part a and b
This commit is contained in:
16
q2/a.md
16
q2/a.md
@@ -1 +1,15 @@
|
||||
WIP
|
||||
## Question 2
|
||||
|
||||
> A relation $\star$ is defined on the set $\mathbb{Q}^{2}$ by
|
||||
> $\left(x_{1},x_{2}\right)\star\left(y_{1},y_{2}\right)$ if and only if $x_{1}y_{2}=x_{2}y_{1}$.
|
||||
>
|
||||
> For each of the questions below, be sure to provide a proof supporting your answer.
|
||||
|
||||
### Part A
|
||||
|
||||
> Is $\star$ reflexive?
|
||||
|
||||
A relation on $\mathbb{Q}^2$ is reflexive if $a\star a$ for all $a\in\mathbb{Q}^2$. To prove reflexivity, assume that $x_{1}x_{2}=y_{2}y_{1}$ is true (let this be $(1)$).
|
||||
|
||||
Given that $x_1 y_2 = x_2 y_1$ is known to be true, we can rearrange for $y_1$ to find that $y_1=\frac{x_1 y_2}{x_2} (2)$. In a similar manner, rearranging for $y_2$ provides $y_2=\frac{x_2 y_1}{x_1} (3)$. Substituting $(2)$ and $(3)$ into $(1)$, it can be found that $x_2=y_2$ and $x_1 = y_1$. Therefore, $x_{1}x_{2}=y_{2}y_{1}=x_1 y_2 = x_2 y_1$. Since every $(x_1,x_2)$ satisfies the requirement that $x_{1}x_{2}=y_{2}y_{1}$, the relation is reflexive.
|
||||
|
||||
|
||||
16
q2/b.md
16
q2/b.md
@@ -1 +1,15 @@
|
||||
WIP
|
||||
## Question 2
|
||||
|
||||
> A relation $\star$ is defined on the set $\mathbb{Q}^{2}$ by
|
||||
> $\left(x_{1},x_{2}\right)\star\left(y_{1},y_{2}\right)$ if and only if $x_{1}y_{2}=x_{2}y_{1}$.
|
||||
>
|
||||
> For each of the questions below, be sure to provide a proof supporting your answer.
|
||||
|
||||
### Part B
|
||||
|
||||
> Is $\star$ symmetric?
|
||||
|
||||
For $\star$ to be symmetric, every set of elements $(x_1,x_2)\in\Q^2,(y_1,y_2)\in\Q^2$ which satisfies $(x_1,x_2)\star(y_1,y_2)$ must also satisfy $(y_1,y_2)\star(x_1,x_2)$.
|
||||
|
||||
Assume $(y_1,y_2)\star(x_1,x_2)$ is true. This means that $y_1 x_2 =y_2 x_1$, which is the same as as $x_{1}y_{2}=x_{2}y_{1}$ (i.e. the original relation). Therefore the requirement is satisfied, and $\star$ is symmetric.
|
||||
|
||||
|
||||
Reference in New Issue
Block a user