Assignment-daixieTM为您提供普利茅斯大学University of Plymouth MATH 1704 Analysis and Group Theory 分析与群论代写代考和辅导服务！
Instructions:
Analysis is a branch of mathematics that deals with the study of real numbers, sequences, limits, continuity, differentiation, and integration. It provides a rigorous foundation for calculus and forms the basis for many other areas of mathematics, such as differential equations, partial differential equations, and probability theory. Analysis also has numerous applications in science, engineering, and economics.
Group theory is a branch of abstract algebra that studies the properties of groups, which are sets equipped with a binary operation that satisfies certain axioms. Groups provide a way to describe and analyze symmetry, and they have many applications in physics, chemistry, cryptography, and computer science. Some of the key concepts in group theory include subgroups, homomorphisms, isomorphisms, normal subgroups, and quotient groups.
Both analysis and group theory involve a rigorous approach to mathematical reasoning and proof, and they require a solid foundation in mathematical concepts such as set theory and algebra. Python can be a useful tool for exploring and visualizing concepts in both areas, and there are many libraries and packages available for this purpose.
Let $A(r, R)$ denote the annulus centered at the origin with inner radius $r$ and outer radius $R$, where $0 \leq r<R \leq \infty$. Find all values of $r$ and $R$ for which $(A(r, R), \cdot)$ is a binary structure.
To show that $(A(r, R), \cdot)$ is a binary structure, we need to define a binary operation $\cdot$ on $A(r, R)$ that satisfies the following three properties:
- Closure: For all $a, b \in A(r, R)$, we must have $a \cdot b \in A(r, R)$.
- Associativity: For all $a, b, c \in A(r, R)$, we must have $(a \cdot b) \cdot c = a \cdot (b \cdot c)$.
- Identity: There must exist an element $e \in A(r, R)$ such that for all $a \in A(r, R)$, we have $a \cdot e = e \cdot a = a$.
Let $z_1 = re^{i\theta_1}$ and $z_2 = Re^{i\theta_2}$ be two complex numbers in $A(r, R)$. We can write $z_1$ and $z_2$ in Cartesian form as $z_1 = r(\cos\theta_1+i\sin\theta_1)$ and $z_2 = R(\cos\theta_2+i\sin\theta_2)$, respectively.
We can define the binary operation $\cdot$ on $A(r, R)$ as multiplication of complex numbers, i.e., $z_1 \cdot z_2 = rR(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))$.
- Closure: Since $0 \leq r<R \leq \infty$, we have $0 \leq rR \leq R^2 < \infty$. Also, $|\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)|=1$ for all $\theta_1$ and $\theta_2$, so $z_1 \cdot z_2 \in A(r, R)$.
- Associativity: Let $z_3 = Te^{i\theta_3}$ be a third complex number in $A(r, R)$. Then,
\begin{align*} (z_1 \cdot z_2) \cdot z_3 &= (rR(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))) \cdot Te^{i\theta_3}\ &= rRT(\cos(\theta_1+\theta_2+\theta_3)+i\sin(\theta_1+\theta_2+\theta_3))\ &= r(z_1 \cdot (z_2 \cdot z_3)), \end{align*}
so the operation is associative.
- Identity: The identity element $e$ must satisfy $z \cdot e = e \cdot z = z$ for all $z \in A(r, R)$. Let $e = e^{i\theta}$ be the identity element. Then, for all $z = Te^{i\theta_1} \in A(r, R)$, we have
$$z \cdot e = Te^{i\theta_1} \cdot e^{i\theta} = Te^{i(\theta_1+\theta)} = Te^{i\theta_1} = z,$$
Let $L$ denote a line passing through the origin in the complex plane. Verify that the multiplication – on the plane is not an induced binary operation on $L$. A binary structure $(S, \star)$ is associative if $x \star(y \star z)=(x \star y) \star z$ for every $x, y, z \in S$. We say that $(S, \star)$ is abelian if $x \star y=y \star x$ for every $x, y \in S$.
Let $z_1$ and $z_2$ be two non-zero complex numbers on the line $L$. Then, $z_1$ and $z_2$ can be written as $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$, where $r_1, r_2 > 0$ and $\theta_1, \theta_2$ are real numbers such that $\theta_1 = \theta_2$ since both $z_1$ and $z_2$ are on the same line passing through the origin.
Now, consider the product $z_1z_2$.
$$z_1z_2 = r_1 e^{i\theta_1} r_2 e^{i\theta_2} = r_1r_2 e^{i(\theta_1 + \theta_2)}$$
Since $\theta_1 = \theta_2$, we have $\theta_1 + \theta_2 = 2\theta_1$, so
$$z_1z_2 = r_1r_2 e^{i 2\theta_1}$$
Notice that $r_1r_2 > 0$ and $e^{i2\theta_1}$ lies on the line $L$ passing through the origin. Hence, $z_1z_2$ also lies on the line $L$.
Now, let us check if the multiplication is an induced binary operation on $L$. Suppose we have two non-zero complex numbers $z_1$ and $z_2$ on the line $L$. Then, $z_1z_2$ lies on $L$, as we have just shown. However, the origin $0$ is also on the line $L$.
If the multiplication is an induced binary operation on $L$, then $z_1z_2$ and $0$ must have a unique product on the line $L$. But this is not the case, since $0z_1 = 0$ for any non-zero complex number $z_1$ on the line $L$. Therefore, the multiplication is not an induced binary operation on $L$.
Note that this also means that the binary structure $(L, \cdot)$ is not a group, since a group requires an associative binary operation that is induced by the group operation.
Let $G$ be a group and let $a \in G$. By $a^2$, we understand $a a$. Inductively, we define $a^n$ for all positive integers $n \geq 2$. Show that $(a b)^n=a^n b^n$ for all $a, b \in G$ if and only if $G$ is abelian.
Suppose $G$ is abelian. Then for any $a, b \in G$, we have $(ab)^n = \underbrace{(ab)(ab)\cdots(ab)}{n \text{ times}} = \underbrace{a a \cdots a}{n \text{ times}} \underbrace{b b \cdots b}_{n \text{ times}} = a^n b^n$. Thus, $(ab)^n = a^n b^n$ for all $a, b \in G$.
Conversely, suppose $(ab)^n = a^n b^n$ for all $a, b \in G$. Let $a, b \in G$. Then we have:
\begin{align*} (ab)(ba) &= a(ba)b \ &= a(ab^{-1}a)b && \text{since } ba = ab^{-1}a \ &= (aa)(bb^{-1}) && \text{using the given condition with } x = a, y = b^{-1} \ &= ab^{-1}ba. \end{align*}
Similarly, we have $(ba)(ab) = ba^{-1}ab$. Thus, $(ab)(ba) = (ba)(ab)$ if and only if $ab^{-1}ba^{-1} = a^{-1}b^{-1}ab$, which is equivalent to $ab^{-1} = b^{-1}a$. This condition holds for all $a, b \in G$ if and only if $G$ is abelian. Thus, $G$ is abelian if and only if $(ab)^n = a^n b^n$ for all $a, b \in G$.
