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I have learnt how to prove expressions by induction based on the use of three assumptions

  1. $n=1$,
  2. $n=k$,
  3. $n=k+1$.

But can someone help me prove that $1+10^{2n-1}$ is divisible by $11$

MattAllegro
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Dr. NGILAZI BANDA
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  • That's not quite right. The way induction works is that: You prove that a statement holds for $1$; You show that "if the statement holds for $k$, it also holds for $k+1$." – Braindead Apr 24 '14 at 18:26
  • The way you prove an implication in mathematics, statements of the form "if P, then Q", is that you try to show that Q holds under the assumption of P. In induction, P is "something works for $k$", and Q is "something works for $k+1$". (Please note that "something works for $n=1$" is not part of the inductive hypothesis, but rather something you need to verify.) – Braindead Apr 24 '14 at 18:29

1 Answers1

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With $n=1$, the expression becomes $1+10^{2\cdot 1-1}$. Is that divisible by $11$?

With $n=k$ and $n=k+1$, the expression becomes $1+10^{2k-1}$ and $1+10^{2k+1}$, respectively. Their difference is $10^{2k+1}-1-(10^{2k-1}-1)=10^{2k-1}\cdot(10^2-1)$, so if one is divisible by $11$, then ...?

Hagen von Eitzen
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