代数|MA20217/MA20219 Algebra 2B代写

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Elementary axiomatic theory of rings. Integral domains, fields, characteristic. Subrings and product of rings. Homomorphisms, ideals and quotient rings. Isomorphism theorems. Fields of fractions. 

这是一份Bath巴斯大学MA20217/MA20219作业代写的成功案

代数|MA20217/MA20219 Algebra 2B代写

Use Cramer’s rule to solve the system
$$
\begin{aligned}
2 x_{1}-x_{2} &=1 \
4 x_{1}+4 x_{2} &=20 .
\end{aligned}
$$
Solution. The coefficient matrix and right-hand-side vectors are
$$
A=\left[\begin{array}{rr}
2 & -1 \
4 & 4
\end{array}\right] \text { and } \mathbf{b}=\left[\begin{array}{r}
1 \
20
\end{array}\right]
$$
so that
$$
\operatorname{det} A=8-(-4)=12
$$
and therefore
$$
x_{1}=\frac{\left|\begin{array}{rr}
2 & 1 \
4 & 20
\end{array}\right|}{\left|\begin{array}{rr}
2 & -1 \
4 & 4
\end{array}\right|}=\frac{36}{12}=3 \quad \text { and } \quad x_{2}=\frac{\left|\begin{array}{rr}
1 & -1 \
20 & 4
\end{array}\right|}{\left|\begin{array}{rr}
2 & -1 \
4 & 4
\end{array}\right|}=\frac{24}{12}=2
$$

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MA20217/MA20219 COURSE NOTES :

Since this holds for all $x$, we conclude that $(f+g)+h=f+(g+h)$, which is the associative law for addition of vectors.

Next, if 0 denotes the constant function with value 0 , then for any $f \in V$ we have that for all $0 \leq x \leq 1$,
$$
(f+0)(x)=f(x)+0=f(x) .
$$
(We don’t write the zero element of this vector space in boldface because it’s customary not to write functions in bold.) Since this is true for all $x$ we have that $f+0=f$, which establishes the additive identity law. Also, we define $(-f)(x)=-(f(x))$ so that for all $0 \leq x \leq 1$,
$$
(f+(-f))(x)=f(x)-f(x)=0,
$$



代数 Algebra II MATH2581-WE01

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这是一份durham杜伦大学MATH2581-WE01作业代写的成功案例

代数 Algebra II MATH2581-WE01
问题 1.

\begin{aligned}
\varphi_{L}(a) \vee \varphi_{L}(b)=\left(a^{\star \uparrow} \vee b^{\star \dagger}\right) \cap D(L) &=\left(a^{\star} \wedge b^{\star}\right)^{\dagger} \cap D(L) \
&=(a \vee b)^{\star \uparrow} \cap D(L) \
&=\varphi_{L}(a \vee b)
\end{aligned}

证明 .


$$
\begin{aligned}
\varphi_{L}(a) \cap \varphi_{L}(b)=a^{\star \uparrow} \cap b^{\star \uparrow} \cap D(L) &=\left(a^{\star} \vee b^{\star}\right)^{\dagger} \cap D(L) \
&=(a \wedge b)^{\star \uparrow} \cap D(L) \
&=\varphi_{L}(a \wedge b) .
\end{aligned}
$$
Since also $\varphi_{L}(0)={1}$ and $\varphi_{L}(1)=D(L)$ it follows that $\varphi_{L}$ is a $(0,1)$-lattice morphism.
Now, by the distributivity, we have $x=x^{\star \star} \wedge\left(x \vee x^{\star}\right)$. It follows that
$$
(\forall x \in L) \quad x^{\dagger}=x^{\star \star \uparrow} \vee\left(x^{\dagger} \cap D(L)\right) \text {. }
$$
In fact, if $t \in x^{\dagger}$ then $t=t \vee x=\left(t \vee x^{\star \star}\right) \wedge\left(t \vee x \vee x^{\star}\right)$ where $x \vee x^{\star} \in D(L)$; and conversely if $t \in x^{\star \star \uparrow} \vee\left(x^{\dagger} \cap D(L)\right)$ then $t \geqslant y \wedge z$ where $y \geqslant x^{\star \star} \geqslant x$ and $z \geqslant x$ and thus $t \geqslant x$.

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MATH2581-WE01 COURSE NOTES :

Proof $\Rightarrow$ : If $L$ is a Heyting algebra follow from the above whereas (3) is immediate from the fact that $x:(x \wedge y)=1$.
$\Leftarrow$ : Conversely, if the identites hold then $x \wedge(y: x)=x \wedge y \leqslant y$; and if $x \wedge z \leqslant y$ then
$$
z \wedge(y: x)=z \wedge[(z \wedge y):(z \wedge x)]=z \wedge[(y \wedge z):(x \wedge y \wedge z)]=z
$$
and so $z \leqslant y: x$. Thus $(L ;:)$ is a Heyting algebra.








代数 Algebra MAT00010C

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这是一份YORK约克大学MAT00001C作业代写的成功案例

代数 Algebra MAT00010C
问题 1.

first box. Since the third box is formed from $x_{1}\left(\begin{array}{l}1 \ 3\end{array}\right)+x_{2}\left(\begin{array}{l}2 \ 1\end{array}\right)=\left(\begin{array}{l}6 \ 8\end{array}\right)$ and $\left(\begin{array}{l}2 \ 1\end{array}\right)$, and the determinant is unchanged by adding $x_{2}$ times the second column to the first column, the size of the third box equals that of the second. We have this.
$$
\left|\begin{array}{ll}
6 & 2 \
8 & 1
\end{array}\right|=\left|\begin{array}{ll}
x_{1} \cdot 1 & 2 \
x_{1} \cdot 3 & 1
\end{array}\right|=x_{1} \cdot\left|\begin{array}{ll}
1 & 2 \
3 & 1
\end{array}\right|
$$

证明 .

Solving gives the value of one of the variables.
$$
x_{1}=\frac{\left|\begin{array}{ll}
6 & 2 \
8 & 1
\end{array}\right|}{\left|\begin{array}{ll}
1 & 2 \
3 & 1
\end{array}\right|}=\frac{-10}{-5}=2
$$
The theorem that generalizes this example, Cramer’s Rule, is: if $|A| \neq 0$ then the system $A \vec{x}=\vec{b}$ has the unique solution $x_{i}=\left|B_{i}\right| /|A|$ where the matrix $B_{i}$ is formed from $A$ by replacing column $i$ with the vector $\vec{b}$. Exercise 3 asks for a proof.
For instance, to solve this system for $x_{2}$
$$
\left(\begin{array}{ccc}
1 & 0 & 4 \
2 & 1 & -1 \
1 & 0 & 1
\end{array}\right)\left(\begin{array}{l}
x_{1} \
x_{2} \
x_{3}
\end{array}\right)=\left(\begin{array}{c}
2 \
1 \
-1
\end{array}\right)
$$

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MAT00010C COURSE NOTES :

Recall the definitions of the complex number addition
$$
(a+b i)+(c+d i)=(a+c)+(b+d) i
$$
and multiplication.
$$
\begin{aligned}
(a+b i)(c+d i) &=a c+a d i+b c i+b d(-1) \
&=(a c-b d)+(a d+b c) i
\end{aligned}
$$