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Okay so I just finished a final in discrete mathematics and I could not figure out how to finish this proof:

"Prove by mathematical induction, that $(a^n - b^n)$ is a multiple of $(a - b)$ when $a$ and $b$ are integers and $n \geq 1$.

$\underline{Base case: n = 1}$

$(a^1 - b^1) = (a - b)x$

$x = 1$

$a - b = a - b$

$\underline{Inductive Hypothesis:}$

$n = k$ for some $k$

$(a^k - b^k) = (a - b)x$

$\underline{Inductive Step:}$

$n = k + 1$

$(a^{k + 1} - b^{k + 1}) = (a - b)x_1$

$(a \cdot a^{k}) - (b \cdot b^{k}) = (a - b) x_1$

$a \cdot [(a - b)x + b^k] + b \cdot [(a - b)x - a^k]= (a-b)x_1$

$(a -b)xa + ab^k + (a - b)xb - ba^k = (a - b)x_1$

$(a -b)xa + (a - b)xb - ab^k + ba^k = (a - b)x_1$

$(a -b)xa + (a - b)xb + ba^k - ab^k = (a - b)x_1$

.....

I don't know if I did it right but I couldn't get any further than this.

N.Luscomb
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  • Are you forced to use induction? – user Dec 14 '18 at 20:49
  • Yes, not strong induction as well. – N.Luscomb Dec 14 '18 at 22:29
  • Why are my posts getting downvoted / removed? I am very new here, but I want to fix whatever it is I am doing wrong. I am posting this here because this post got -2 votes and I don't know what is wrong? – N.Luscomb Dec 14 '18 at 22:38
  • @N.Luscomb: It may be because the tags may be incorrect or maybe some feel its too basic. – Yadati Kiran Dec 14 '18 at 22:52
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    @N.Luscomb The question seems fine to me. You've given context, made the question clear, shown what you tried, said how far you got. (Maybe someone didn't like the title being worded as an instruction or something—the site gets bombarded with homework questions. I'm just guessing though.) – timtfj Dec 15 '18 at 00:50
  • Your question is much better than a lot of other questions flooding this site. I don't think it deserves to be downvoted, but just be aware that such basic questions have almost surely been posted and answered before, so a quick search for it would be preferable. But I commend you on wanting to finish your exam even after it is over. =) – user21820 Dec 17 '18 at 06:34
  • @user21820 Agreed. And yet, it is a real pity that none of the six answers below is addressing the main problem the OP seems to have, which is how induction proofs actually work. As we know, the induction step of a proof by induction that P(n) holds for every n, simply cannot start with "Assume P(n+1) then..." and yet this is what the OP does above, which leads them to a deadend, honestly signalled by "I don't know if I did it right but I couldn't get any further than this". As long as the OP does not identify this misconception, they will fail to "build" any coherent proof by induction. – Did Dec 18 '18 at 08:26
  • @Did: Right. And none of the answers even did the induction in a logically cogent manner... N.Luscomb: To make it clear, induction for any property $P$ asserts that from $P(0)$ and $\forall n \in \mathbb{N}\ ( P(n) ⇒ P(n+1) )$ you can deduce $\forall n \in \mathbb{N}\ ( P(n) )$. So before you can even use induction on $P$, you must prove the two statements that it requires. If you do not know how that works, you would need to learn basic logic (which clearly your course failed to teach)... – user21820 Dec 18 '18 at 08:49

5 Answers5

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Assumed $a^k-b^k=(a-b)m,m\in\Bbb Z\implies a^k=b^k+(a-b)m$

We have $a^{k+1}-b^{k+1}=a\times a^k-b^{k+1}=a[b^k+(a-b)m]-b^{k+1}\\=am(a-b)+ab^k-b^{k+1}\\=am(a-b)+b^k(a-b)\\=(am+b^k)(a-b)$

Shubham Johri
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1)$n=1 : a-b ✓$

Hypothesis: $a^n-b^n$ is a multiple of $(a-b)$.

Step $n+1$:

$a^{n+1}-b^{n+1} =a a^n -b b^n= $

$aa^n-ba^n+ba^n -bb^n=$

$(a-b)a^n +b(a^n -b^n)=$

$(a-b)a^n +bk(a-b)=$

$ (a-b)(a^n +bk).$

(Since by hypothesis: $a^n-b^n =k(a-b)$).

Peter Szilas
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Firstly we prove the statement for $n=0,1$ (or $n=1,2$ if you prefer, but the statement is true for $n\geq0$).

Case $n=0$ : $a^0-b^0 = 0 = 0*(a-b)$

Case $n=1$ : $a^1-b^1 = 1*(a-b)$

Now we prove the induction case. Suppose the statement is true for $k\leq n$, and let's prove it for $n+1$: $$ (a^{n+1}-b^{n+1})=(a^n-b^n)(a+b) - a^nb + b^na = (a^n-b^n)(a+b)- ab(a^{n-1} - b^{n-1}) $$

Since both $(a^n-b^n)$ and $(a^{n-1} - b^{n-1})$ are divisible by $(a-b)$ by induction, so is $(a^{n+1}-b^{n+1})$ since it is a sum of their multiples.

Kolja
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  • Note that we needed to put two cases in the basis case ($n=0$ AND $n=1$), since we use strong induction. In other words to prove the statement for $n+1$ we use both $n$ and $n-1$, so we need two $n$'s for the basis case. – Kolja Dec 14 '18 at 21:01
  • This is how I wanted to solve it, but it says specifically mathematical induction.(The next question asked for strong mathematical induction) – N.Luscomb Dec 14 '18 at 22:32
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    Strong induction is not needed here, simply write $$a^{n+1}-b^{n+1}=a^{n+1}-a^nb+a^nb-b^{n+1}=(a-b)a^n+b(a^n-b^n)$$ hence if $a-b$ divides $a^n-b^n$, then $a-b$ also divides $(a-b)a^n+b(a^n-b^n)$, qed. – Did Dec 18 '18 at 08:21
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Can you do one step of a long division?

$$ \frac{a^{k+1}-b^{k+1}}{a-b} = a^k + b\frac{(a^k-b^k)}{a-b} $$

with the remainder being a polynomial in $a$ and $b$ by induction hypothesis.

Matthias
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$$ a^n-b^n=(a-b)\sum_{m=0}^{n-1}a^m\cdot b^{n-1-m}$$

Samvel Safaryan
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