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I think I proved this but I am not to confident in my proof. I am also not that good at formating my proofs so any feedback would be appreciated :)

Let $P(n)=9|10^n-1$

Base Case: $9|10^1-1\implies9|9\checkmark$ So we have proven the base case.

Induction Step: $$9|10^n-1\implies9k=10^n-1\implies10^n=9k+1$$ $$9|10^{n+1}-1\implies9|(9k+1)\cdot10-1\implies9|90k+9$$ Every term in the above has a common factor of 9 hence the left divides the right. We have thus shown that $P(n)\implies P(n+1)$ with a base case. Therefore by the principle of induction $P(n)$ holds $\forall n\in\mathbb{N}$.

Edit: The second line of the induction step was flawed as it didn't really prove anything. The fixed version goes as follows: $$10^{n+1}=(9k+1)\cdot10\implies10^{n+1}-1=90k+10-1\implies10^{n+1}-1=90k+9$$ We can see that this $10^{n+1}-1=90k+9$ is divisible by 9. We have thus shown that $P(n)\implies P(n+1)$ with a base case. Therefore, by the principle of induction, $P(n)$ holds $\forall n\in\mathbb{N}$.

Nicojwn
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    Your proof is not good because you suppose that $P(n+1)$ is true... – Surb Dec 30 '21 at 09:50
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    Your implies arrows are going the wrong way in the second line to show that $P(n) \implies P(n+1)$. An easier way might be to say that $9|(10^n-1) \implies 9|((10^n-1)\cdot 10)\implies 9|((10^n-1)\cdot 10 + 9) \implies 9|(10^{n+1}-1)$. – Varun Vejalla Dec 30 '21 at 09:55
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    You have the right idea, except you wrote the arrows in the wrong direction in the 2nd displayed inference in your induction step. What you meant was: the inductive hypothesis implies $,10^{n+1}-1 = 9(10k+1),$ so $9$ divides $10^{n+1}-1,,$ which completes the proof of the inductive step. To make the induction step completely obvious I highly recommend using the the method in this answer in the linked dupe. – Bill Dubuque Dec 30 '21 at 10:54
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    Dupe of this and this and hundreds of others. Please don't reopen dupes of FAQs – Bill Dubuque Dec 30 '21 at 12:11
  • You can prove statements of this kind much easier with modular arithmetic. In this case, the induction is not very difficult, but in more complicated situations it can quickly get cumbersome. In the given case , there is an even easier way if you think of the decimal repsentation of $10^n-1$ , only nines are occuring. – Peter Dec 30 '21 at 12:13
  • @BillDubuque No, it is not a duplicate. This question is about checking a proof, not about how to prove something. – José Carlos Santos Dec 30 '21 at 12:15
  • @JoséCarlosSantos But induction proofs of this kind (to be checked) have been posted many many times, so I have to agree Bill. – Peter Dec 30 '21 at 12:22
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    @Peter This question was originally closed as a dupe of this question and of this one, none of which is a proof-verification question. – José Carlos Santos Dec 30 '21 at 12:26
  • @JoséCarlosSantos That would mean that questions should be reopened because they were closed for the wrong reason , even if there is a suitable reason to close them. Not my philosophy ! – Peter Dec 30 '21 at 12:33
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    @Jose A proper PV question is a question about some specific point of the proof where the author has doubts, along with context explaining why those doubts exist - not an open ended "please check my entire proof". Often such questions can be posed in comments on an answer in a dupe (and they help clarify the answer). If we allow open ended "check my whole proof" questions then we will have hundreds of matches to basic questions that will swamp search results with matches that are mostly useless (showing only answers addressing trivial oversights etc vs. the real proof). – Bill Dubuque Dec 30 '21 at 13:04
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    @jose Usually the best way to handle questions like this is to close them as dupes, then leave a comment pointing out the (usually trivial) oversight (and if the oversight is common enough to be helpful to others there might be a a post to link to on that particular type of error), then eventually delete the question to stop it from cluttering search results on the actual proof. But in many cases once the OP reads the dupes the answer becomes immediately clear (and if not a further comment or two usually resolves it, and may help to polish rough points in answers in the dupes). – Bill Dubuque Dec 30 '21 at 13:13
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    ... with the goal being that eventually the (canonical) dupe tagets will converge to "proofs from the book" after enough iterations and feedback from readers. But "the book" is not helpful if its index is so cluttered with unhelpful references that you cannot easily locate the proofs (which is precisely what is happening here for FAQs like this question). – Bill Dubuque Dec 30 '21 at 13:17
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    @BillDubuque What you wrote makes a lot of sense. But I don't think that it justifies closing this question as a duplicate of two other questions that contain other proofs of the same statement. We (as a group) could adopt a policy according to which it would be inappropriate to post question of the sort “Please check my proof of […]” unless the author stated the specific points where they have doubts. Before that happens, I don't think that this question should be closed (unless we find another similar attempted proof). – José Carlos Santos Dec 30 '21 at 14:18
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    @JoséCarlosSantos I'm very disappointed to hear that, since it essentially destroys attempts at site organization (which are already difficult enough as it is w/o this complication). Others (including at least one mod - Xander) have espoused similar views to mine on SV posts. The dupes don't (only) contain "other" proofs - they contain the same (common) proof too and, furthermore, they give much elaboration that likely helps one to avoid such oversights. – Bill Dubuque Dec 30 '21 at 14:23
  • @amWhy What you wrote that is relevant is not true, and what that is true is not relevant. – Bill Dubuque Dec 30 '21 at 18:55
  • Eight comments from Bill. Half of all comments here. I wrote only one, and have since deleted it, because I'm sick said user's offensiveness; it was both relevant and true. – amWhy Dec 30 '21 at 19:01

2 Answers2

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I think you're almost all the way there-- the main feedback I'd give is around your final statement:

$$9|10^{n+1}-1\implies9|(9k+1)\cdot10-1\implies9|90k+9$$

Above you're assuming what you want to prove. Try working backwards starting with what you know to get to what you want to prove. Here, try the following to get started:

$$10^n = 9k + 1 \implies 10^{n+1} = (9k + 1) \cdot 10 \implies 10^{n+1} - 1 = ... $$

N.B. Also be sure to declare what $k$ is when you use it.

I hope this helps!

DanZimm
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When you write $9\mid10^n-1\implies9k=10^n-1$, you should say what that $k$ is. Something like “$9k=10^n-1$” for some non-negative integer $k$.

And then your aim is to prove that $9\mid10^{n+1}-1$. That's not what you do. You start from $9\mid10^{n+1}-1$ and then you see what you can conclude from that.

José Carlos Santos
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