The task is to find examples for the following relations on the set of 
(and prove its correctness) :
1: antisymmetric and transitiv
2: antisymmetric and not transitiv (intransitiv)
3: not antisymmetric and transitiv
4: not antisymmetric and not transitive (intransitiv)
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So far I got the following examples:
1: Relation: ≤ (smaller than or equal to)
Proof of antisymmetry through a truth tabel.
Proof of transitivity: not sure: definition: aRb and bRc => aRc
Question: Is this allowed: a = b = α and c = β : αRα and αRβ => αRβ ? ––––––––-–––––––––––––-–––––––––––––-–––––––––––––-–––––––––––––-––––– 2: Relation: "is direct predecessor of" (e.g.: nR(n+1) -> n is direct predecessor of (n+1))
Proof of antisymmetry through truth table
Proof of intransitivity through definition (
)
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3: Relation: /, divided by (e.g.: aRb -> a divided by b
Proof of "not antisymmetric" by giving one example that does not hold for antisymmetry. (4/-4)
Proof of Transitivity: not sure how to prove it for everything. Maybe a truth table? ––––––––-–––––––––––––-–––––––––––––-–––––––––––––-–––––––––––––-––––– 4: Did not come up with a relation.
Question: Any tips that guide me in the right direction?
$1\require{cancel}$. Correct example. But I don't know what you mean by proving antisymmetry with a truth table. Both properties are almost too obvious to prove for relation $\leq$.
For any $a,b\in \mathbb{R}$ suppose we have both $a \leq b$ and $b \leq a$. This can only be true if $a=b$. Hence $\leq$ is antisymmetric.
For any $a,b,c\in \mathbb{R}$ suppose we have both $a \leq b$ and $b \leq c$. This implies $a\leq c$. Hence $\leq$ is transitive.
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$2$. Correct example, but you shouldn't call it "direct predecessor of" because the set is $\mathbb{R}$, not $\mathbb{Z}$. And $n$ is not a good variable name for a typical real number; $x,y,a,b$ are better.
Antisymmetry holds simply because there is no $a,b \in \mathbb{R}$ such that $aRb$ and $bRa$: if $a+1=b$ then $b+1=a+2 \neq a$.
Usually, proving a property doesn't hold is best done with a specific counterexample. So here, $0R1$ and $1R2$ because $0+1=1$ and $1+1=2$. But $0+1 \neq 2$ so $0\cancel{R}2$. Hence, $R$ is not transitive.
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$3$. I assume you mean "divides by" rather than "divided by", which is a binary operation, not a relation. But anyway, you seem to be thinking of integers rather than reals. The $|$ relation doesn't really make sense in $\mathbb{R}$ because every number divides every number except that $0$ doesn't divide any number.
Nevertheless, it is a valid example here, if you take $a|b$ on $\mathbb{R}$ to mean "$\frac{a}{b}$ is defined", which means $a\mid b$ for all $a,b\in\mathbb{R}$ where $a \neq 0$.
Then, $1|2$ and $2|1$ but $1\neq 2$ so "$|$" is not antisymmetric.
Now, suppose $a|b$ and $b|c$. Then we know $a\neq 0$, so $a|c$. Hence, "$|$" is transitive.
Having said that it's a valid example, still, you might want to find another because of the unorthodox usage of "$|$".
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$4$. You could use a simple example such as $R=\{(1,2),\; (2,1),\; (2,3)\}$. Or, perhaps a "better" example is the relation "is the negative of". That is, $R=\{(a,-a) \mid a\in\mathbb{R}\}$.
Neither of these relations are antisymmetric or transitive, which can be easily proved with simple counterexamples.