Relations

For symmetry, ((3,2)) must accompany ((2,3)). The other required reverse pairs are already present.

From ((1,2)) and ((2,3)), ((1,3)) is present, so the needed transitive condition holds. But ((1,1),(2,2),(3,3)) are absent, so it is not reflexive.

Every present pair has its reverse, so the relation is symmetric. But ((1,2)) and ((2,3)) are present while ((1,3)) is not, so it is not transitive.

Since (a-a=0\in\mathbb{Z}), (a-b\in\mathbb{Z}) implies (b-a\in\mathbb{Z}), and the sum of integers is an integer. Hence it is an equivalence relation.

Because (7\equiv2\pmod{5}), all integers with the same remainder are in its class. Always form an equivalence class from the relation condition.

The remainders (0,1,2,3) form (4) different classes. For a modulo relation, the number of classes equals the modulus.

By same parity, the odd class is ({1,3}) and the even class is ({2,4}). A partition is the collection of equivalence classes.

Since (a^2=a^2), equality is symmetric, and (a^2=b^2), (b^2=c^2) imply (a^2=c^2). Hence (R) is an equivalence relation.

From (|x|=|-3|=3), (x=-3) or (x=3). In an absolute value relation, opposite signs may lie in the same class.

Symmetry: since $\gcd(a,b)=\gcd(b,a)$, if $aRb$ then $bRa$, so $R$ is symmetric. Not transitive: take $a=2$, $b=3$, $c=4$. We have $\gcd(2,3)=1$ so $2R3$, and $\gcd(3,4)=1$ so $3R4$. But $\gcd(2,4)=2\ne 1$, so $2\not R4$. Hence transitivity fails. Therefore, $R$ is symmetric but not transitive.

Because (a+b=b+a), reversing the pair keeps the same sum. But not every (a+a) is prime, so reflexivity is not guaranteed.

Every (a\leq a) is true, and (a\leq b\leq c) implies (a\leq c). But (2\leq3) does not imply (3\leq2).

Every (A\subseteq A), and the subset relation is antisymmetric and transitive. Do not treat it like equality because it is generally not symmetric.

A relation on $A$ is reflexive if every $(a,a)$ is in $R$. Here, $(1,1)$, $(2,2)$ and $(3,3)$ are not present in $R$ (for example, $(2,2)\notin R$). Hence reflexive property is false.

If ((a,b)) with (a\neq b) occurs, symmetry forces ((b,a)), contradicting antisymmetry. Hence only diagonal pairs can occur.

In the order ( \leq ), the greatest element is one that every element is less than or equal to. Here (a\leq4) for every (a\in A), so (4) is greatest.

A least element must divide every element. Here neither (2\mid3) nor (3\mid2), so no least element exists.

(1) divides every element, and every element divides (8). In divisibility order, identify least and greatest by the direction of divisibility.

In the universal relation, every possible ordered pair is present. Hence reflexive, symmetric, and transitive properties all hold automatically.

The empty relation has no counterexample, so symmetry and transitivity are vacuously true. But ((1,1)\notin R), so it is not reflexive.

The identity relation is reflexive, symmetric, and transitive, so it is an equivalence relation. It is also reflexive, antisymmetric, and transitive, so it is a partial order.

A reflexive (R) contains every ((a,a)), and the inverse of ((a,a)) is again ((a,a)). Therefore (R^{-1}) remains reflexive.

In a symmetric relation, ((a,b)\in R) implies ((b,a)\in R), so the inverse has the same pairs. Hence (R=R^{-1}).

The classes are ({1,4},{2},{3}), so the number of pairs is (2^2+1^2+1^2=6). Add the squares of the sizes of the equivalence classes.

Since (|a-a|=0\leq1) and (|a-b|=|b-a|), it is reflexive and symmetric. But (1R2) and (2R3) hold while (1R3) does not.