Question

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

Answer 1

Answer:

a) 4(9$x$+2)

b)6(7$x$ +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

∴  $3^{2k } -1$  = 4$x$(equation 1)

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

$9^{k} -1$ = 4$x$ (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 (4$x$) +8 (from equation 1)

= 36$x$ + 8

= 4(9$x$+2)

if (9$x$+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(6$x$ )+6 ( from equation 2)

= 42$x$ + 6

= 6(7$x$ +1)

if 6(7$x$ +1) =  p

then = 6p

Since$7^{k+1} -1$ = 6p evenly divisible by  6

therefore given statement is true