, I claim supremums preserve inequalities (at least for bounded sets). Let $x \leq y$ for all $x \in X$ with bounded $X \subset \mathbb{R}$. Then suppose $\sup X > y$, but $y$ is an upper bound for $X$ that is smaller. Therefore $\sup X \leq y$. \\
Third, to show $\sup A + \sup B \leq \sup (A+B)$ consider any $b \in B$. Then for any $a \in A$, I have by definition $a + b \leq \sup (A+B) \implies a \leq \sup(A+B) - b$. Then taking the supremum over $A$ and then over $B$ I get $\sup A \leq \sup (A+B) - \sup B$. By a symmetric argument $\sup B \leq \sup(A+B) - \sup A$\\
Then note that $\sup A \geq a$ and $\sup B \geq b$ for all $a,b$ and therefore $\sup A + \sup B \geq a + b$ and therefore $\sup A + \sup B \geq \sup (A+B)$.
    \item Consider $-A,-B$, the sets where all elements are negated. Then $\inf (A+B) = -\sup (-A + -B) = -\sup(-A) - \sup(-B) = \inf A + \inf B$
\end{enumerate}

\section*{7.5}
\begin{enumerate}[(a)]
    \item 0, as multpliying and dividing by the conjugate gives $\frac{1}{\sqrt{n^2+1}+n}$ which clearly approaches $0$
    \item $1/2$, by the same method as before.
    \item $1/4$ symmetric to the above.
\end{enumerate}

\end{document}