Let $a,b\in R$ a ring with unity where $a,b$ commutes. Show that $a^{n+1}-b^{n+1}=(a-b)\sum_{k=0}^n a^kb^{n-k}$

Question

Let $a,b\in R$ a ring with unity where $a,b$ commutes and $n\in \Bbb N_0$.

1) Show that $a^{n+1}-b^{n+1}=(a-b)\sum_{k=0}^n a^kb^{n-k}$

2) Show that $a^{n+1}-1=(a-1)\sum_{k=0}^n a^k$

Check my proof please.

1) Using induction we have that the base case for $n=0$ holds, i.e. $a-b=(a-b)(a^0 b^0)$ where I assume that $x^0=1$ for any $x\in R$ (but I dont have real reasons to believe that).

The induction step

$$(a-b)\sum_{k=0}^{n+1}a^k b^{n+1-k}=(a-b)\left(b^{n+1}+a\sum_{k=1}^{n+1}a^{k-1}b^{n+1-k}\right)=\\=(a-b)\left(b^{n+1}+a\sum_{k=0}^n a^kb^{n-k}\right)=(a-b)b^{n+1}+(a^{n+1}-b^{n+1})a=a^{n+2}-b^{n+2}$$

If $a$ and $b$ are divisors of zero we can check that it still holds, so I assume my proof is concluded at this point.

2) It is a particular case of 1) for $b=1$.

My main problem is how to justify that $x^0=1$ for a ring with unity, some advice?

I dont see the circle that you said @AlexM. I assume the previous step for $n$, just change the side of LHS and RHS in the formula in the question 1). — Masacroso, Sep 20 '16 at 22:13
You assume $a^{n+1}-b^{n+1}=(a-b)\sum_{k=0}^n a^k b^{n-k}$. Then why do you begin the main step in your proof by writing that $a^{n+2}-b^{n+2}=(a-b)\sum_{k=0}^{n+1}a^k b^{n+1-k}$? This is precisely what you have to prove! — Alex M., Sep 20 '16 at 22:15
@AlexM. Induction is: prove the base step (I did for $n=0$). Now I assume the case for $n$ and prove the case for $n+1$. — Masacroso, Sep 20 '16 at 22:16
Exactly! Now it's fine! :) Also notice that it is completely irrelevant whether $a$, $b$ are divisors of zero, you haven't used this anywhere. — Alex M., Sep 20 '16 at 22:22

score 1 · Accepted Answer · answered Sep 20 '16 at 22:37

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You're doing the proof backwards! You should start from the other end: \begin{align} a^{n+2}-b^{n+2} &=ab^{n+1}-b^{n+2}+a^{n+2}-b^{n+1}a &&\text{(commutativity)}\\ &=(a-b)b^{n+1}+(a^{n+1}-b^{n+1})a &&\text{(collect terms)}\\ &=(a-b)b^{n+1}+(a-b)\biggl(\sum_{k=0}^n a^kb^{n-k}\biggr)a &&\text{(induction hypothesis)}\\ &=(a-b)\biggl(b^{n+1}+\biggl(\sum_{k=0}^n a^kb^{n-k}\biggr)a\biggr) &&\text{(collect terms)}\\ &=(a-b)\biggl(b^{n+1}+a\biggl(\sum_{k=0}^n a^kb^{n-k}\biggr)\biggr) &&\text{(commutativity)}\\ &=(a-b)\biggl(b^{n+1}+\sum_{k=0}^n a^{k+1}b^{n-k}\biggr) &&\text{(distributivity)}\\ &=(a-b)\sum_{k=0}^{n+1} a^{k}b^{n+1-k} &&\text{(change indices)} \end{align}

About $x^0=1$, don't worry. We're doing abstract algebra and it's so by definition.

You should actually prove that $ab^m=b^ma$, which is also easy and also the other passage where commutativity is used is not obvious, but easy as well.

answered Sep 20 '16 at 22:37

egreg

238,574

Yes, I ignore the proof for the commutativity of $ab^m$. I did "backwards" but, well, you think is a problem a proof from RHS to LHS in an equality? – Masacroso Sep 20 '16 at 22:40
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@Masacroso In a well written proof you have to justify the steps; besides, you had another term in the proof by induction, which I based my “backwards” statement on. Doing the way it's written now in your question is good (but the justification are missing). There's no problem with zero divisors, though, so no separate proof is needed. If you don't trust in $x^0=1$ (but you should), do separately the cases $a=0$ or $b=0$. – egreg Sep 20 '16 at 22:44
But, how I can do the cases for $a=0$ and $b=0$? I mean, in any case I must justify things like $0_R^0=1$. In this context it is supposed that the $0_R$ dont need to represent a natural number or so because is abstract algebra, right? – Masacroso Sep 20 '16 at 22:50
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@Masacroso The quarrel about $0^0$ will never end. This is a very good example showing that defining it to be $1$ has no problem. Really: it is an empty product, so it must be the neutral element of the multiplication. – egreg Sep 21 '16 at 08:04
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@Masacroso: You may convince yourself here. – Vincenzo Oliva Sep 21 '16 at 11:16
Thank you @VincenzoOliva but I get a better reason related directly to abstract algebra. If we define $c^ac^b=c^{a+b}$ then for $c^1c^0=c^{1+0}$, and then we get that $c^0=1$ for any $c$. – Masacroso Sep 21 '16 at 12:04
@Masacroso: Well yeah, but do note $0^{1+0}=0,$ so if you defined $0^0=0$ you'd still get $0^1\cdot0^0=0^{1+0}.$ While your argument is good for $c\ne0$, it's not really convincing for $c=0$ (since $0^c=0$ for all $c>0$). – Vincenzo Oliva Sep 21 '16 at 13:02

Let $a,b\in R$ a ring with unity where $a,b$ commutes. Show that $a^{n+1}-b^{n+1}=(a-b)\sum_{k=0}^n a^kb^{n-k}$

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