Proof by Induction

  • Thread starter roam
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  • #1
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Homework Statement



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


Homework Equations




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 | (10k+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,

10k+2-10(k+1)-10=10 × 10k+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...
 
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Answers and Replies

  • #2
rock.freak667
Homework Helper
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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.
 
  • #3
1,267
11
That is how you'd arrive at your answer.

It's a bit awkward... are you sure that there is not a better way?
 
  • #4
1,101
3
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]
 
  • #5
1,267
11
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 10k+1 by 81m+9k-10, since [tex]10^{k+1}-9k-10=81m \Rightarrow 10^{k+1} = 81m+9k+10[/tex]
 
  • #6

Homework Statement



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 | (10k+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,

10k+2-10(k+1)-10=10 × 10k+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:
10k+2-9(k+1)-10=10 × 10k+1-9(k+1)-10
=10(81m+9k+10)-9k-9-10

Then keep going.
 
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  • #7
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3
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.
 
  • #8
rock.freak667
Homework Helper
6,223
31
[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 HN+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.
 

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