Confusion about Reflexive, symmetric, anti-symmetric and transitivity So some equations are easy to understand and figure out. But, my lack of understanding of the basic info makes a lot of questions very hard:
for reflexive its pretty easy all i have to do is change y to x and see if both sides are equal. however, for example for X is the set of prime numbers greater than 2 and xRy (x+y)/2 is prime so i need to prove reflexive but x+y need to be odd+even so that its divisible of 2 is prime so taking x as odd proves it right but taking x as even gets us even makes me confused.
for transitive all i have to do is prove there is a relation between x and c through y and c and x and y but a lot of answers say 1R0 0R1 implies 1R1 isn't transitive and i do not get it when to use this way for example Let X be a nonempty set of positive natural numbers and let R be
the binary relation defined by xRy ∃n ∈ ℕ* y=x^n
1. prove that R is an order relation
so i should do z=x^n^n thus proving it transitive or should i say z=x^n proving it not transitive.  
for symmetric and anti symmetric its i should switch them x and y with each other and if they were the same in all cases then its symmetric. if there was one case then that means its anti symmetric like x=y is the only way we will get the same answer.
so basically, unless i memorized the way of solving of each type of question ill get lost in what to do and how to prove it. hope someone can clear up my misunderstanding about this topic.


 A: The relation you use in your question is defined on the set $X$ of prime numbers greater than 2, so $2\notin X$, and it says $x$ is related to $y$, which we will denote by $xRy$ if and only if $\frac{x+y}2$ is a prime number.
To show that this relation IS reflexive you have to show that for all $x\in X$, $x$ is related to itself. This is not hard to prove using the definition since $$xRx \iff \frac{x+x}2 \text{ is prime} \iff x \text{ is prime}.$$
Since $x\in X$, $x$ is prime, so $x$ is related to itself.
Now we show that this relation IS symmetric, this means that if two elements $x,y\in X$ are related, that is, $xRy$, then $yRx$ holds. This is easy to do here because addition is commutative. Assume $xRy$, $$xRy \iff \frac{x+y}2 \text{ is prime} \iff \frac{y+x}2 \text{ is prime} \iff yRx.$$
For antisymmetry you have to show that whenever you have $xRy$ and $yRx$ this implies $y=x$. This relation is NOT antisymmetric since $3R7, 7R3$ but $3\neq 7$.
$$3R7 \text{ because } \frac{3+7}2 = 5 \text{ which is prime.}$$
For transitivity you have to show that $xRy$, $yRz$ implies $xRz$. Your relation is NOT transitive since $3R7$, $7R67$ but 3 is not related to 67.
Hope this helps you understand the properties of relations better.
Now we prove the second relation is an order relation.
Reflexive: Let $x\in X$, $xRx \iff \exists n \text{ natural s.t. } x=x^n$. Clearly $x = x^1$, so such an $n$ exists, $xRx$.
Antisymmetric: Assume $xRy$ and $yRx$, there exist $m,n \in \mathbb{N}$ such that $y=x^n, x=y^m$, this implies $x = x^{mn}$ and $y= y^{mn}$. From this you can get $1 = x^{mn-1}$ so $mn-1 = 0$. Then, $mn = 1$ and this implies $m = n = 1$. Hence, $x = y$.
Transitive: Assume $xRy$ and $yRz$, then there exist $m,n \in \mathbb{N}$ such that $y=x^n, z=y^m$, this implies $z = x^{mn}$. Hence, $mn\in \mathbb{N}$ verifies $z = x^{mn}$ and this means $x Rz$.
Therefore, $R$ is an order relation.
