Let x$,$ y $\backslash$epsi Z be given such that $($x$,$ y$)\backslash$epsi R iff y $>1-$ x with x$,$ y $\backslash$epsi Z$.$

R is a symmetric relation. Which one of the following alternatives can be used to prove that R is a symmetric relation?

Select one:

a$.$

Let x$,$ y $\backslash$epsi Z be given such that $($x$,$ y$)\backslash$epsi R and $($y$,$ x$)\backslash$epsi R$,$

then y $>1-$ x $=$ x $>1-$ y

i$.$e$.$ y $>1-$ x $=$ x $-1>-$y

i$.$e$.$ y $>1-$ x $=1-$ x $<$ y $($multiply by $-1,$ which changes $>$ to $<$ as well$)$

b$.$

R is not a symmetric relation. We give a counterexample:

Ordered pairs $(5,-4)$ and $(-4,5)$ are not ordered pairs in R$.$

We substitute the ordered pair $(5,-4)$ for $($x$,$ y$)$:

$-4>1-5$

i$.$e$.-4>-4,$ which is false.

We substitute the ordered pair $(-4,5)$ for $($y$,$ x$)$:

$5>1-(-4)$

i$.$e$.5>5,$ which is also false.

We have therefore proved that R is not symmetric.

c$.$

$(5,-3)\backslash$epsi R and $(-3,5)\backslash$epsi R$,$ therefore R is symmetric.

We substitute these pairs for the proof:

y $>1-$ x and x $>1-$ y

i$.$e$.-3>1-5$ and $5>1-(-3)$

i$.$e$.-3>-4,$ which is true and $5>4,$ which is also true.

d$.$

Let x$,$ y $\backslash$epsi Z be given such that $($x$,$ y$)\backslash$epsi R$.$

We have to proof that $($y$,$ x$)\backslash$epsi R$,$ i$.$e$.$ x $>1-$ y

If $($x$,$ y$)\backslash$epsi R then

y $>1-$ x

i$.$e$.$ y $-1>-$x

i$.$e$.1-$ y $<$ x $($multiply by $-1,$ which changes $>$ to $<)$

i$.$e$.$ x $>1-$ y

thus $($y$,$ x$)\backslash$epsi R$.$