Relations and Functions: questions with answers

Membership is decided by substituting into the condition in the given order. Putting $a = 4$ and $b = 3$ into $a + b = 7$ gives the sum is still 7, so the reverse pair is in $R$. The order in an ordered pair matters: $(3, 4)$ and $(4, 3)$ are different elements of $\mathbf{N} \times \mathbf{N}$, and each has to be tested separately.

For $y$ to be a natural number we need $y = 53 - 2x \geq 1$, that is $2x \leq 53 - 1$, so $x \leq 26$. Since $x$ is a natural number the domain is $\{1, 2, \dots, 26\}$. Taking $x \leq 27$ would force $y = 0$ or a negative value, which is not in $\mathbf{N}$.

Reflexivity ranges over the whole of $S$, so all four self-pairs $(S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4)$ are needed -- including those for $S_3$, which $R$ never mentions. Adding self-pairs only for the elements that already appear in $R$ leaves $S_3$ without one, and reversing the existing pairs repairs symmetry instead.

A one-one function must give the four elements of $S$ four different images, and $J$ offers only three values to choose from, so two elements of $S$ are always forced to share an image. Plenty of functions from $S$ to $J$ do exist -- $3^4 = 81$ of them -- so it is not that no function exists; and unequal sizes are no obstacle in the other direction, where one-one functions from a 3-element set to a 4-element set are plentiful.

Reflexivity needs $(a, a) \in R$ for every element of $A$, and $A$ has 5 elements, so all 5 self-pairs have to be added. The empty relation is symmetric and transitive vacuously, but reflexivity demands that pairs be present, so $\phi$ is not already reflexive.

Reflexivity asks for $(a, a) \in R$ for every $a \in A$, so all three of $(1,1)$, $(2,2)$ and $(3,3)$ must be present. The pair $(2, 2)$ is missing. Extra pairs beyond the self-pairs never disturb reflexivity, and an element of $A$ that appears in no pair of $R$ at all still has to be checked.

Symmetry asks that every pair present in $R$ has its reverse present. The pair $(2, 3)$ is in $R$ while $(3, 2)$ is not. A pair $(a, a)$ is its own reverse, so self-pairs never obstruct symmetry, and finding one matched pair proves nothing about the others.

A chain is a pair whose second component begins another pair. The chain $(2, 1)$ and $(1, 3)$ needs the shortcut $(2, 3)$, which is not in $R$, so it disproves transitivity. The chain $(1, 2)$ and $(2, 2)$ does have its shortcut $(1, 2)$ in $R$, so it proves nothing. A pair together with its reverse is a symmetry observation, not a chain.

Distinct equivalence classes are always disjoint: if $c$ lay in both $A_i$ and $A_j$ then everything in $A_i$ would be related to everything in $A_j$ by transitivity, making them the same class. So $A_i \cap A_j \neq \phi$ for $i \neq j$ is the false statement. The other three are exactly the partition properties the chapter states.

The assertion is false: the chain $(1,2),(2,1)$ needs $(1,1)$, which is absent. The reason is true: that is Definition 3(iii) exactly. A correct definition is offered for a claim that is false, so the reason is true and the assertion is not.

A counterexample has to clear three bars at once: the two inputs must be different, both must lie in the stated domain, and their images must be equal. Here the third bar fails: an odd power keeps the sign, so the two images differ and the pair is not a counterexample. The arithmetic being right is not the point; and a single valid pair really is enough, because one-one is a claim about every pair and a single counterexample defeats it.

Every element of $X$ sends its arrow to a different element of $Y$, so no target is hit twice and $f$ is one-one. But $Y$ has four elements and only three arrows arrive, so one element of $Y$ -- namely $4$ -- receives none, and $f$ is not onto. Onto has to be checked target by target; it is not enough to note that a target set was written down.

Onto is decided by solving $f(x) = y$ and asking whether the solution lands inside the stated domain for every $y$ of the stated codomain. The solution is an integer only when $y - 3$ is a multiple of $4$, and it is positive only when $y > 3$, so $y = 1$ is missed. Notice that the formula alone settles nothing: the same rule is onto for one pair of sets and not for another. Nor does checking a couple of convenient targets settle it, and writing down that the range equals the codomain is the claim, not the argument.

(R) is true: $3x_1 = 3x_2$ gives $x_1 = x_2$. (A) is false, because a bijection must also be onto and no natural number outside the multiples of $3$ is $3x$ for a natural $x$ -- $4$, for instance, is never reached. So being one-one, true as it is, does not make (A) true -- which is the whole point of the (A)-false, (R)-true verdict.

The first square root requires $x\ge1$ and the second requires $x\le4$. Both must exist to add their values, so the sum’s domain is their intersection, $[1,4]$. The assertion’s larger set is false; the intersection rule in the reason is true.