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I've been trying to prove using mathematical induction the following statement :

"Prove that for all integers n ≥ 1 we have 8|9^n − 1 ."

What I did and where I struggled :

Verify :

if n=1 
9^1 -1 = 8 TRUE

Assumption : 

9^n -1 = 8m 
9^(n+1) - 1 = 9*9^n - 1

When I arrived I did not know what to do anymore, I checked the solution, but I cannot understand how come there are a "-1" and "-1" that pop up for apparently no reason ; solution :

9*9^n-1 +0 = 9*9^n-1+1-1 = 9(9^n-1)+8 = 9(8m) + 8 = 9*8m+8 = 8(9m+1)

So my questions are :

  1. Why did he added the +1-1 ? (and how come we can add numbers like that)
  2. why did he use 8m to define 9^n -1 , couldn't it have been another variable, i.e : only the letter "k"
  3. Where does the "8" come from ?

In conclusion I do not understand the whole procedure...

Pedro Gomes
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  • Try rewriting 9^n as 8m+1, which it is, in the last line of your second box and expanding, you will see an expression that is a multiple of 8. – Andrew Lazarus Sep 22 '17 at 20:56
  • Adding $+1-1$ is the same as adding $0$, and you can always add $0$ without changing the value of anything. It's a common trick that really helps in some arguments, like this one. Being able to spot when to use it is a matter of practice. – Arthur Sep 22 '17 at 20:57
  • I recommend you read this answer, which is a great explanation of induction. – Ross Millikan Sep 22 '17 at 20:59
  • @Arthur that was what I was looking for. Thank you. – Pedro Gomes Sep 22 '17 at 21:02

3 Answers3

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The equation is not wrong at all, but truly confusing. What we need is to add $-9+9$ since it is zero and you get $$ 9\cdot 9^n-1 = 9\cdot 9^n\underbrace{-9+9}_{=0}-1=\underbrace{9\cdot 9^n-9}_{=9\cdot(9^n-1)}\underbrace{+9-1}_{=+8}=9\cdot(9^n-1)+8. $$

Mundron Schmidt
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  1. because you did one thing to part of a side not the whole part, ( second line in your assumption is wrong)
  2. sure he could have ( as long as you don't confuse them you can use any variable names for example $$\begin {align}\\A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z\\a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z\\\alpha,\beta,\gamma,\delta,\epsilon,\eta,\theta,\pi,\omega,\chi,\phi,\psi,\omicron,\mu,\nu\end {align}$$ and more
  3. you have to cancel out the -9 that occurs if you do it fully $9\cdot(9^n-1) =9^{n+1}-9$ so you have to add 8 to get back to $9^{n+1}-1$
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Hint: $9\cdot 9^n-1=(8+1)9^n-1=8\cdot 9^n+9^n-1$

Note, that the first term is a multiple of 8 and the second term is a multiple of 8 because of your induction assumption.

MrYouMath
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