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vivado [14]
2 years ago
3

Exercise 8.5.1: Proving divisibility results by induction. About Prove each of the following statements using mathematical induc

tion. (a) Prove that for any positive integer n, 4 evenly divides 32n-1. (b) Prove that for any positive integer n, 6 evenly divides 7n - 1.
Mathematics
1 answer:
Step2247 [10]2 years ago
3 0

Answer:

a) 4(9x+2)

b)6(7x +1)

Step-by-step explanation:

a) Prove that for any positive integer n, 4 evenly divides 32n-1

checking whether the statement is correct or not

∴  n = 1;

= 3^{2n} -1

= 3^{2\times1} -1

= 9 - 1

= 8

hence it is divisible by 4

Let the statement is for n = k

∴  3^{2k } -1  = 4x(equation 1)

(3^{2})^{k} -1 = 4x

9^{k} -1 = 4x (equation 1)

Now, we have to proof the statement is true for n = k+1

= 3^{2(k+1)} -1

= (3^{2k} \times 3^{2} ) -1 (x^{a+b} = x^{a} \times x^{b})

Adding  & Subtracting 8

= (3^{2k} \times 3^{2} ) -1 +8 -8

= 9^{k} \times 9 -9 + 8

taking common 9

= 9(9^{k} -1)+8

= 9 (4x) +8 (from equation 1)

= 36x + 8

= 4(9x+2)

if (9x+2) =  p

then = 4p

Since 3^{2(k+1)} -1 = 4p evenly divisible by  4

therefore given statement is true

b)Prove that for any positive integer n, 6 evenly divides 7^{n} - 1

checking whether the statement is correct or not

∴  n = 1;

7^{n} - 1

7 - 1

6

6 is divisible by  6

hence the given  statement is true for n = 1

let it  also true  for n = k

7^{k} - 1 = 6x (equation 2)

Now we have to proof the statement is true for n = k+1

7^{k+1} - 1

7^{k}\times7  - 1

Adding  & Subtracting 6

7^{k}\times7  - 1 +6 - 6

7^{k}\times7  - 7 +6

7(7^{k}\times  - 1) +6

7(6x )+6 ( from equation 2)

= 42x + 6

= 6(7x +1)

if 6(7x +1) =  p

then = 6p

Since7^{k+1} -1 = 6p evenly divisible by  6

therefore given statement is true

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