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Prove that number of subsets of a set containing n distinct elements is 2n, for all n ϵ N. |
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Answer» To prove; Number of subsets of a set containing n distinct elements is 2n. For a null set there is only one element Φ and therefore only one subset. ⇒ at n = 0 number of subsets = 1 = 20 By considering a set with 1 element there will be 2 subsets with 1 and Φ ⇒ at n = 0 number of subsets = 2 = 22 By considering a set Sk with k element Let at n = k number of subsets = 2k be true. For a set Sk+1 containing k+1 element The extra one element in set Sk+1 when compared with Sk will form an extra collection of subsets by combing with the existing 2k subsets and as a result an extra 2k subsets are formed. ⇒ at n = k+1 number of subsets = 2k + 2k = 2.2k = 2k+1 ⇒ P(k+1) is true. ∴ By Mathematical Induction number of subsets of a set containing n distinct elements is 2n, for all n ϵ N. |
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