Answer:
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.
