Question

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

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

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.

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.

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.

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): \((2m+1) + 2k\)

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.

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).

Key concepts

Odd Integers

Odd integers are numbers that cannot be divided exactly by 2. These numbers have a remainder of 1 when divided by 2. They are typically represented in the form of \(2m+1\), where \(m\) is any integer.Odd integers include numbers like 1, 3, 5, 7, and so on.

• When you add 2 to an odd integer, you get another odd integer.
• Multiplying two odd integers also results in an odd integer.
• The sum of an odd and an even integer is odd.

Understanding odd integers is crucial when dealing with mathematical induction problems, where proving the oddness of expressions is often required.

Base Case

The base case in mathematical induction is the initial step where you prove that a statement holds true for a starting value, usually \(n=1\) or the smallest value considered in the context.In our problem, the initial attempt for a base case was \(n=1\), but this resulted in 2, an even number, which was not suitable. Trying \(n=2\), proved successful because the calculation and result yielded 5, an odd integer.By confirming the base case for this smaller value, we establish a foundation for further inductive proofs to build upon.

• Choosing the correct base case is crucial as it sets the precedent for the logical steps that follow.
• Misidentifying the base case can lead to incorrect conclusions in the induction process.

Inductive Step

The inductive step is critical in proving statements for all numbers in a given set using mathematical induction.It involves showing that if the statement holds for some arbitrary integer \(k\), it must also hold for \(k+1\).
In this exercise, we assume \(P(k)\) is true, meaning that \(k^2 + 1\) is odd. We prove that \((k+1)^2 + 1\) is also odd.

• The equation for \(P(k)\) was expressed as \(k^2 + 1 = 2m + 1\).
• We expanded \((k+1)^2 + 1\) to find it equivalent to \((2m+1) + 2k\), simplifying to an odd integer form \(2m'+1\).

This step ensures the pattern continues indefinitely, reinforcing the truth of the statement across all applicable integers.

Integer Sequences

Integer sequences are ordered lists of whole numbers. They can follow various patterns, such as arithmetic sequences where each term is generated by adding a constant value to the previous term.In the context of this exercise, we analyzed a sequence defined by the expression \(n^2 + 1\) to determine the oddness of each term.
Using induction to prove properties of integer sequences involves checking individual terms against a conjecture or property.

• Start with a base case, proving the property for the initial term of the sequence.
• Proceed with the inductive step to ensure the property holds for subsequent terms.

For the given problem, the sequence of interest was derived from squaring an integer \(n\) and adding 1, thus forming an infinite series of numbers where our task was to confirm the oddness for values \(n > 2\).