Question

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

Answer 1

Solution :

a). $P(0)$ is true

Then ,$P(0+2)=P(2)$ is true.

         $P(2+2)=P(4)$ is true

          $P(4+2)=P(6)$ is true.

Therefore, we see that $P(n)$ is true for all the even integers : $\{0, 2,4,6,...\}$

b). $P(0)$ is true

Then ,$P(0+3)=P(3)$ is true.

         $P(3+3)=P(6)$ is true

          $P(6+3)=P(9)$ is true.

Therefore, we see that $P(n)$ is true for all the multiples of 3 : $\{0, 3,6,9,12,...\}$

c). $P(0)$ and $P(1)$ is true, then $P(0+2)=P(2)$ is true

$P(1)$ and $P(2)$ is true, then $P(1+2)=P(3)$ is true.

$P(2)$ and $P(3)$ is true, then $P(2+2)=P(4)$ is true.

So, we observe that  $P(n)$ is true for all the non- negative integers : $\{0, 1,2,3,4,5,6,...\}$.

d). $P(0)$ is true,

   So, $P(0+2)$ and $P(0+3)$ is true or $P(2)$ and $P(3)$ is true.

   Now,   $P(2)$ is true.

Again, $P(2+2)$ and $P(2+3)$ is true or $P(4)$ and $P(5)$ is true.

   Now, $P(3)$ is true.

Again, $P(3+2)$ and $P(3+3)$ is true or $P(5)$ and $P(6)$ is true.

Thus,

$P(n)$ is true for all the non- negative integers except 1 : $\{0, 2,3,4,5,6,...\}$.