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I am familiar with induction proofs on the set of the natural numbers. However now I have to prove a statement for a subset $K = \{k \in N |k = 12x+9\}$ with $x \in N$.

So basically $K = \{9, 21, 33, 45, 57,...\}$.

Now I have to write an induction proof for this set and I'm not entirely sure how to go about it. Is the base case $k = 9$ and the induction step $k \rightarrow k+12$? Obviously it works that way but I am not sure if it's mathematically correct as it doesn't look very "clean".

Would I have to prove additionally that $k=9$ is the first element and, and that all the elements are $12$ apart from each other or something?

Thanks in advance

math-physicist
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cassnx
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  • Yes, that's correct. You essentially do induction on $x$ where $k = 12x + 9$. – Anon May 29 '22 at 01:02
  • You could also do induction on $x$ with $x=0, 1, 2,...$. Then your $k=12x+9$ runs through the desired values $9, 21, 33, ...$. Can you say specifically what you are trying to prove? – paw88789 May 29 '22 at 01:08

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You can think of it as induction on $\Bbb N$, where the statement you're proving about each $n\in\Bbb N$ is something about the number $12n+9$.

Karl
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