Question

QUESTION 8 Consider the following propositional functions. Choose the true statements. If we know that p(1) is true and we know that the truth of p(n) implies the truth of p(n + 1), then we know that pn) is true for all positive integers n. If we know that p(1) is true, and we know that for all positive integers, if p(n) is true then pln+2) is true, then p(n) is true for all n 2 1. If we know that p(1) and p(2) are true and we know that if p(n) and p(n + 1) are true then p(n + 2) is also true, then we can conclude that p(n) is true for all n 2 1. If we know that p(1) is true, and we know that the truth of p(n) implies the truth of p(2n), then we can conclude that p(n) is true for all even n. QUESTION 9 Consider encryption algorithms where p = letter's position in our 26- character alphabet and a is assigned to o a) f(p)= (p+3) mod 26 b) f(p) = (p+13) mod 26 c) f(p) = (3p + 7) mod 26 which of the following strings is the encrypted version of do not pass go? QX UXM AHJJ ZX MULHR YNPP SV QB ABG CNFF TB GR QRW SDWJR

Answer 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