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Exercise 2.2 · Q3

Q.A={1,2,3,5}A = \{1, 2, 3, 5\} and B={4,6,9}B = \{4, 6, 9\}. Define a relation RR from AA to BB by R={(x,y):the difference between x and y is odd; x∈A, y∈B}R = \{(x, y) : \text{the difference between }x\text{ and }y\text{ is odd};\ x \in A,\ y \in B\}. Write RR in roster form.

Bihar BsebTextbookSubjective· 2mImportance★★★★★est
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The relation RR consists of all ordered pairs (x,y)(x,y) from A×BA \times B where x−yx-y is an odd number. Checking each pair gives 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)\}.

The core idea here is simple: we need to find every pair (x,y)(x,y) with xx from set AA and yy from set BB such that the difference x−yx-y is an odd integer. "Difference" here means subtraction — and we only care whether the result is odd, not whether it's positive or negative.

A quick intuition: an odd number is one that is not divisible by 2. When you subtract two integers, the parity (odd/even nature) of the result depends only on whether the two numbers have the same parity or different parity. If both are even or both are odd, their difference is even. If one is even and the other odd, their difference is odd. So we're really looking for pairs where xx and yy have opposite parity.

Let's check the parity of each element:

  • A={1,2,3,5}A = \{1, 2, 3, 5\}: 1 (odd), 2 (even), 3 (odd), 5 (odd)
  • B={4,6,9}B = \{4, 6, 9\}: 4 (even), 6 (even), 9 (odd)

So the condition "difference is odd" means we pair an odd xx with an even yy, or an even xx with an odd yy.

Now let's list all possible pairs systematically.

  1. Take x=1x = 1 (odd). It must pair with an even yy from BB: y=4y = 4 and y=6y = 6. Check: 1−4=−31-4 = -3 (odd), 1−6=−51-6 = -5 (odd). So (1,4)(1,4) and (1,6)(1,6) are in RR.

  2. Take x=2x = 2 (even). It must pair with an odd yy from BB: only y=9y = 9. Check: 2−9=−72-9 = -7 (odd). So (2,9)(2,9) is in RR.

  3. Take x=3x = 3 (odd). Pair with even yy: y=4y = 4 and y=6y = 6. Check: 3−4=−13-4 = -1 (odd), 3−6=−33-6 = -3 (odd). So (3,4)(3,4) and (3,6)(3,6) are in RR.

  4. Take x=5x = 5 (odd). Pair with even yy: y=4y = 4 and y=6y = 6. Check: 5−4=15-4 = 1 (odd), 5−6=−15-6 = -1 (odd). So (5,4)(5,4) and (5,6)(5,6) are in RR.

What about x=2x = 2 with y=4y = 4 or 66? 2−4=−22-4 = -2 (even), 2−6=−42-6 = -4 (even) — not included. What about x=1,3,5x = 1,3,5 with y=9y = 9? 1−9=−81-9 = -8 (even), 3−9=−63-9 = -6 (even), 5−9=−45-9 = -4 (even) — not included.

Watch out

A common mistake is to think "difference" means absolute difference. But the problem says "the difference between xx and yy" — in standard set theory and relations, this means x−yx-y, not ∣x−y∣|x-y|. However, since parity of x−yx-y is the same as parity of ∣x−y∣|x-y| (oddness doesn't depend on sign), the result would be the same either way here. Still, be precise.

So collecting all valid pairs, we get:

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)\}

✓Final answer

The relation in roster form is 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)\}.

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