Question

Use the principle of mathematical induction to show that 8 is a factor of 9^n - 1.

Answer 1

I already did some of the work so I'll just post what I've got so far (:

Answer 2

yeah you may take a screenshot and attach the pic instead of typing everything by hand

Answer 3

\[9^n-1=(8+1)^n-1\]\[(8^n+n8^{n-1}+..+1)-1\]\[(8^n+n8^{n-1}+..+1)=8k\]

Answer 4

Let P\[_{n}\] denote the statement that 9^n -1. We want to show that Pn is true for all natural numbers n.
I. We must check that P1 is true, where P1 is the statement that...
9-1=8 (which is a factor of 8)
Thus P1 is true.
II. Assuming that Pk is true, we must show that Pk+1 is true. Thus we can assume that 8 is a factor of 9^k -1 and we must show that 8 is a factor os 9^(k+1) -1 = 9^k*9-1

Answer 5

@amirreza1870 Could you explain how you got the second line? Thank you (:

Answer 6

well prove its true for n = 1

9^1 - 1 = 8

so true for n = 1

assume its true for n = k

then 9^k - 1 = 8M

the show for n = k + 1

start with

\[9^{k + 1} -1 = 9 \times 9^k - 1\]

so for the key... let -1 = -9 + 8

so and subsitute so you get

\[9 \times 9^k - 9 + 8 = 9(9^k - 1) + 8\]

substitute the n = k term ... 8M

then you have

\[9(8M) + 8 = 8(9M + 1)\]

so proven... etc

Answer 7

9^(k+1) -1 = 9^k*9-1
= 9^k(8+1) - 1
= 9^k*8 + 9^k - 1

Answer 8

Or you could go like this :

9^k - 1 = 8M

multiply 9 both sides
9^(k+1) - 9 = 8(9M)
9^(k+1) -1 -8 = 8(9M)
9^(k+1)-1 = 8(9M+1)

QED.

Answer 9

@campbell_st Ok awesome! But I still do not understand what you mean by "substitute the n = k term ... 8M" will you be able to specify on that? Thank you so much~

Answer 10

Oh and @rational Thank you too but I do not understand how the last equation that you made "9^k*8 + 9^k - 1" proves that it is a factor of 8.

Answer 11

Oh and for the very previous problem that you sent me.. I do not think it will work due to the fact that you are implying that you already think that both sides are indeed equal when multiplying both sides by 9. @rational

Answer 12

"9^k*8 + `9^k - 1`"

8 is a factor of `9^k - 1` from ur induction assumption, right ?

Answer 13

well when you assume its true for the kth term...

so

\[9^k - 1\]

you're assuming its true ... that is... 8 is a factor... you don't care what the other factor is... making it something arbitrary...I used M

so I'm saying the kth term

\[9^k - 1 = 8 \times M...or...8M\]

which is needed to substitute into the term n = k + 1

Answer 14

OH I see!! So adding two factors of 8 will obviously equal to another factor of 8! Thanks @rational

Answer 15

Yes! the very last proof also works. go thru it again, thats the recommended method in induction proofs : start with P(k) assumption and arrive at P(k+1) conclusion

Answer 16

@campbell_st That is a really interesting way to solve the problem! Thank you :DD

Answer 17

well when you substitute into the n = k + 1 term if you do it the long way

\[9\times 9^k - 9 + 8 = 9(9^k - 1) + 8\]

now subsitute 8M

\[9(8M) + 8 = 72M + 8\]

remove the common factor or 8 and you get

\[8(9M + 1)\]

since you have a factor of 8... you've proved the case...

Answer 18

@rational Awesome! That solution really helped, thank you (:

Answer 19

your induction assumption for n=k case is an equation :

9^k - 1 = 8M

since it is an equation, you can multiply anything u wish both sides

Answer 20

the good thing to multiply both sides here is `9` because that makes the exponent `k+1`