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Question
Prove by method of induction, for all n ∈ N:
(23n − 1) is divisible by 7
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Solution
23n − 1 is divisible by 7 if and only if 23n − 1 is a multiple of 7.
Let P(n) = 23n − 1 = 7m, where m ∈ N.
Step 1:
For n = 1, 23n − 1 = 23 − 1 = 7, which is divisible by 7.
∴ P(1) is true.
Step 2:
Let us assume that for some k ∈ N, P(k) is true,
i.e., 23k − 1 = 7a, where a ∈ N
∴ 23k = 7a + 1 ...(1)
Step 3:
To prove that P(k + 1) is true, i.e., to prove that
`2^(3("k"+1)) − 1` is a multiple of 7,
i.e., 23k+3 − 1 = 7b, where b ∈ N
Now, 23k+3 − 1 = 23k·23 − 1
= (7a + 1)8 − 1 ...[By (1)]
= 56a + 7
= 7(8a + 1)
= 7b, where b = (8a + 1) ∈ N.
∴ P(k + 1) is true
Step 4:
From all the above steps and by the principle of mathematical induction, P(n) is true for all n ∈ N,
i.e., (23n − 1) is divisible by 7, for all n ∈ N.
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