# 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.

• what does xRx mean exactly and how should i prove it in a general case @TonyK – oma May 6 at 21:39
• It looks like you know that $xRy$ means that $x$ relates to $y$ under the relation $R$. Then $xRx$ just means that $x$ relates to itself under $R$, and if this holds for all $x$, then $R$ is reflexive. For your example with prime numbers, saying that $R$ is reflexive means that $(x+x)/2$ is always prime. But $(x+x)/2=x$, and your relation is on the set of all prime numbers greater than $2$, so $x$ is prime, hence $xRx$ for all $x$ and $R$ is reflexive. I don't see why you would ever have to consider even values of $x$, since your relation is only on odd primes. – Kevin Long May 6 at 21:50

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.

• i get the reflexive a lot simpler now. transitive can get me confused in some other ones but its quite easier in most cases now. the thing i do not get about symmetric is that in a general case how do i prove what i said is right and anti symmetric gets me even more confused because i do not understand what am i even trying to prove , i need to prove that it stands for x=y? so something like $x^2-y^2=x-y$ is anti symmetric by nature. – oma May 6 at 22:12
• For symmetric you have to assume $xRy$ and show $yRx$ and for antisymmetric you have to assume $xRy$ and $yRx$ and show $x=y$. – Manuel DaGeo May 6 at 22:14
• i also would like to add that the question i got y=x^n from said that i need to prove an order relation but if its not transitive i can not prove it any more( i can add the full question to my post if you need) – oma May 6 at 22:23
• In your question there are two distinct relations, the relation $xRy \iff \exists n \in \mathbb{N} s.t. y = x^n$ is indeed an order relation. If you edit the question and add your work I can help you prove it. – Manuel DaGeo May 6 at 22:27
• added the question – oma May 6 at 22:49