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Exercise 1.2 · Q5

Q.On the set of natural numbers let RR be the relation defined by aRbaRb if 2a+3b=302a+3b=30. Write down the relation by listing all the pairs. Check whether it is

(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
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Step 1. Solve 2a+3b=302a+3b=30 for a,b∈Na,b\in N: b=30−2a3b=\dfrac{30-2a}3 must be a positive integer, so 30−2a30-2a must be a positive multiple of 3, forcing aa to be a multiple of 3 (since 2a≡0(mod3)  ⟺  a≡0(mod3)2a\equiv0\pmod3\iff a\equiv0\pmod3). Checking a=3,6,9,12a=3,6,9,12 (with a=15a=15 giving b=0∉Nb=0\notin N, excluded): (3,8),(6,6),(9,4),(12,2)(3,8),(6,6),(9,4),(12,2).

Step 2 (Reflexive). Need (a,a)∈R(a,a)\in R for every a∈Na\in N. Only (6,6)(6,6) has that form; e.g. (1,1)∉R(1,1)\notin R. Not reflexive.

Step 3 (Symmetric). (3,8)∈R(3,8)\in R but (8,3)∉R(8,3)\notin R (check: 2(8)+3(3)=16+9=25≠302(8)+3(3)=16+9=25\ne30). Not symmetric. …

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