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Chapter 6: Problem 478

Let \(\mathrm{P}(\mathrm{n}): \mathrm{n}^{2}+1\) is an odd integer, if it is assumed that \(\mathrm{P}(\mathrm{k})\) is true \(\Rightarrow \mathrm{P}(\mathrm{k}+1)\) is true. Therefore, \(\mathrm{P}(\mathrm{n})\) is true (a) for \(\mathrm{n}>1\) (b) for all \(\mathrm{n} \in \mathrm{N}\) (c) for \(\mathrm{n}>2\) (d) None of these

Short Answer

Expert verified
The statement P(n): \(n^2 + 1\) is an odd integer holds for the base case n=2 and, by induction, it also holds for any n > 2 whenever P(k) is true, so the statement is true for \(n > 2\). Therefore, the answer is \(\textbf{(c)}\).

Step by step solution

01

The base case of the statement P(n)

Let's consider the base case where n=1. If P(1) is true, we have: \(1^2 + 1 = 2\) Since 2 is an even integer, we need to find a different base case.
02

Searching for another base case

Let's consider n=2 as a possible base case: \(2^2 + 1 = 5\) Since 5 is odd, we can use n=2 as our new base case to tackle the problem.
03

Inductive step for P(k) to P(k+1)

If the statement P(k) is true, then we have: \(k^2 + 1 = (2m+1)\), where m is an integer (odd integer definition). We want to prove that P(k+1) is true meaning that: \((k+1)^2 + 1\) is an odd integer too.
04

Expressing P(k+1) in terms of P(k)

Expanding \((k+1)^2 + 1\), we have: \((k^2 + 2k + 1) + 1\) Recall that our P(k) statement is: \(k^2+1=(2m+1)\). Now, we can rewrite the P(k+1) expression in terms of P(k): \((2m+1) + 2k\)
05

Proving P(k+1) is also an odd integer

Recalling that P(k) is an odd integer, we need to show that the expression \((2m+1)+2k\) is also odd. Let's denote \(m' = m+k\). Then the expression becomes \((2m'+1)\). As the expression \((2m'+1)\) perfectly fits the definition of odd integer (where m' is also an integer), we have proved that P(k+1) is also an odd integer when P(k) is true.
06

Conclusion

Since the statement P(n): n^2 + 1 is an odd integer holds for n=2 (our base case) and if P(k) is true, P(k+1) is also true for all k > 2. So the statement is true for all n > 2, which corresponds to option (c).

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Most popular questions from this chapter

If matrix \(\mathrm{A}=\left|\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right|\) and \(\mathrm{I}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|\) then which one of the following holds for all \(n \in N\). (use principle of mathematical Induction) (a) \(\mathrm{A}^{\mathrm{n}}=\mathrm{n} \cdot \mathrm{A}-(\mathrm{n}-1) \mathrm{I}\) (b) \(\mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1} \cdot \mathrm{A}+(\mathrm{n}-1) \mathrm{I}\) (a) \(\mathrm{A}^{\mathrm{n}}=\mathrm{n} \cdot \mathrm{A}+(\mathrm{n}-1) \mathrm{I}\) (b) \(A^{n}=2^{n-1} \cdot A-(n-1) I\)

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