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.

The direct answer is A: symmetric but not reflexive. The rule is
aRb\iff\gcd(a,b)=1
. The greatest common divisor does not change when the two numbers are exchanged:
\gcd(a,b)=\gcd(b,a)
. Thus, whenever
aRb
, we also have
bRa
, so the relation is symmetric. Reflexivity would require
\gcd(a,a)=1
for every
a
in the set. But
\gcd(2,2)=2
, and also
\gcd(3,3)=3
, so the relation is not reflexive. Option A is correct. Option B is wrong because reflexivity fails. Option C, equivalence relation, is wrong because an equivalence relation must be reflexive, symmetric and transitive; it is not reflexive, and transitivity also fails, for example
\gcd(2,3)=1
and
\gcd(3,4)=1
, but
\gcd(2,4)=2\ne1
. Option D is wrong because the relation is not reflexive. Memory cue: coprime relation is symmetric, but a number is not coprime with itself unless it is 1.

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.

The direct answer is A: symmetric is the false property. A relation is reflexive when every diagonal pair
(a,a)
is present. Here
(1,1),(2,2),(3,3)
are all listed, so it is reflexive. It is antisymmetric because the only non-diagonal pair is
(1,2)
, while
(2,1)
is absent; there is no distinct pair occurring in both directions. It is transitive: diagonal pairs cause no problem, and the only possible chain involving the extra pair is
(1,1),(1,2)
or
(1,2),(2,2)
, both leading to
(1,2)
, which is present. However, it is not symmetric because
(1,2)\in R
but
(2,1)\notin R
. Therefore A is the false statement. Option B is true due to all diagonal pairs. Option C is true by the definition of antisymmetry. Option D is true after checking the possible chains. A common mistake is to assume that having
(1,2)
automatically requires
(2,1)
; that requirement belongs to symmetry, not every relation.

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.