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# inductive proof of binomial theorem

We prove the theorem for a ring. We do not assume a unit for the ring. We do not need commutativity of the ring, but only that $a$ and $b$ commute.

When $n=1$, the result is clear.

For the inductive step, assume it holds for $m$. Then for $n=m+1$,

$\displaystyle(a+b)^{{m+1}}$ | $\displaystyle=$ | $\displaystyle(a+b)(a+b)^{m}$ | ||

$\displaystyle=$ | $\displaystyle(a+b)(a^{m}+b^{m}+\sum_{{k=1}}^{{m-1}}\binom{m}{k}a^{{m-k}}b^{k})% \text{ by the inductive hypothesis}$ | |||

$\displaystyle=$ | $\displaystyle a^{{m+1}}+b^{{m+1}}+ab^{m}+ba^{m}+\sum_{{k=1}}^{{m-1}}\binom{m}{% k}a^{{m-k+1}}b^{k}+\sum_{{k=1}}^{{m-1}}\binom{m}{k}a^{{m-k}}b^{{k+1}}$ | |||

$\displaystyle=$ | $\displaystyle a^{{m+1}}+b^{{m+1}}+\sum_{{k=1}}^{m}\binom{m}{k}a^{{m-k+1}}b^{k}% +\sum_{{k=0}}^{{m-1}}\binom{m}{k}a^{{m-k}}b^{{k+1}}\text{ by combining terms}$ | |||

$\displaystyle=$ | $\displaystyle a^{{m+1}}+b^{{m+1}}+\sum_{{k=1}}^{m}\binom{m}{k}a^{{m-k+1}}b^{k}% +\sum_{{j=1}}^{m}\binom{m}{j-1}a^{{m+1-j}}b^{j}\text{ let j=k+1 in second sum}$ | |||

$\displaystyle=$ | $\displaystyle a^{{m+1}}+b^{{m+1}}+\sum_{{k=1}}^{m}\left[\binom{m}{k}+\binom{m}% {k-1}\right]a^{{m+1-k}}b^{k}\text{ by combining the sums}$ | |||

$\displaystyle=$ | $\displaystyle a^{{m+1}}+b^{{m+1}}+\sum_{{k=1}}^{m}\binom{m+1}{k}a^{{m+1-k}}b^{% k}\text{ from Pascal's rule}$ |

as desired.

Keywords:

number theory

Major Section:

Reference

Type of Math Object:

Proof

Parent:

## Mathematics Subject Classification

05A10*no label found*

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## Recent Activity

Jul 5

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

## Comments

## Sups and infs

I wanted some hints on this problem. Let f be a bounded function on interval I, prove that

sup({-f(x): x in I}) = -inf({f(x): x in I}).

## Re: Sups and infs

We had a problem similar to this in our honors calculus class last year; The way I found best was to start by contradiction, that they are not equal.

That is, if f is a function f: I -> A, then there is some element a in A such that a < -inf({f(x) : x in I}) with a an upper bound for {-f(x): x in I}.

What then, can you say about -a and its relation to {f(x) : x in I}

Well, at least something like that.