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# System prepended metadata

title: Abstract Algebra I-solution

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# Abstract Algebra I-solution
## Quiz1 
### 1.Consider if the following relation on $\mathbb{R}$
a) a ~ b if |a -- b| < 1

$\forall$ a $\in$ $\mathbb{R}$, |a -- a| = 0 < 1 i.e a ~ a  $\Rightarrow$ reflexivity.

Let  a ~ b i.e.  |a -- b| < 1.   |b -- a| = |a -- b| < 1 i.e. b ~ a $\Rightarrow$ symmetry.

Let a=1 b=1.5 c=2, satisfying |a -- b|<1 and |b -- c|<1 
We have a ~ b and b ~ c ,but |a -- c|=1 i.e. a $\not\sim$ c. 
$\Rightarrow$ not transitivity.

b) a ~ b if a $\le$ b 
$\forall$ a $\in$ $\mathbb{R}$, a $\le$ a i.e. a ~ a $\Rightarrow$ it is reflexivity. 

Let  0 $\le$ 1  satisfying 0 ~ 1. But $1\not \le0$ i.e. 1 $\not\sim0$  $\Rightarrow$  not symmetry.

Let a ~ b and b ~ c
i.e. a $\le$ b  and b $\le$ c 
We have a $\le$ c  i.e a ~ c $\Rightarrow$  transitivity.
### 2.Define a relation on $\mathbb{R^2}$ as follows. For any $\vec{v}$=$(v_1,v_2)$,$\vec{u}$=$(u_1,u_2)$ $\in$ $\mathbb{R^2}$, $\vec{v}$ ~ $\vec{u}$ if  $v_1$ -- $u_1$ = $v_2$ -- $u_2$
a) 
$\forall$ $\vec{v}$ $\in$ $\mathbb{R^2}$, $\vec{v}$=$(v_1,v_2)$
$v_1$ -- $v_1$ = $0$ = $v_2$ -- $v_2$ i.e. $\vec{v}$ ~ $\vec{v}$ $\Rightarrow$ reflexivity.

Let $\vec{v}$  ~ $\vec{u}$ $\vec{v}$=$(v_1,v_2)$  $\vec{u}$=$(u_1,u_2)$ i.e. $v_1$ -- $u_1$ = $v_2$ -- $u_2$
$u_1$ -- $v_1$ = -- ($v_1$ -- $u_1$)  = -- ($v_2$ -- $u_2$)  = $u_2$ -- $v_2$
i.e. $\vec{u}$ ~ $\vec{v}$ $\Rightarrow$symmetry 

Let $\vec{v}$  ~ $\vec{u}$ , $\vec{u}$  ~ $\vec{w}$ 
$\vec{v}$=$(v_1,v_2)$  $\vec{u}$=$(u_1,u_2)$ $\vec{w}$=$(w_1,w_2)$
i.e $v_1$ -- $u_1$ = $v_2$ -- $u_2$ and $u_1$ -- $w_1$ = $u_2$ -- $w_2$

$v_1$ -- $w_1$ = $v_1$ + ( -- $u_1$ + $u_1$) -- $w_1$ = ($v_2$ -- $u_2$) + ($u_2$ -- $w_2$) = $v_2$ -- $w_2$
i.e. $\vec{v}$ ~ $\vec{w}$ $\Rightarrow$ transitivity

Thus, ~ is an equivalence relation.

b)
\{ (0,x) | x $\in$ $\mathbb{R}$ }
for any  $\vec{v}$=$(v_1,v_2)$ $\in$ $\mathbb{R^2}$
let x=$v_2$ -- $v_1$ 
$v_2$ -- x = $v_2$ -- ($v_2$ -- $v_1$) = $v_1$ -- 0
 $\Rightarrow$ $\vec{v}$ $\in$ \{ (0,x) | x $\in$ $\mathbb{R}$ }

Thus, \{ [(0,x)] | x $\in$ $\mathbb{R}$ } is a complete set of eqivanlence classes. 

c) Let $\vec{v}$  ~ $\vec{u}$ write $\vec{v}$=$(v_1,v_2)$  $\vec{u}$=$(u_1,u_2)$
We have $v_1$ -- $u_1$ = $v_2$ -- $u_2$
We want to show f([$\vec{v}$]) = f([$\vec{u}$])



f([$\vec{v}$]) = 2$v_1$ -- 2$v_2$ = 2$v_1$ -- (2$u_1$ - 2$u_1$) -- 2$v_2$ = (2$v_1$ -- 2$u_1$) + 2$u_1$ -- 2$v_2$ = 2$v_2$ -- 2$u_2$ + 2$u_1$ -- 2$v_2$ 
= 2$u_1$ -- 2$u_2$ = f([$\vec{u}$])
Thus,f is well-defined.
