Homework Help: Proof by Induction

roam · Aug 23, 2009

1. The problem statement, all variables and given/known data

http://img200.imageshack.us/img200/7097/99175506.gif [Broken]

2. Relevant equations

3. The attempt at a solution

For [tex]n \in N[/tex] let P(n) be the statement: "81 | (10ⁿ⁺¹-9n-10)"

Base Case: when n=1: 10ⁿ⁺¹-9n-10 = 81 = 81 × 1

So P(1) is true

Inductive Step: let [tex]k \in N[/tex] and suppose [tex]P(k)[/tex] is true, that is 81 | (10^k+1-9k-10) is true. Then [tex]10^{k+1}-9k-10=81m[/tex] for some [tex]m \in Z[/tex]. Then [tex]10^{k+1}=(81m+9k+10)[/tex].

So,

10^k+2-10(k+1)-10=10 × 10^k+1-10(k+1)-10

=10 × (81m+9k+10)-10k+10-10

= 10 × (81m+9k+10-k)

I'm stuck here and I don't know how to factor things out and bring the 81 forth to get the expression in the form: "81(something here)" to show that it's divisible by 81 for all n. Any help is appreciated.

P.S.
I might be able to do it this way:

10 × (81m+9k+10-k) = 810m+90k+100-10k
=> 81(10m+(90/81)k+(100/81)-(10/81)k)

But I don't feel that this is the right way...

rock.freak667 · Aug 23, 2009

roam said: ↑

P.S.
I might be able to do it this way:

10 × (81m+9k+10-k) = 810m+90k+100-10k
=> 81(10m+(90/81)k+(100/81)-(10/81)k)

But I don't feel that this is the right way...

That is how you'd arrive at your answer.

roam · Aug 23, 2009

rock.freak667 said: ↑

That is how you'd arrive at your answer.

It's a bit awkward... are you sure that there is not a better way?

snipez90 · Aug 23, 2009

You have a few typos in your inductive step (should be 9(k+1), not 10(k+1)).

Note that
[tex]10^{(k+1)+1} - 9(k+1) - 10 = 10\cdot10^{k+1} - 9k - 9 - 10 = 10(10^{n+1} - 9n - 10) + 81n + 81[/tex]

roam · Aug 23, 2009

How did you get from [tex]10\cdot10^{k+1} - 9k - 9 - 10[/tex] to [tex]10(10^{n+1} - 9n - 10) + 81n + 81[/tex]? Could you please explain :)

Yes that was a typo but I thought it would be a good way to replace 10^k+1 by 81m+9k-10, since [tex]10^{k+1}-9k-10=81m \Rightarrow 10^{k+1} = 81m+9k+10[/tex]

mathie.girl · Aug 23, 2009

roam said: ↑

1. The problem statement, all variables and given/known data

http://img200.imageshack.us/img200/7097/99175506.gif [Broken]
Inductive Step: let [tex]k \in N[/tex] and suppose [tex]P(k)[/tex] is true, that is 81 | (10^k+1-9k-10) is true. Then [tex]10^{k+1}-9k-10=81m[/tex] for some [tex]m \in Z[/tex]. Then [tex]10^{k+1}=(81m+9k+10)[/tex].

So,

10^k+2-10(k+1)-10=10 × 10^k+1-10(k+1)-10

=10 × (81m+9k+10)-10k+10-10

= 10 × (81m+9k+10-k)

I'm stuck here and I don't know how to factor things out and bring the 81 forth to get the expression in the form: "81(something here)" to show that it's divisible by 81 for all n. Any help is appreciated.

The above looks good, without some typos. Should be:
10^k+2-9(k+1)-10=10 × 10^k+1-9(k+1)-10
=10(81m+9k+10)-9k-9-10

Then keep going.

snipez90 · Aug 23, 2009

roam said: ↑

How did you get from [tex]10\cdot10^{k+1} - 9k - 9 - 10[/tex] to [tex]10(10^{n+1} - 9n - 10) + 81n + 81[/tex]? Could you please explain :)

You always want to apply the inductive hypothesis as easily as possible. If the inductive step gives you
[tex]10^{(k+1)+1} - 9(k+1) - 10 = 10\cdot10^{(k+1)} - 9(k+1) - 10 ,[/tex]
then the fact that you can pull out the very first 10 factor on the right hand side suggests multiplying the expression [tex]10^{(k+1)} - 9k - 10[/tex] in our inductive hypothesis by 10 and relating it to the inductive step somehow.

rock.freak667 · Aug 23, 2009

[tex]H_N: 10^{N+1}-9N-10=81m[/tex]

x10

[tex]10^{N+2}-90N-100=810m[/tex]

[tex]10^{N+2}-90N-90-10=810m[/tex]

[tex]10^{N+2}-90(N+1)-10=810m \Rightarrow 10^{N+2}-9(N+1)-10=810m+81(N+1)[/tex]

Left side happens to be H_N+1 and the right side is divisible by 81 no matter what N is.....

so even if it is awkward, once you show it is divisible by 81, you have proven it.