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The set whose elements are 3, 6, 9, 12, and 15. *

Sagot :

Answer:

The elements of the given set are the multiples of 3:

3=3×1

6=3×2

363=3×121.

Hence,there exists a bijection between sets A={3,6,9,12,…,363} and B={1,2,3,4,…,121}.

Now,clearly the cardinality of the set B is 121.

Hence,due to bijection between B and A, the cardinality of set A is also 121.

Step-by-step explanation: