# 抽象代数代考| THE INTEGERS作业代写

Problem 1.

When $a, b, c \in \mathbb{Z}$, prove: $a+c>b+c$ if and only if $a>b$.

Proof .

Take $a \leftrightarrow[s, m], b \leftrightarrow[t, n]$, and $c \leftrightarrow[u, p] .$ Suppose first that
$$a+c>b+c \quad \text { or } \quad([s, m]+[u, p])>([t, n]+[u, p])$$
Now this implies
which in tum implies $(s+u)+((n+p)>(t+t)+(m+p)]$
which, in turn, implies $(s+u)+(n+p)>(t+u)+(m+p)$
Then, by Theorem II’, Chapter $3,(s+n)>(t+m)$ or $[s, m]>[t, n]$ and $a>b$, as required.
Suppose next that $a>b$ or $[s, m]>[t, n] ;$ then $(s+n)>(t+m) .$ Now to compare

Problem 2.

Prove the Trichotomy Law: For any $m, n \in \mathbb{N}$, one and only one of the following is true:
When $a, b, c \in \mathbb{Z}$, prove: If $c<0$, then $a \cdot cb$.

Proof .

Take $a \leftrightarrow[s, m], b \leftrightarrow[t, n]$, and $c \leftrightarrow[u, p]$ in which ut+m \ a &>b \end{aligned}$(b) Suppose$a>b$. By simply reversing the steps in$(a)$, we obtain$a \cdot c<b \cdot c$, as required. Problem 3. Prove:$|a \cdot b|=|a| \cdot|b|$for all$a, b \in \mathbb{Z}$. Proof . Suppose$a>0$and$b>0$; then$|a|=a$and$|b|=b$. Then$|a \cdot b|=a \cdot b=|a| \cdot|b|$. Suppose$a<0$and$b<0$; then$|a|=-a$and$|b|=-b$. Now$a \cdot b>0$; hence,$|a \cdot b|=a \cdot b=(-a) \cdot(-b)=|a| \cdot|b|$. Suppose$a>0$and$b<0$; then$|a|=a$and$|b|=-b$. Since$a \cdot b<0,|a \cdot b|=-(a \cdot b)=a \cdot(-b)=|a| \cdot|b|$. The case$a<0$and$b>0\$ is left as an exercise.

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