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Relations and Functions - Relations

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Ordered Pairs: An ordered pair consists of two elements in a fixed order, written as (a,b)(a, b). Two ordered pairs are equal if and only if their corresponding first and second elements are equal, i.e., (a,b)=(x,y)(a, b) = (x, y) if and only if a=xa = x and b=yb = y. Visually, these are represented as coordinates on a 2D Cartesian plane where the first element is the xx-coordinate and the second is the yy-coordinate.

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Cartesian Product of Sets: For any two non-empty sets AA and BB, the Cartesian product AtimesBA \\times B is the set of all ordered pairs (a,b)(a, b) such that ainAa \\in A and binBb \\in B. If AA has pp elements and BB has qq elements, then n(AtimesB)=pqn(A \\times B) = pq. Visually, this can be described as a grid of points where each point corresponds to a pairing from set AA and set BB.

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Definition of a Relation: A relation RR from a set AA to a set BB is a subset of the Cartesian product AtimesBA \\times B. This subset is formed by linking elements of AA and BB based on a specific rule. Visually, relations are often represented by arrow diagrams, where an arrow points from an element in the source set AA to its related element in the target set BB.

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Domain of a Relation: The domain of a relation RR is the set of all first elements of the ordered pairs belonging to RR. It is a subset of set AA. In an arrow diagram, the domain consists of all elements in set AA that have at least one outgoing arrow.

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Range of a Relation: The range of a relation RR is the set of all second elements of the ordered pairs belonging to RR. It is a subset of set BB. Visually, in an arrow diagram, the range is the set of elements in the second oval (set BB) that have at least one arrow pointing toward them.

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Codomain: In a relation RR from AA to BB, the entire set BB is called the codomain. It is important to remember that the Range is always a subset of the Codomain (RangesubseteqCodomainRange \\subseteq Codomain). Visually, while the range is only the 'hit' elements, the codomain is the entire destination set.

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Total Number of Relations: If n(A)=pn(A) = p and n(B)=qn(B) = q, the total number of possible relations from AA to BB is 2pq2^{pq}. This is because a relation is any subset of AtimesBA \\times B, and a set with pqpq elements has 2pq2^{pq} possible subsets (the power set).

📐Formulae

AtimesB=(a,b):ainA,binBA \\times B = \\{(a, b) : a \\in A, b \\in B\\}

n(AtimesB)=n(A)timesn(B)n(A \\times B) = n(A) \\times n(B)

Total relations from AA to B=2n(A)cdotn(B)B = 2^{n(A) \\cdot n(B)}

(a,b)=(x,y)iffa=xtextandb=y(a, b) = (x, y) \\iff a = x \\text{ and } b = y

Range(R)=binB:(a,b)inRtextforsomeainARange(R) = \\{b \\in B : (a, b) \\in R \\text{ for some } a \\in A\\}

💡Examples

Problem 1:

If (fracx3+1,y−frac23)=(frac53,frac13)(\\frac{x}{3} + 1, y - \\frac{2}{3}) = (\\frac{5}{3}, \\frac{1}{3}), find the values of xx and yy.

Solution:

Step 1: Since the ordered pairs are equal, equate the corresponding elements. fracx3+1=frac53\\frac{x}{3} + 1 = \\frac{5}{3} y−frac23=frac13y - \\frac{2}{3} = \\frac{1}{3} Step 2: Solve the first equation for xx: fracx3=frac53−1\\frac{x}{3} = \\frac{5}{3} - 1 fracx3=frac23\\frac{x}{3} = \\frac{2}{3} x=2x = 2 Step 3: Solve the second equation for yy: y=frac13+frac23y = \\frac{1}{3} + \\frac{2}{3} y=frac33y = \\frac{3}{3} y=1y = 1 Final Answer: x=2x = 2 and y=1y = 1.

Explanation:

This solution relies on the fundamental property of ordered pairs: two pairs are identical if and only if their first components are equal and their second components are equal.

Problem 2:

Let A=1,2,3,5A = \\{1, 2, 3, 5\\} and B=4,6,9B = \\{4, 6, 9\\}. Define a relation RR from AA to BB by R=(x,y):x−ytextisodd,xinA,yinBR = \\{(x, y) : x - y \\text{ is odd}, x \\in A, y \\in B\\}. Write RR in roster form and find its domain.

Solution:

Step 1: Test the 'difference is odd' condition for all pairs (x,y)(x, y). Recall that a difference is odd if one number is even and the other is odd. Step 2: Check for x=1x=1 (odd): 1−4=−31-4=-3 (odd), 1−6=−51-6=-5 (odd), 1−9=−81-9=-8 (even). Pairs: (1,4),(1,6)(1,4), (1,6). Step 3: Check for x=2x=2 (even): 2−4=−22-4=-2 (even), 2−6=−42-6=-4 (even), 2−9=−72-9=-7 (odd). Pair: (2,9)(2,9). Step 4: Check for x=3x=3 (odd): 3−4=−13-4=-1 (odd), 3−6=−33-6=-3 (odd), 3−9=−63-9=-6 (even). Pairs: (3,4),(3,6)(3,4), (3,6). Step 5: Check for x=5x=5 (odd): 5−4=15-4=1 (odd), 5−6=−15-6=-1 (odd), 5−9=−45-9=-4 (even). Pairs: (5,4),(5,6)(5,4), (5,6). Step 6: Write RR in roster form: R=(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6).R = \\{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\\}. Step 7: Find the domain (set of all first elements): Domain=1,2,3,5.Domain = \\{1, 2, 3, 5\\}.

Explanation:

To solve this, we systematically verify the arithmetic condition (x−ytextisodd)(x-y \\text{ is odd}) for every possible pairing in the Cartesian product AtimesBA \\times B, then extract the unique first elements to define the domain.