What is the inverse relation of R 
Determine the inverse relation $R^{−1}$ for the relation $R = \{(x,y) : x + 4y \text{ is odd}\}$ defined on $\mathbb{N}$.

does this mean that $R^{-1} = \{(y,x):y+4x \text{ is odd}\}$ ?
 A: Don't double switch.
Variables are just place holders.
$x R y$ when $x + 4y$ is odd and $R = \{(x,y)|x+4y$ is odd$\}$.
And $x R^{-1}y$ when $y Rx$ or when $y + 4x$ is odd. 
So $R^{-1} = \{(x,y)||y + 4x$ is $\}$.
If we want to "cheat" and take the easy way out.
If $R = \{(x,y)|$ something to do with $x$ and $y$$\}$ (In this case "something" is $x+4y$ is odd.)
$R^{-1} = \{(y,x)|$ something to do with $x$ and $y$$\}$.  (In this case "somethings" is the same thing.  $x + 4y$ is odd.)
But for aesthetic and conventional reasons we like to list ordered pairs as $x$ in the first positions so it'd be bet relable $x_{new}= y_{old}$ and $y_{new}=x_{old}$
So if 
$R = \{(x_{old}, y_{old})|$ some thing to do with $x_{old}$ and $y_{old}$}$.
The 
$R^{-1} = \{(x_{new}=y_{old}, y_{new}=x_{old}$ the same thing to do with $x_{old}$ (which is now $y_{new}$) and $y_{old}$ (which is now $x_{new}$$\}$.
So if $R = \{(x_{old},y_{old})| x_{old}+4y_{old}$ is odd$\}$ 
The $R^{-1} = \{(x_{new},y_{new})| y_{new}+4x_{new}$ is odd$\}$
A: *

*$(x,y) \in R^{-1} \iff (y,x)\in R$ ( by the definition of " inverse relation") 

*But $(y,x) \in R \iff y+4x$ is odd 
For , R is the relation such that the pair  (first element, second element ) belongs to $R$ iff 

first element + (4$\times$ second element) is odd



*

*So $R^{-1} = \{(x,y) | \space y+4x$ is odd $\}$
A: It means 
$$R^{-1}=\{(x,y): y+4x\text{ is odd}\}$$
