'For all natural numbers N, if P(n) is a statement about n and P(k+1) is true if P(k) is true for an arbitrary natural number k, then P(n) is always true.' State true or false.

True

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False

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Solution

The correct option is B
False

For the proof by mathematical induction to work, the statement P(n) must be true for a specific instance of a natural number.

Hence, if P(m) is true, where m is a specific natural number and P(k+1) is true if P(k) is true for an arbitrary natural number k, then, $P (n) i s t r u e \forall n \geq m$

Without the base case P(m), we cannot say that P(n) is true. Hence, the statement is false.

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Similar questions

Which of the following illustrates the inductive step to prove a statement P(n) about natural numbers n by mathematical induction, where k is an arbitrary natural number?

Q. If

P (n)

is statement such that

P (3)

is true. Assuming P(k) is true

\Rightarrow

P (k + 1)

is true for all

k

\geq

2

, then

P (n)

is true.

Q. Let

P (n) : n^{2} + n

is an odd integer

P (k) \Rightarrow P (k + 1)

is true
Then

P (n)

is true for all

'If P(n) is a statement about natural numbers, mathematical induction may be used to prove P(n) only if P(1) is true.' State true or false.

$P (n) : 1.2 + 2.3 + 3.4 + . . . + n (n + 1) = \frac{(n + 1) (n + 2)}{3}$

The statement P(n) is

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