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Question
Prove by method of induction, for all n ∈ N:
`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N
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Solution
Let P(n) ≡ `[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]`
Step 1:
For n = 1,
L.H.S. = `[(1, 2),(0, 1)]^1 = [(1, 2),(0, 1)]`
R.H.S. = `[(1, 2),(0, 1)]`
∴ L.H.S. = R.H.S.
∴ P(1) is true.
Step 2:
Let us assume that for some k ∈ N, P(k) is true,
i.e., `[(1, 2),(0, 1)]^"k" = [(1, 2"k"),(0, 1)]` ...(1)
Step 3:
To prove that P(k+ 1) is true, i.e., to prove that
`[(1, 2),(0, 1)]^("k" + 1) = [(1, 2("k" + 1)),(0, 1)]`
Now, L.H.S. = `[(1, 2),(0, 1)]^("k" + 1)`
= `[(1, 2),(0, 1)]^"k" [(1, 2),(0, 1)]`
= `[(1, 2"k"),(0, 1)] [(1, 2),(0, 1)]` ...[By (1)]
= `[(1 + 0, 2 + 2"k"),(0 + 0, 0 + 1)]`
= `[(1, 2("k" + 1)),(0, 1)]`
= R.H.S.
∴ 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., `[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]`, ∀ n ∈ N
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