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

Solution :

a). is true

Then , is true.

          is true

          is true.

Therefore, we see that is true for all the even integers :

b). is true

Then , is true.

          is true

          is true.

Therefore, we see that is true for all the multiples of 3 :

c). and is true, then is true

and is true, then is true.

and is true, then is true.

So, we observe that   is true for all the non- negative integers : .

d). is true,

   So, and is true or and is true.

   Now,   is true.

Again, and is true or and is true.

   Now, is true.

Again, and is true or and is true.

Thus,

is true for all the non- negative integers except 1 : .

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

5+2x=a

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