Relations

A relation from set \(A\) to set \(B\) is defined as any subset of \(A\times B\), so option A is correct. Not every relation is a function: a function must assign exactly one image in \(B\) to each element of \(A\). Option D is incorrect because an identity relation is generally defined on the same set. Exam tip: remember ‘any subset’ for a relation and ‘exactly one image for every element’ for a function.

In a Cartesian product, the first component comes from the first set and the second from the second set. Therefore (p\in A) and (q\in B).

If (a+b) is odd, then (b+a) is also odd, so it is symmetric. But (a+a) is even, so it is not reflexive.

It is reflexive because ((1,1)) and ((2,2)) are present. The pair ((1,2)) causes no violation of transitivity, so it is also transitive.

Both ((1,2)) and ((2,1)) are present, so it is symmetric. But ((1,1)) and ((2,2)) are absent, so it is not reflexive.

For transitivity, whenever \((a,b)\in R\) and \((b,c)\in R\), the pair \((a,c)\) must also belong to R. Here, \((1,2)\in R\) and \((2,3)\in R\), so \((1,3)\) must be in R; however, it is absent. Therefore, R is not transitive. Exam tip: diagonal pairs such as \((1,1),(2,2),(3,3)\) do not compensate for a missing required pair.

All elements in the given set are non-negative, so (a^2=b^2) gives (a=b). Therefore only (5) diagonal pairs exist.

For the pair \((-2,2)\), \((-2)^2=2^2=4\), so it satisfies the condition \(a^2=b^2\) and hence \((-2,2)\in R\). In the other options, the squares of the two numbers are unequal. Exam tip: when a relation involves squares, check numbers with opposite signs as well.

Every (a) divides itself and (a\mid b,\ b\mid c) imply (a\mid c). Hence it is reflexive and transitive.

The definition of equivalence relation includes reflexive, symmetric, and transitive properties. In exams, check all three separately.

For the ordered pair \((4,2)\), \(a+b=4+2=6\). The relation \(R\) contains only those ordered pairs for which \(a+b\le 5\), but \(6\le 5\) is false. Therefore, \((4,2)\notin R\). Option B incorrectly treats the inequality condition as satisfied. Exam tip: To test whether an ordered pair belongs to a relation, substitute its two components directly into the defining condition.

Direct answer: option A, symmetric but not reflexive. The rule is R = {(a,b): |a-b|=1} on A={1,2,3,4,5}. Symmetric means that whenever (a,b) belongs, (b,a) also belongs. Since |a-b|=|b-a|, this always happens. For example, (1,2) is present and (2,1) is also present. Reflexive means every (a,a) must belong. But |a-a|=0, not 1, so no pair such as (1,1) belongs. Therefore it is not reflexive. C, “transitive,” is false: (1,2) and (2,3) are present, but (1,3) is absent because |1-3|=2. D, equivalence, is false because an equivalence relation must be reflexive, symmetric and transitive; this relation fails reflexivity and transitivity. B is false because it claims reflexivity. Memory cue: distance exactly 1 gives two-way pairs, but never self-pairs.

To be reflexive, both ((1,1)) and ((2,2)) are needed. ((1,1)) is already present, so ((2,2)) must be added.

The domain of a relation is the set of first components of its ordered pairs. The first components here are 1, 2, and 1; removing the repetition gives the domain \(\{1,2\}\). Option B contains the second components and therefore represents the range. Exam tip: for the domain, list the first element of every ordered pair and remove duplicates.

The range of a relation consists of all distinct second components of its ordered pairs. Here, the second components are 2, 3, and 3, so the range is \(\{2,3\}\). Options A and C use incorrect combinations of components, while option D also includes the first component 1. Exam tip: To find the range, list the second component of every ordered pair and remove repetitions.

For a relation R from A to B, the codomain is the entire target set B, whether or not every element of B is actually used. Hence, the codomain is {x,y}. The set {1,3} is the domain, while the range is also {x,y} here because both x and y occur as second components. Exam tip: identify the codomain from the stated target set B, but find the range from the actual second components of the ordered pairs.

The range of a relation consists of the actual output values, whereas the codomain is the set in which those outputs are allowed to lie. Therefore, the range is always a subset of the codomain. Here, \({p,q}\subset {p,q,r}\); the element \(r\) belongs to the codomain but not to the range, so the two sets are not equal. Exam tip: range is always contained in the codomain, but it need not be the whole codomain.

A relation from A to B is any subset of the Cartesian product \(A\times B\). The largest possible relation is obtained when every ordered pair in that product is included. The number of ordered pairs in a Cartesian product is found by multiplying the number of elements in the first set by the number in the second set.

Here \(|A|=2\) and \(|B|=5\), so \(|A\times B|=2\times5=10\). A relation cannot contain more than all ten pairs because it is a subset of the product. It can contain fewer pairs, including none, but the maximum is reached by taking \(R=A\times B\). Therefore, option C, 10 ordered pairs, is correct; 25 would incorrectly square or multiply the wrong quantities.

Using the condition b = a + 3, a = 1 gives (1, 4), a = 2 gives (2, 5), and a = 3 gives (3, 6). Thus, R = {(1, 4), (2, 5), (3, 6)} contains 3 ordered pairs. The value 6 is the number of pairs in the Cartesian product A × B, not in this restricted relation. Exam tip: Apply the defining condition to each element before counting the ordered pairs.

For \((1,4)\), we have \(a=1\) and \(b=4\), so \(ab=1\times4=4\). Hence, \((1,4)\) belongs to \(R\). The products for the other options are \(6,3\), and \(8\), respectively, so they do not satisfy the relation. Exam tip: To test whether an ordered pair belongs to a relation, verify both the allowed elements and the defining condition.

It contains all pairs of (A\times A), so it is universal. It is also reflexive, symmetric, and transitive.

In an antisymmetric relation, for distinct (a) and (b), both ((a,b)) and ((b,a)) cannot occur together. Since (3\ne5), ((5,3)) cannot occur.

All diagonal pairs are present, so it is reflexive. Since ((2,1)) is absent, ((1,2)) does not break antisymmetry.

Because (a\ge a) is true, and (a\ge b,\ b\ge c) imply (a\ge c). Also (a\ge b) and (b\ge a) imply (a=b).

Since (n(A\times B)=3\times 2=6), the number of relations is (2^6=64). In exams, first count the elements of (A\times B).