Introduction to Relations

Step 1: (|a-b|=0) only when (a=b). Step 2: So only pairs like ((1,1)), ((2,2)), and ((3,3)) occur. Step 3: A relation based on equality is the identity relation.

Step 1: (|a-a|=0) is even, so it is reflexive. Step 2: (|a-b|=|b-a|), so it is symmetric. Step 3: Even difference means same parity, which also gives transitivity.

Step 1: From ((1,2)) and ((2,3)), transitivity requires ((1,3)). Step 2: This pair is not given in the relation. Step 3: Quickly identify the third pair formed by two connected pairs.

Step 1: For (A={1,2}), (A\times A) has four pairs. Step 2: The given relation contains all four pairs. Step 3: Hence it equals (A\times A), the universal relation.

Step 1: A relation from (A) to (B) is a subset of (A\times B). Step 2: (A\times B) has (3\times3=9) pairs. Step 3: Hence the number of relations is (2^9).

Step 1: All self-pairs are present, so it is reflexive. Step 2: Both ((1,4)) and ((4,1)) are present, so it is symmetric. Step 3: The formed transitive chains are completed by existing self-pairs, so it is an equivalence relation.

Step 1: Symmetry requires ((2,1)) along with ((1,2)). Step 2: ((2,1)) is not in the relation. Step 3: A cyclic-looking relation is not necessarily symmetric.

Step 1: From ((1,2)) and ((2,3)), transitivity requires ((1,3)). Step 2: ((1,3)) is not in the relation. Step 3: Therefore, the relation is not transitive.

Step 1: It contains all self-pairs, so it is reflexive. Step 2: Every pair has its reverse, so it is symmetric. Step 3: Since all possible pairs are present, transitivity also holds.

Step 1: (a\equiv b \pmod{2}) means same parity. Step 2: The odd elements like (1) are (1) and (3). Step 3: Hence the class of (1) is ({1,3}).

Step 1: Dividing (2) by (3) gives remainder (2). Step 2: Dividing (5) by (3) also gives remainder (2). Step 3: Therefore, the class of (2) is ({2,5}).

Step 1: This condition says that two distinct elements cannot be related in both directions. Step 2: This is the identity of antisymmetry. Step 3: Understand symmetry and antisymmetry separately through reverse pairs.

Step 1: Every number is less than or equal to itself, so it is reflexive. Step 2: If (a\le b) and (b\le a), then (a=b), so it is antisymmetric. Step 3: The (\le) chain is also transitive.

Step 1: Antisymmetry fails when both ((a,b)) and ((b,a)) exist for distinct (a,b). Step 2: Here ((1,2)) is present but ((2,1)) is not. Step 3: So antisymmetry is not violated, and the relation is antisymmetric.

Step 1: In antisymmetry, pairs in both directions should not occur between distinct elements. Step 2: Here (1) and (2) are distinct, yet both ((1,2)) and ((2,1)) are present. Step 3: Therefore, antisymmetry fails.

Step 1: A partial order relation needs three special properties. Step 2: These are reflexivity, antisymmetry and transitivity. Step 3: Equivalence relation uses symmetry, while partial order uses antisymmetry.

Step 1: In divisibility, the least element is the one that divides every element. Step 2: (1) divides (1), (2), and (3). Step 3: In divisibility questions, check (1) carefully.

Step 1: A least element must divide every element in the set. Step 2: (2) does not divide (3), and (3) does not divide (2). Step 3: Hence no element divides all elements of this set.

Step 1: A greatest element is divisible by every element of the set. Step 2: (1), (2), (3), and (6) all divide (6). Step 3: Therefore, (6) is the greatest element.

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: No reverse pair exists for distinct elements, so it is antisymmetric. Step 3: ((1,2)) and ((2,3)) require ((1,3)), which is present, so it is transitive.

The correct answer is A, the empty relation. By definition, (x,z) belongs to R∘R only when there exists an intermediate element y such that (x,y)∈R and (y,z)∈R. Since R is empty, it contains no first pair and no second pair. Thus no intermediate y can satisfy both conditions, and no ordered pair can enter the composition. Therefore R∘R=∅. B and D are incorrect because diagonal pairs require pairs such as (x,x) to be available in R, but none are. C is incorrect because the universal relation contains every possible pair, whereas the composition contains none. The fact that A is non-empty does not change the result; it only means there are elements available in A, not that the empty relation contains pairs. Memory cue: composition needs a two-link path, and an empty relation has no link at all.

The composite function \((h\circ g\circ f)(x)=h(g(f(x)))\), so the functions must be applied from right to left. First, \(f(1)=1+1=2\); then \(g(2)=2\times2=4\); finally, \(h(4)=4^2=16\). Therefore, the correct answer is 16. Option 8 results from reversing the order of \(g\) and \(h\), while 4 is obtained by stopping after the second step. Exam tip: in a function composition, apply the rightmost function first.

In a composition of functions, evaluate the inner function first and then apply the outer function. Thus, \((g\circ f)(4)=g(f(4))\). We have \(f(4)=4-2=2\), so \(g(2)=2^2+1=5\). Therefore, option A is correct. Option B, 17, is obtained by calculating \(g(4)\) directly without first applying \(f\), which is the common error. Exam tip: in \((g\circ f)(x)\), apply \(f\) first and \(g\) second.

Step 1: (f^{-1}(23)) means the value of (x) for which (f(x)=23). Step 2: From (4x+7=23), we get (4x=16), so (x=4). Step 3: To find an inverse value, equate the original function to the given value and solve.

Step 1: Write (f(x+1)=(x+1)^2-4(x+1)). Step 2: Simplifying gives (x^2-2x-3). Step 3: Now (x^2-2x-3-(x^2-4x)=2x-3).