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5.1: CHAPTER 1
https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/05%3A_Solutions_and_Hints_for_Selected_Exercises/5.01%3A_Section_1-
Using $frac{\varepsilon}{2}$ in part $(2^{\prime})$ of Proposition 1.5.1 applied to the sets $A$ and $B$ , there exits $a \in A$ and $b \in B$ such that \[\sup A-\frac{\varepsilon}{2}<a \te...Using $frac{\varepsilon}{2}$ in part $(2^{\prime})$ of Proposition 1.5.1 applied to the sets $A$ and $B$ , there exits $a \in A$ and $b \in B$ such that $\sup A-\frac{\varepsilon}{2}<a \text { and } \sup B-\frac{\varepsilon}{2}<b .$ It follows that $\sup A+\sup B-\varepsilon<a+b .$ This proves condition $(2^{\prime})$ of Proposition 1.5.1 applied to the set $A + B$ that $\sup A + \sup B = \sup (A + B)$ as desired
5.5: Division in $\mathbb{Z}_{b}$
https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/05%3A_Modular_Arithmetic_and_Primes/5.05%3A_New_Page
It tells us that there is a unique element $a^{-1}$ such that $aa^{-1} = _{b} 1$ if and only if $a$ is in the reduced set of residues (modulo $b$ ). For every $a \in \mathbb{Z}_{p}$ there is ...It tells us that there is a unique element $a^{-1}$ such that $aa^{-1} = _{b} 1$ if and only if $a$ is in the reduced set of residues (modulo $b$ ). For every $a \in \mathbb{Z}_{p}$ there is a unique $a′ = _{p} -a$ such that $a+a′ = _{p} 0$ . For every $a \in \mathbb{Z}_{p}$ and $a \ne 0$ , there is a unique $a′ = a^{-1}$ so that $aa′ = _{p} 1$ .
1.2: Rational and Irrational Numbers
https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/01%3A_A_Quick_Tour_of_Number_Theory/1.02%3A_Rational_and_Irrational_Numbers
Note that any non-empty set $S$ of integers with a lower bound can be transformed by addition of a integer $b \in N_{0}$ into a non-empty $S+b$ in $N_{0}$ . The advantage of expressing a ration...Note that any non-empty set $S$ of integers with a lower bound can be transformed by addition of a integer $b \in N_{0}$ into a non-empty $S+b$ in $N_{0}$ . The advantage of expressing a rational number as the solution of a degree 1 polynomial, however, is that it naturally leads to Definition 1.12. The crux of the following proof is that we take an interval and scale it up until we know there is an integer in it, and then scale it back down.
4.6: Exercise
https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/04%3A_Number_Theoretic_Functions/4.06%3A_New_Page
Show that $\varphi (n) = |S-\bigcup_{i=1}^{r} A_{i}$ . (Hint: any number that is not co-prime with $n$ is a multiple of at least one of the $p_{i}$ .) Show that the principle of inclusion-exclusio...Show that $\varphi (n) = |S-\bigcup_{i=1}^{r} A_{i}$ . (Hint: any number that is not co-prime with $n$ is a multiple of at least one of the $p_{i}$ .) Show that the principle of inclusion-exclusion implies that $|S-\bigcup_{i=1}^{r} A_{i}| = n+n \sum_{l=1}^{r} (-1)^{l} \sum_{I_{l} \subseteq R} \prod_{i \in I_{l}} \frac{1}{p_{i}}$ . Show that $n+n \sum_{l=1}^{r} (-1)^{l} \sum_{I_{l} \subseteq R} \prod_{i \in I_{l}} \frac{1}{p_{i}} = n \prod_{i=1}^{r} (1-\frac{1}{p_{i}})$ .
5.2: CHAPTER 2
https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/05%3A_Solutions_and_Hints_for_Selected_Exercises/5.02%3A_Section_2-
Define $\alpha_{n}=\sup _{k \geq n}\left(a_{n}+b_{n}\right), \beta_{n}=\sup _{k \geq n} a_{k}, \gamma_{n}=\sup _{k \geq n} b_{k} .$ By the definition, \[\limsup _{n \rightarrow \infty}\left(a_{n}+b_...Define $\alpha_{n}=\sup _{k \geq n}\left(a_{n}+b_{n}\right), \beta_{n}=\sup _{k \geq n} a_{k}, \gamma_{n}=\sup _{k \geq n} b_{k} .$ By the definition, $\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\lim _{n \rightarrow \infty} \alpha_{n}, \limsup _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \beta_{n}, \limsup _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} \gamma_{n} .$ By Exercise 2.5.3, \[\alpha_{n} \leq \beta_{n}+\gamma_{n} \text { for all } n \in \math…
4.5: Taylor's Theorem
https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/04%3A_Differentiation/4.05%3A_Section_5-
Consider the function $g(t)=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(t)}{k !}(x-t)^{k}-\frac{\lambda}{(n+1) !}(x-t)^{n+1} .$ Then \[g(\bar{x})=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(\bar{x})}{k !}(x-\bar{x})^{k...Consider the function $g(t)=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(t)}{k !}(x-t)^{k}-\frac{\lambda}{(n+1) !}(x-t)^{n+1} .$ Then $g(\bar{x})=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(\bar{x})}{k !}(x-\bar{x})^{k}-\frac{\lambda}{(n+1) !}(x-\bar{x})^{n+1}=f(x)-P_{n}(x)-\frac{\lambda}{(n+1) !}(x-\bar{x})^{n+1}=0 .$ and $g(x)=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(x)}{k !}(x-x)^{k}-\frac{\lambda}{(n+1) !}(x-x)^{n+1}=f(x)-f(x)=0 .$ By Rolle's theorem, there exists $c$ in between $\bar{x}$ and $x$ such that…
1.5: Exercises
https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/01%3A_A_Quick_Tour_of_Number_Theory/1.05%3A_Exercises
Use an pictorial argument similar to that of Figure 2 to show that $\mathbb{N} \times \mathbb{N}$ (the set of lattice points $(n, m)$ with $n$ and $m$ in $\mathbb{N}$ ) is countable. a) Given...Use an pictorial argument similar to that of Figure 2 to show that $\mathbb{N} \times \mathbb{N}$ (the set of lattice points $(n, m)$ with $n$ and $m$ in $\mathbb{N}$ ) is countable. a) Given a set $A$ , show that there is an injection $f : A \rightarrow P(A)$ . (Hint: for every element $a \in A$ there is a set $\{a\}$ .) Show that addition respects this equivalence relation. (Hint: If $a+b = c, a \sim a′$ , and $b \sim b′$ , then $a′+b′ =c′$ with $c \sim c′$ .)
2: Sequences
https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/02%3A_Sequences
More precisely, a sequence of elements of a set $A$ is a function $f: \mathbb{N} \rightarrow A$ . We will denote the image of $n$ under the function with subscripted variables, for example, \(a_{...More precisely, a sequence of elements of a set $A$ is a function $f: \mathbb{N} \rightarrow A$ . We will denote the image of $n$ under the function with subscripted variables, for example, $a_{n}=f(n)$ . We will also denote sequences by $\left\{a_{n}\right\}_{n=1}^{\infty}$ , $\left\{a_{n}\right\}_{n}$ , or even $\left\{a_{n}\right\}$ . Each value $a_{n}$ is called a term of the sequence, more precisely, the $n$ -th term of the sequence.
2.4: Corollaries of the Fundamental Theorem of Arithmetic
https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/02%3A_The_Fundamental_Theorem_of_Arithmetic/2.04%3A_New_Page
$\begin{array}{ccc} {a = \prod_{i=1}^{s} p_{i}^{k_{i}}}&{and}&{b = \prod_{i=1}^{s} p_{i}^{l_{i}}} \end{array} \nonumber$ On the other hand, any integer greater than $m$ has a unique factorization ... $\begin{array}{ccc} {a = \prod_{i=1}^{s} p_{i}^{k_{i}}}&{and}&{b = \prod_{i=1}^{s} p_{i}^{l_{i}}} \end{array} \nonumber$ On the other hand, any integer greater than $m$ has a unique factorization that either contains a prime not in the list $P$ and therefore divides neither $a$ nor $b$ , or, if not, at least one of the primes in $P$ in its factorization has a power greater than $m_{i}$ .
Back Matter
https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/zz%3A_Back_Matter
3.6 Exercises
https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/03%3A_Linear_Diophantine_Equations/3.6_Exercises
Define $\gcd(p_{1}, p_{2})$ as the highest degree polynomial that is a divisor of both $p_{1}$ and $p_{2}$ . (Remark: if you do this very carefully, you will realize that in this problem you real...Define $\gcd(p_{1}, p_{2})$ as the highest degree polynomial that is a divisor of both $p_{1}$ and $p_{2}$ . (Remark: if you do this very carefully, you will realize that in this problem you really need to consider equivalence classes of polynomials, in the sense that two polynomials $p$ and $q$ are equivalent if there is a non-zero constant $c \in \mathbb{R}$ such that $p = cq$ .

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