If {N_a} = { an : n in N } , then {N_3} cap { | vedclass

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(B) Given that ${N_a} = \{ an : n \in N \}$.This represents the set of all multiples of $a$.Therefore,${N_3} = \{ 3, 6, 9, 12, 15, 18, 21, 24, \dots \

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${N_{12}}$

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(B) Given that ${N_a} = \{ an : n \in N \}$.This represents the set of all multiples of $a$.Therefore,${N_3} = \{ 3, 6, 9, 12, 15, 18, 21, 24, \dots \}$ and ${N_4} = \{ 4, 8, 12, 16, 20, 24, \dots \}$.The intersection ${N_3} \cap {N_4}$ consists of elements that are common to both sets,which are the common multiples of $3$ and $4$.The least common multiple of $3$ and $4$ is $12$.Thus,${N_3} \cap {N_4} = \{ 12, 24, 36, \dots \} = {N_{12}}$.In general,${N_a} \cap {N_b} = {N_{	ext{lcm}(a, b)}}$.

If ${N_a} = \{ an : n \in N \} ,$ then ${N_3} \cap {N_4} = $

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