The binomial theorem is introduced, the existences of nth roots of real numbers is explored, the binomial coefficient is defined, and a theorem providing a formula for the sum of a geometric progression is included.
There are two special algebraic identities that hold in
$R$ (in fact in any field
$F$ whatsoever)
that we emphasize. They are bothproved by mathematical induction. The first is the
formula for the sum of a geometric progression.
Geometric progression
Let
$x$ be a real number, and let
$n$ be a natural number. Then,
 If
$x\ne 1,$ then
$$\sum _{j=0}^{n}{x}^{j}=\frac{1{x}^{n+1}}{1x}.$$
 If
$x=1,$ then
$$\sum _{j=0}^{n}{x}^{j}=n+1.$$
The second claim is clear, since there are
$n+1$ summands and each is equal to 1.
We prove the first claim by induction.
Thus, if
$n=1,$ then the assertion is true,
since
$$\sum _{j=0}^{1}{x}^{j}={x}^{0}+{x}^{1}=1+x=(1+x)\frac{1x}{1x}=\frac{1{x}^{2}}{1x}.$$
Now, supposing that the assertion is true for the natural number
$k,$ i.e., that
$$\sum _{j=0}^{k}{x}^{j}=\frac{1{x}^{k+1}}{1x},$$
let us show that the assertion holds for the natural number
$k+1.$ Thus
$$\begin{array}{ccc}\hfill \sum _{j=0}^{k+1}{x}^{j}& =& \sum _{j=0}^{k}{x}^{j}+{x}^{k+1}\hfill \\ & =& \frac{1{x}^{k+1}}{1x}+{x}^{k+1}\hfill \\ & =& \frac{1{x}^{k+1}+{x}^{k+1}{x}^{k+2}}{1x}\hfill \\ & =& \frac{1{x}^{k+1+1}}{1x},\hfill \end{array}$$
which completes the proof.
The second algebraic formula we wish to emphasize is the
Binomial Theorem. Before stating it, we must introduce some useful notation.

Let
$n$ be a natural number.
As earlier in this chapter, we define
$n!$ as follows:
$$n!=n\times (n1)\times (n2)\times ...\times 2\times 1.$$
For later notational convenience, we also define
$0!$ to be 1.
If
$k$ is any integer for which
$0\le k\le n,$ we
define the
binomial coefficient
$\left(\genfrac{}{}{0pt}{}{n}{k}\right)$ by
$$\left(\genfrac{}{}{0pt}{}{n}{k}\right)=\frac{n!}{k!(nk)!}=\frac{n\times (n1)\times (n2)\times ...\times (nk+1)}{k!}.$$
 Prove that
$\left(\genfrac{}{}{0pt}{}{n}{0}\right)=1,\phantom{\rule{0.166667em}{0ex}}\left(\genfrac{}{}{0pt}{}{n}{1}\right)=n\phantom{\rule{0.166667em}{0ex}}$ and
$\left(\genfrac{}{}{0pt}{}{n}{n}\right)=1.$
 Prove that
$$\left(\genfrac{}{}{0pt}{}{n}{k}\right)\le \frac{2{n}^{k}}{{2}^{k}}$$
for all natural numbers
$n$ and all integers
$0\le k\le n.$
 Prove that
$$\left(\genfrac{}{}{0pt}{}{n+1}{k}\right)=\left(\genfrac{}{}{0pt}{}{n}{k}\right)+\left(\genfrac{}{}{0pt}{}{n}{k1}\right)$$
for all natural numbers
$n$ and all integers
$1\le k\le n.$
If
$x,y\in R$ and
$n$ is a natural number, then
$${(x+y)}^{n}=\sum _{k=0}^{n}\left(\genfrac{}{}{0pt}{}{n}{k}\right){x}^{k}{y}^{nk}.$$
We shall prove this theorem by induction.
If
$n=1,$ then the assertion is true, for
${(x+y)}^{1}=x+y$ and
$$\sum _{k=0}^{1}\left(\genfrac{}{}{0pt}{}{1}{k}\right){x}^{k}{y}^{1k}=\left(\genfrac{}{}{0pt}{}{1}{0}\right){x}^{0}{y}^{1}+\left(\genfrac{}{}{0pt}{}{1}{1}\right){x}^{1}{y}^{0}=x+y.$$
Now, assume that the assertion holds for the natural number
$j;$ i.e.,
$${(x+y)}^{j}=\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{jk},$$
and let us prove that the assertion holds
for the natural number
$j+1.$ We will make
use of part (c) of
[link] . We have that
$$\begin{array}{ccc}\hfill {(x+y)}^{j+1}& =& (x+y){(x+y)}^{j}\hfill \\ & =& (x+y)\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{jk}\hfill \\ & =& x\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{jk}+y\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{jk}\hfill \\ & =& \sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k+1}{y}^{jk}+\sum _{k=0}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j+1k}\hfill \\ & =& \sum _{k=0}^{j1}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k+1}{y}^{jk}+\left(\genfrac{}{}{0pt}{}{j}{j}\right){x}^{j+1}{y}^{0}\hfill \\ & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}& +\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j+1k}+\left(\genfrac{}{}{0pt}{}{j}{0}\right){x}^{0}{y}^{j+1}\hfill \\ & =& {x}^{j+1}+\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j}{k1}\right){x}^{k}{y}^{j+1k}\hfill \\ & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}& +\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j}{k}\right){x}^{k}{y}^{j+1k}+{y}^{j+1}\hfill \\ & =& {x}^{j+1}+\sum _{k=1}^{j}(\left(\genfrac{}{}{0pt}{}{j}{k1}\right)+\left(\genfrac{}{}{0pt}{}{j}{k}\right)){x}^{k}{y}^{j+1k}+{y}^{j+1}\hfill \\ & =& {x}^{j+1}+\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j+1}{k}\right){x}^{k}{y}^{j+1k}+{y}^{j+1}\hfill \\ & =& \left(\genfrac{}{}{0pt}{}{j+1}{j+1}\right){x}^{j+1}{y}^{0}+\sum _{k=1}^{j}\left(\genfrac{}{}{0pt}{}{j+1}{k}\right){x}^{k}{y}^{j+1k}+\left(\genfrac{}{}{0pt}{}{j+1}{0}\right){x}^{0}{y}^{j+1}\hfill \\ & =& \sum _{k=0}^{j+1}\left(\genfrac{}{}{0pt}{}{j+1}{k}\right){x}^{k}{y}^{j+1k},\hfill \end{array}$$
which shows that the assertion of the theorem holds for the
natural number
$j+1.$ This completes the proof.
The next exercise is valid in any ordered field,
but, since we are mainly interested in the order field
$R,$ we state everything in terms of that field.
 If
$x$ and
$y$ are positive real numbers, and if
$n$ and
$k$ are natural numbers with
$k\le n,$ show that
${(x+y)}^{n}\ge \left(\genfrac{}{}{0pt}{}{n}{k}\right){x}^{k}{y}^{nk}.$
 For any positive real number
$x$ and natural number
$n,$ show that
${(1+x)}^{n}\ge 1+nx.$
 For any real number
$x>1$ and natural number
$n,$ prove that
${(1+x)}^{n}\ge 1+nx.$ HINT: Do not try to use the binomial theorem as in part (b); it won't work because the terms are not all positive;
prove this directly by induction.
There is one more important algebraic identity,
which again can be proved by induction.It is actually just a corollary of the geometric progression formula.
If
$x,y\in R$ and
$n$ is a natural number, then
$${x}^{n}{y}^{n}=(xy)(\sum _{j=0}^{n1}{x}^{j}{y}^{n1j}.$$
If
$n=1$ the theorem is clear.
Suppose it holds for a natural number
$k,$ and let us
prove the identity for the natural number
$k+1.$ We have
$$\begin{array}{ccc}\hfill {x}^{k+1}{y}^{k+1}& =& {x}^{k+1}{x}^{k}y+{x}^{k}y{y}^{k+1}\hfill \\ & =& (xy){x}^{k}+y({x}^{k}{y}^{k})\hfill \\ & =& (xy){x}^{k}+y(xy)\left(\sum _{j=0}^{k1}{x}^{j}{y}^{k1j}\right)\hfill \\ & =& (xy){x}^{k}+(xy)(\sum _{j=0}^{k1}{x}^{j}{y}^{kj}\hfill \\ & =& (xy)({x}^{k}{y}^{kk}+\sum _{j=0}^{k1}{x}^{j}{y}^{kj})\hfill \\ & =& (xy)\left(\sum _{j=0}^{k}{x}^{j}{y}^{kj}\right)\hfill \end{array}$$
,
which shows that the assertion holds for the natural number
$k+1.$ So, by induction, the theorem is proved.
Let
$x$ and
$y$ be real numbers.
 Let
$n$ be an odd natural number; i.e.,
$n=2k+1$ for some natural number
$k.$ Show that
$${x}^{n}+{y}^{n}=(x+y)(\sum _{j=0}^{n1}{(1)}^{j}{x}^{j}{y}^{n1j}.$$
HINT: Write
${x}^{n}+{y}^{n}={x}^{n}{(y)}^{n}.$
 Show that
${x}^{2}+{y}^{2}$ can not be
factored into a product of the form
$(ax+by)(cx+dy)$ for any choices of real numbers
$a,b,c,$ and
$d.$
Using the Binomial Theorem together with the preceding theorem, we may
now investigate the existence of
$n$ th roots of real numbers.
This next theorem is definitely not valid in any ordered field, forit again depends on the completeness property.
Let
$n$ be a natural number and let
$x$ be a positive real number. Then there exists
a unique positive real number
$y$ such that
${y}^{n}=x;$ i.e.,
$x$ has a unique positive
$n$ th root.
Note first that if
$0\le t<s,$ then
${t}^{n}<{s}^{n}.$ (To see this, argue by induction, and use part (e) of
[link] .)
Using this, we mimic the proof of
[link] .
Thus, let
$S$ be the set of all positive real numbers
$t$ for which
${t}^{n}\le x.$ Then
$S$ is nonempty and bounded above.
Indeed, if
$x\ge 1,$ then
$1\in S,$ while
if
$x<1,$ then
$x$ itself is in
$S.$ Therefore,
$S$ is nonempty.
Also, using part (b) of
[link] , we see that
$1+(x/n)$ is an upper bound for
$S.$ For, if
$t>1+x/n,$ then
$${t}^{n}>{(1+(x/n))}^{n}\ge 1+n(x/n)>x.$$
Now let
$y=supS,$ and let us show that
${y}^{n}=x.$ We rule out the other two possibilities.
First, if
${y}^{n}>x,$ let
$\u03f5$ be the positive number
${y}^{n}x,$ and define
${\u03f5}^{\text{'}}$ to be the positive number
$\u03f5/\left(n{y}^{n1}\right).$ Then, using
[link] , choose
$t\in S$ so that
$y{\u03f5}^{\text{'}}<t\le y.$ (
[link] is where the completeness of the ordered field
$R$ is crucial.)
We have
$$\begin{array}{ccc}\hfill \u03f5& =& {y}^{n}x\hfill \\ & =& {y}^{n}{t}^{n}+{t}^{n}x\hfill \\ & \le & {y}^{n}{t}^{n}\hfill \\ & =& (yt)\left(\sum _{j=0}^{n1}{y}^{j}{t}^{n1j}\right)\hfill \\ & \le & (yt)\left(\sum _{j=0}^{n1}{y}^{j}{y}^{n1j}\right)\hfill \\ & =& (yt)(\sum _{j=0}^{n1}{y}^{n1}\hfill \\ & <& {\u03f5}^{\text{'}}n{y}^{n1}\hfill \\ & =& \u03f5,\hfill \end{array}$$
and this is a contradiction. Therefore,
${y}^{n}$ is not greater than
$x.$
Now, if
${y}^{n}<x,$ let
$\u03f5$ be the positive number
$x{y}^{n},$ and choose a
$\delta >0$ such that
$\delta <1$ and
$\delta <\u03f5/{(y+1)}^{n}.$ Then, using the Binomial Theorem, we have that
$$\begin{array}{ccc}\hfill {(y+\delta )}^{n}& =& \sum _{k=0}^{n}\left(\genfrac{}{}{0pt}{}{n}{k}\right){y}^{k}{\delta}^{nk}\hfill \\ & =& {y}^{n}+\sum _{k=0}^{n1}\left(\genfrac{}{}{0pt}{}{n}{k}\right){y}^{k}{\delta}^{nk}\hfill \\ & =& {y}^{n}+\delta \sum _{k=0}^{n1}\left(\genfrac{}{}{0pt}{}{n}{k}\right){y}^{k}{\delta}^{n1k}\hfill \\ & <& {y}^{n}+\delta \sum _{k=0}^{n}\left(\genfrac{}{}{0pt}{}{n}{k}\right){y}^{k}{1}^{nk}\hfill \\ & =& {y}^{n}+\delta {(y+1)}^{n}\hfill \\ & =& x\u03f5+\delta {(y+1)}^{n}\hfill \\ & <& x\u03f5+\u03f5\hfill \\ & =& x,\hfill \end{array}$$
implying that
$y+\delta \in S.$ But this is a contradiction,
since
$y=supS.$ Therefore,
${y}^{n}$ is not less than
$x,$ and so
${y}^{n}=x.$
We have shown the existence of a positive
$n$ th root of
$x.$ To see the uniqueness, suppose
$y$ and
${y}^{\text{'}}$ are two positive
$n$ th roots of
$x.$ Then
$$\begin{array}{ccc}\hfill 0& =& {y}^{n}{{y}^{\text{'}}}^{n}\hfill \\ & =& (y{y}^{\text{'}})(\sum _{j=0}^{n1}{y}^{j}{{y}^{\text{'}}}^{nj1},\hfill \end{array}$$
which implies that either
$y{y}^{\text{'}}=0$ or
${\sum}_{j=0}^{n1}{y}^{j}{{y}^{\text{'}}}^{nj1}=0.$ Since this latter sum consists of positive terms, it cannot be 0,
whence
$y={y}^{\text{'}}.$ This shows that there is but one positive
$n$ th root of
$x,$ and the theorem is proved.
 Show that if
$n=2k$ is an even natural number,
then every positive real number has exactly two distinct
$n$ th roots.
 If
$n=2k+1$ is an odd natural number,
show that every real number has exactly one
$n$ th root.
 If
$n$ is a natural number greater than 1,
prove that there is no rational number whose
$n$ th power equals 2, i.e.,
the
$n$ th root of 2 is not a rational number.