QUESTION 8 Consider the following propositional functions. Choose the true statements. If we know that p(1)...

QUESTION 8:

1) TRUE:

This is true because we know that p(1) is true also p(n) implies p(n+1) i.e. p(1) implies p(2). Hence, p(2) is also true. Now, we put 2 in place of n and we get that p(2) implies p(3) so, p(3) is also true. Now, we put n=3 and so on.

2) FALSE:

This is false. It is given that p(1) is true and also that if p(n) is true then p(n+2) is true so if we put n=1 then we get that if p(1) is true then p(3) is also true. Hence p(3) is also true. Now we miss out p(2) and we can never conclude that p(2) is true. So, we miss out on some of the elements to be true in the set of positive integers.

3) TRUE:

This is true because we know that p(1) and p(2) are true also p(n) and p(n+1) implies p(n+2) i.e. if p(1) and p(2) are true then p(3) is true. Hence, p(3) is also true. Now, we put 2 in place of n and we get that p(2) and p(3) implies p(4) so, p(4) is also true. Now, we put n=3 and so on.

4) FALSE:

This is false. It is given that p(1) is true and also that if p(n) is true then p(2n) is true so if we put n=1 then we get that if p(1) is true then p(2) is also true. Hence p(2) is also true. Now we put n=2 and get that if p(2) then p(4) is also true. Hence, P(4) is true. Now we put n=4 and similarly get that p(8) is also true. Now we miss out p(6) and 6 is also an even integer.

QUESTION 9:

I'm considering the first case where f(p) = (p+3)mod26

d:

so, p=3

Now, f(3) = (3+3)mod26 = 6mod26 =6 = G

o:

so, p=14

Now, f(14) = (14+3)mod26 = 17mod26 =17 = R

n:

so, p=13

Now, f(13) = (13+3)mod26 = 16mod26 =16 = Q

t:

so, p=19

Now, f(19) = (19+3)mod26 = 22mod26 =22 = W

p:

so, p=15

Now, f(15) = (15+3)mod26 = 18mod26 =18 = S

a:

so, p=0

Now, f(0) = (0+3)mod26 = 3mod26 =3 = D

s:

so, p=18

Now, f(18) = (18+3)mod26 = 21mod26 =21 = V

g:

so, p=6

Now, f(6) = (6+3)mod26 = 9mod26 =9 = J

So, the answer is:-

GR QRW SDVV JR