Reflexive and symmetric of the relation “xRy iff x^2 + y^2 is odd or y < 0”

Question

We have a relation $R$ that is defined on the integer set $\mathbb{Z}$ such that $$xRy \Leftrightarrow (\exists k \in \mathbb{Z}: x^2 + y^2 = 2k – 1) \vee (y < 0)$$

How can we prove that it is not reflexive, nor symmetric?

I concluded that the relation $R$ is not reflexive as $\{\exists x \in \mathbb{Z}: \neg xRx\}$ and if $x=2$ then $2^2 + 2^2 = 8$ is even and $2 > 0$ which falsifies both the $x^2 + y^2$ being an odd number and $y$ (or $x$ in this case) being less than 0.

To prove that the relation is not symmetric, the predicate needs to be proven where $\{\exists x,y \in \mathbb{Z}: \neg (xRy \Leftrightarrow yRx)\}$. $x^2+y^2$ is always the same as $y^2+x^2$ (whether it is odd or even) in $R$ so how is it that this relation is not symmetric then? Of course, the $x^2+y^2$ do not have to be odd all the time but we are talking?

Answer 1

==== new answer ====

Note: There are two ways we can have $xR y$.

Case 1: $y < 0$. Then $anything$ OR $y< 0$ is true. So $xRy$ for any $x$ and any $y< 0$.

Case 2: $y\ge 0$ and $x^2 + y^2$ is odd. then $x^2 + y^2$ is odd OR $anything$ is true. So $xRy$ anytime $x^2 + y^2$ is odd

... But there is only one way $xRy$ is false:

Case 3: $y \ge 0$ and $x^2 + y^2$ is even. Then $x^2 + y^2$ is odd is false and $y < 0$ is false so $x\not R y$.

So to have $xRy$ and $y\not R x$ we must have either 1) $y< 0$ or 2) $x^2 + y^2$ is odd, and we must have 3) $x\ge 0$ and $y^2 + x^2$ is even

But we can't have both $y^2 + x^2$ be even and $x^2 + y^2$ be odd.

So to have $xRy$ and $y\not Rx$ we must have:

$y< 0, x\ge 0, y^2 + x^2$ is even. In that case we have $xRy$ but $y\not Ry$.

====old answer ====

1) To prove it is not reflexive simply find an $x$ where

[($x^2 + x^2$ is odd) OR ($x < 0$)] is false.

In order for an OR statement to be false we must find an $x$ where

$x^2 + x^2$ is odd, is false. ANd where $x< 0$ is false.

For $x^2 + x^2$ is odd to be false we must have $x^2 + x^2$ be even.

And for $x < 0$ to be false we must have $x\ge 0$.

....

So to show $R$ is not reflexive we have to find an $x$ where:

a) $x^2 + x^2 = 2x^2$ is even.

b) $x\ge 0$.

Can you find such an $x$.

.....

2)

TO prove it is not symmetric find an $x$ and $y$ where

($x^2 + y^2$ is odd or $y < 0$) is true. But ($y^2 + x^2$ is odd or $x< 0$) is false.

For ($y^2 + x^2$ is odd or $x< 0$) to be false we must have $y^2 + x^2$ to be even AND for $x \ge 0$.

But we must have $x^2 + y^2$ odd, of $y< 0$.

So $x^2 + y^2 = y^2 +x^2$ and we must have $y^2 + x^2$ even we cant have $x^2 + y^2$ being odd.

So for ($x^2 + y^2$ is odd or $y < 0$) to be true we must have $y< 0$.

.....

So can we find $x,y$ where $x^2 + y^2$ is even, $x \ge 0$ and $y < 0$?

========

If $x = 1$ then $1^2 + 1^2 =2$ is not odd. And $1\ge 0$ so $1 \not < 0$ so $1 \not R 1$ so $R$ is not reflexive. Indeed, for any $x$ we have $x^2+x^2=2x^2$ is not odd. And so long as $x\ge 0$ then $x \not < 0$. So for any non-negative $x$ we have $x \not R x$

.

If $x = 1$ and $y= -3$ then $y < 0$ and so $1R-3$. But $x\ge 0$ and $x^2 + y^2 = 1+9 =10$ is even so $-3\not R 1$ so $R$ is not symmetric. In fact so long as $x\ge 0$ and $y< 0$ and $x^2 + y^2$ is even (which happens whenever $x,y$ are the same parity we have $y < 0$ so $xRy$ and $x\ge 0$ and $x^2 + y^2$ is even and $y\not Rx$. So if you ever have $x$ not negative and even/odd. And $y< 0$ and even/odd. We will always have $xRy$ but $y\not Rx$.