Suppose that P (n) is a propositional function. Determine for which nonnegative integers n the statement P (n) must be true if a) P (0) is true; for all nonnegative integers n, if P (n) is true, then P (n + 2) is true. b) P (0) is true; for all nonnegative integers n, if P (n) is true, then P (n + 3) is true. c) P (0) and P (1) are true; for all nonnegative integers n, if P (n) and P (n + 1) are true, then P (n + 2) is true. d) P (0) is true; for all nonnegative integers n, if P (n) is true, then P (n + 2) and P (n + 3) are true.

Question

Maeve Gaines

Mathematics

10 July, 04:41

Suppose that P (n) is a propositional function. Determine for which nonnegative integers n the statement P (n) must be true if a) P (0) is true; for all nonnegative integers n, if P (n) is true, then P (n + 2) is true. b) P (0) is true; for all nonnegative integers n, if P (n) is true, then P (n + 3) is true. c) P (0) and P (1) are true; for all nonnegative integers n, if P (n) and P (n + 1) are true, then P (n + 2) is true. d) P (0) is true; for all nonnegative integers n, if P (n) is true, then P (n + 2) and P (n + 3) are true.

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Paityn · Answer 1

a) It must be true for all natural numbers that divide 2 and also for 0.

b) It must be true for all positive numbers that divide 3 and also for 0.

c) The proposition is true for all nonnegative integers.

d) P (n) is true for any nonnegative integer as long as n ≠1.

Step-by-step explanation:

a) We know alredy that P (0) is true. if the propiety is true for n, then it must be true for n+2. Then it must be true for 2 = 0+2. Inductively, you can easily see that it must be true for all positive even numbers (and for 0).

b) P (0) is true. It is also true for 3 = 0+3, and also for any positive number that is a multiple of 3: we alredy know that it is true for 3*1. If it were true for 3*n, then it should be true for 3n+3 = 3 * (n+1), then p (3 (n+1)) is also true. Thus, for induction, P (k) must be true for k=0 and for any positive multiple of 3.

c) The proposition is true for all nonnegative integers. We can prove it by induction. The base cases P (0) and P (1) are given as true. If we have that both P (n) and P (n+1) are true, then P (n+2) should also be true. Then, for induction P (n) is true for any nonnegative integer n.

d) The proposition is true for n=0. Also we can see that it must be true for 2 = 0+2 and for 3 = 0+3. For the value which it is not neccesarily true is for n=1. However, it should be true for any value n > 1; we can see this with inclusion.

We alredy know that P (2) and P (3) are true. If both P (n) and P (n+1) are true, then we want to know if P (n+2) and P (n+3) are also true, however, both numbers are obtained from adding 2 and 3 from n respectively, thus, the proposition must be true for those value according to the hypothesis given. Thus, P (n) is true for any nonnegative integer as long as n ≠1.