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I'm doing some prove by mathematical induction practice problems. I got to this last problem but I don't know how to approach it.

This is the question

For all positive integers $n$ and any distinct real numbers $x$ and $y$, $x - y$ divides $x^n - y^n$.

ForgotALot
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user3023185
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3 Answers3

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Using strong induction:

Base cases $n=1$ and $n=2$ hold trivially $(x-y)\mid (x-y)$ and $(x-y)\mid (x^2-y^2)$ since $x^2-y^2=(x+y)(x-y)$

Assume for our induction hypothesis that the claim $(x-y)\mid (x^k-y^k)$ holds for all $k$ up to $n$ for some $n$.

Consider $n+1$

$x^{n+1}-y^{n+1}=(x+y-y)x^n-(y+x-x)y^n$

$=(x+y)(x^n-y^n)-yx^n+xy^n=(x+y)(x^n-y^n)-xy(x^{n-1}-y^{n-1})$

Each term of which is divisible by $x-y$ by our induction hypothesis, thus the hypothesis is validated and the claim is true for all positive integers $n$.

JMoravitz
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  • Excuse me, I understood that in strong induction the only thing that changes is the inductive hypothesis. Is it really necessary to evaluate two base cases? – ESCM May 03 '20 at 04:59
  • @EduardoS. in the way I wrote my answer, yes, since we apply the inductive hypothesis both to the $x^n-y^n$ that appears as well as the $x^{n-1}-y^{n-1}$ that appears when trying to show that $x^{n+1}-y^{n+1}$ is divisible by $x-y$. So, in the way the inductive argument I laid out above, we need to know the statement is true for both of the previous two values. – JMoravitz May 03 '20 at 05:01
  • You can, of course, prove this a different way. Andre's answer for instance succeeds at avoiding needing to apply the induction hypothesis more than once in his induction argument. If I were to have reworded my induction argument above to a completely different one then yes we could have gotten away without showing the second base case. But, if wishing to apply the inductive hypothesis to multiple expressions simultaneously with different indices (like how I apply both to $x^n-y^n$ as well as to $x^{n-1}-y^{n-1}$) then you need appropriate base cases shown to match. – JMoravitz May 03 '20 at 05:06
  • My strong induction concept is: In the inductive hypothesis, you assume that property holds for all numbers less than a certain n. Suppose that you were able to demonstrate the inductive step, which, with the assumed information, will demonstrate that "If all the naturals preceding a natural number (n + 1) fulfill a property, then (n + 1) fulfills it"). – ESCM May 03 '20 at 18:16
  • But we did not know if what we assumed in the inductive hypothesis was true, we only assumed it. So here the base case comes into play, since we evaluate and obtain that for $n = 1$ it holds, so we have the first case that holds the property.(this is why base case exists) – ESCM May 03 '20 at 18:17
  • Since we prove that "If we choose a number (n+1) and all its previous numbers holds the property, then (n+1) it holds it", then if we choose any number, it is enough to see if all its previous numbers holds the property, so that this number fulfills it. We could see that "all the numbers before n = 2" are only the base case and therefore n = 2 complies, then for n = 3 it complies and so on. "It is enough that the first piece of dominoes fall so that the others fall" So why do you show 2? This isn't necessary, according to my understanding of how strong induction works. Thanks for you reading – ESCM May 03 '20 at 18:20
  • I showed $1$ directly. I showed $2$ directly. From then on, I showed it holds for $n+1$ because it holds both for $n$ and for $n-1$... that is to say, I did not manually show that it worked for $3$, but it followed from the inductive argument and because it was true for both $1$ and for $2$. I could not show that it was true for $2$ because of the inductive argument with the way I worded it, since that would have needed knowledge that it worked for both $1$ and for $0$... but here we are ignoring $0$ and it doesn't work there anyways. – JMoravitz May 03 '20 at 18:40
  • But as I explained to you, n = 2 should be deduced from the same inductive argument, like a domino. – ESCM May 03 '20 at 18:42
  • In the way I worded my argument... it is not the weight of a single domino which causes the next domino to fall, but rather the weight of the two previous dominos together which cause the next to fall. – JMoravitz May 03 '20 at 18:43
  • Yes, it is exactly what I said. We proved that n = 1 was true. Now why prove that n = 2 is true? The "2" domain will fall, since all its previous domains are reduced to one (n = 1). And we already proved that it was true. So I don't understand why you test case n = 2. – ESCM May 03 '20 at 18:50
  • We know that 'domino' $1$ holds,was prove in the base case.

    To see if 'domino' 2 holds, we see that all the ones before it (only n = 1) holds, but that is the base case, therefore n = 2 holds.

    To see if dominos 3 holds, we see that all previous to it (n = 1, n = 2) holds, therefore n = 3 holds .

    and so...

     – ESCM May 03 '20 at 18:53
  • No no no... My argument... to show that domino $n+1$ falls requires both of the dominos $n$ and $n-1$ to fall. If I wanted to show that domino $3$ falls, that requires both of the dominos $2$ and $1$ to have fallen, and the argument the way that I worded it does not show that domino $2$ falls just because domino $1$ fell... the argument I used only works to show the domino $n+1$ falls because both of the dominos $n$ and $n-1$ were known to have fallen. In the case of $2$ we do not know that both of the previous two dominos had fallen because there aren't two dominos before it to fall – JMoravitz May 03 '20 at 18:58
  • Do not confuse the phrase "The previous two cases held" with the phrase "All previous case/cases held" – JMoravitz May 03 '20 at 19:00
  • I understand, your demo requires that the previous two comply. So why assume the entire range (k to n) instead of n and (n-1)? That is, would it not be clearer just to assume what you are going to use? – ESCM May 03 '20 at 19:03
  • If you still don't get it... then how about thinking about the Fibonacci Sequence. If all I told you about the sequence was that the first term was $f(0)=1$ and then I told you that to find $f(n+1)$ you can use $f(n+1)=f(n)+f(n-1)$... then that is not enough information... the Lucas Numbers also follow the same pattern. To find the next fibonacci number or to find the next lucas number, we need to know the previous two numbers to add. – JMoravitz May 03 '20 at 19:04
  • It can be reworded to very specifically refer to only $n$ and $n-1$... sure... but why bother when it is just as easily worded as strong induction and strong induction is used so prevalently already that it should be recognizable and easy to understand. – JMoravitz May 03 '20 at 19:05
  • Okay, so I think I understood. The proof used the assumption that there were 2 previous cases for which they were true, however, for "2" there were not 2 previous cases with the property. – ESCM May 03 '20 at 19:08
  • So in general, if the argument of our proof will assume(and uses them) $n$ previous terms that holds the property I will have to evaluate $n$ base cases. – ESCM May 03 '20 at 19:08
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Compute $(x-y) \sum_{k=0}^n x^k y^{n-k}$.

copper.hat
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Hint: $$x^{n+1}-y^{n+1}=x^{n+1}-x^ny+x^ny-y^{n+1}=x^n(x-y)+y(x^n-y^n).$$

Remark: You don't really want to say any distinct real numbers, for any non-zero real divides any real number, so there is nothing to prove. Integers is fine.

André Nicolas
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  • While this is indeed true, I've always wondered about intuition behind this. I would make an attempt to write x^(n+1) - y^(n+1) as something like (x^n-y^n) * Z to make use of the hypothesis but figuring out I also "need" (for some definition thereof) the (x-y) * W term is a different matter altogether... – ScarletAmaranth May 02 '17 at 17:04