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The Real and Complex Numbers: The Geometric Progression and the Binomial Theorem

Module by: Lawrence Baggett. E-mail the author

Summary: 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 RR (in fact in any field FF whatsoever) that we emphasize. They are both proved by mathematical induction. The first is the formula for the sum of a geometric progression.

Theorem 1: Geometric Progression

Let xx be a real number, and let nn be a natural number. Then,

  1. If x≠1,x≠1, then
    ∑j=0nxj=1-xn+11-x.∑j=0nxj=1-xn+11-x.
    (1)
  2. If x=1,x=1, then
    ∑j=0nxj=n+1.∑j=0nxj=n+1.
    (2)

Proof

The second claim is clear, since there are n+1n+1 summands and each is equal to 1.

We prove the first claim by induction. Thus, if n=1,n=1, then the assertion is true, since

∑ j = 0 1 x j = x 0 + x 1 = 1 + x = ( 1 + x ) 1 - x 1 - x = 1 - x 2 1 - x . ∑ j = 0 1 x j = x 0 + x 1 = 1 + x = ( 1 + x ) 1 - x 1 - x = 1 - x 2 1 - x .
(3)

Now, supposing that the assertion is true for the natural number k,k, i.e., that

∑ j = 0 k x j = 1 - x k + 1 1 - x , ∑ j = 0 k x j = 1 - x k + 1 1 - x ,
(4)

let us show that the assertion holds for the natural number k+1.k+1. Thus

∑ j = 0 k + 1 x j = ∑ j = 0 k x j + x k + 1 = 1 - x k + 1 1 - x + x k + 1 = 1 - x k + 1 + x k + 1 - x k + 2 1 - x = 1 - x k + 1 + 1 1 - x , ∑ j = 0 k + 1 x j = ∑ j = 0 k x j + x k + 1 = 1 - x k + 1 1 - x + x k + 1 = 1 - x k + 1 + x k + 1 - x k + 2 1 - x = 1 - x k + 1 + 1 1 - x ,
(5)

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.

Definition 1:

Let nn be a natural number. As earlier in this chapter, we define n!n! as follows:

n ! = n × ( n - 1 ) × ( n - 2 ) × ... × 2 × 1 . n ! = n × ( n - 1 ) × ( n - 2 ) × ... × 2 × 1 .
(6)

For later notational convenience, we also define 0!0! to be 1.

If kk is any integer for which 0≤k≤n,0≤k≤n, we define the binomial coefficientnknk by

n k = n ! k ! ( n - k ) ! = n × ( n - 1 ) × ( n - 2 ) × ... × ( n - k + 1 ) k ! . n k = n ! k ! ( n - k ) ! = n × ( n - 1 ) × ( n - 2 ) × ... × ( n - k + 1 ) k ! .
(7)

Exercise 1

  1. Prove that n0=1,n1=nn0=1,n1=n and nn=1.nn=1.
  2. Prove that
    nk≤2nk2knk≤2nk2k
    (8)
    for all natural numbers nn and all integers 0≤k≤n.0≤k≤n.
  3. Prove that
    n+1k=nk+nk-1n+1k=nk+nk-1
    (9)
    for all natural numbers nn and all integers 1≤k≤n.1≤k≤n.

Theorem 2

If x,y∈Rx,y∈R and nn is a natural number, then

( x + y ) n = ∑ k = 0 n n k x k y n - k . ( x + y ) n = ∑ k = 0 n n k x k y n - k .
(10)

Proof

We shall prove this theorem by induction. If n=1,n=1, then the assertion is true, for (x+y)1=x+y(x+y)1=x+y and

∑ k = 0 1 1 k x k y 1 - k = 1 0 x 0 y 1 + 1 1 x 1 y 0 = x + y . ∑ k = 0 1 1 k x k y 1 - k = 1 0 x 0 y 1 + 1 1 x 1 y 0 = x + y .
(11)

Now, assume that the assertion holds for the natural number j;j; i.e.,

( x + y ) j = ∑ k = 0 j j k x k y j - k , ( x + y ) j = ∑ k = 0 j j k x k y j - k ,
(12)

and let us prove that the assertion holds for the natural number j+1.j+1. We will make use of part (c) of Exercise 1. We have that

( x + y ) j + 1 = ( x + y ) ( x + y ) j = ( x + y ) ∑ k = 0 j j k x k y j - k = x ∑ k = 0 j j k x k y j - k + y ∑ k = 0 j j k x k y j - k = ∑ k = 0 j j k x k + 1 y j - k + ∑ k = 0 j j k x k y j + 1 - k = ∑ k = 0 j - 1 j k x k + 1 y j - k + j j x j + 1 y 0 + ∑ k = 1 j j k x k y j + 1 - k + j 0 x 0 y j + 1 = x j + 1 + ∑ k = 1 j j k - 1 x k y j + 1 - k + ∑ k = 1 j j k x k y j + 1 - k + y j + 1 = x j + 1 + ∑ k = 1 j ( j k - 1 + j k ) x k y j + 1 - k + y j + 1 = x j + 1 + ∑ k = 1 j j + 1 k x k y j + 1 - k + y j + 1 = j + 1 j + 1 x j + 1 y 0 + ∑ k = 1 j j + 1 k x k y j + 1 - k + j + 1 0 x 0 y j + 1 = ∑ k = 0 j + 1 j + 1 k x k y j + 1 - k , ( x + y ) j + 1 = ( x + y ) ( x + y ) j = ( x + y ) ∑ k = 0 j j k x k y j - k = x ∑ k = 0 j j k x k y j - k + y ∑ k = 0 j j k x k y j - k = ∑ k = 0 j j k x k + 1 y j - k + ∑ k = 0 j j k x k y j + 1 - k = ∑ k = 0 j - 1 j k x k + 1 y j - k + j j x j + 1 y 0 + ∑ k = 1 j j k x k y j + 1 - k + j 0 x 0 y j + 1 = x j + 1 + ∑ k = 1 j j k - 1 x k y j + 1 - k + ∑ k = 1 j j k x k y j + 1 - k + y j + 1 = x j + 1 + ∑ k = 1 j ( j k - 1 + j k ) x k y j + 1 - k + y j + 1 = x j + 1 + ∑ k = 1 j j + 1 k x k y j + 1 - k + y j + 1 = j + 1 j + 1 x j + 1 y 0 + ∑ k = 1 j j + 1 k x k y j + 1 - k + j + 1 0 x 0 y j + 1 = ∑ k = 0 j + 1 j + 1 k x k y j + 1 - k ,
(13)

which shows that the assertion of the theorem holds for the natural number j+1.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,R, we state everything in terms of that field.

Exercise 2

  1. If xx and yy are positive real numbers, and if nn and kk are natural numbers with k≤n,k≤n, show that (x+y)n≥nkxkyn-k.(x+y)n≥nkxkyn-k.
  2. For any positive real number xx and natural number n,n, show that (1+x)n≥1+nx.(1+x)n≥1+nx.
  3. For any real number x>-1x>-1 and natural number n,n, prove that (1+x)n≥1+nx.(1+x)n≥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.

Theorem 3

If x,y∈Rx,y∈R and nn is a natural number, then

x n - y n = ( x - y ) ( ∑ j = 0 n - 1 x j y n - 1 - j . x n - y n = ( x - y ) ( ∑ j = 0 n - 1 x j y n - 1 - j .
(14)

Proof

If n=1n=1 the theorem is clear. Suppose it holds for a natural number k,k, and let us prove the identity for the natural number k+1.k+1. We have

x k + 1 - y k + 1 = x k + 1 - x k y + x k y - y k + 1 = ( x - y ) x k + y ( x k - y k ) = ( x - y ) x k + y ( x - y ) ( ∑ j = 0 k - 1 x j y k - 1 - j ) = ( x - y ) x k + ( x - y ) ( ∑ j = 0 k - 1 x j y k - j = ( x - y ) ( x k y k - k + ∑ j = 0 k - 1 x j y k - j ) = ( x - y ) ( ∑ j = 0 k x j y k - j ) x k + 1 - y k + 1 = x k + 1 - x k y + x k y - y k + 1 = ( x - y ) x k + y ( x k - y k ) = ( x - y ) x k + y ( x - y ) ( ∑ j = 0 k - 1 x j y k - 1 - j ) = ( x - y ) x k + ( x - y ) ( ∑ j = 0 k - 1 x j y k - j = ( x - y ) ( x k y k - k + ∑ j = 0 k - 1 x j y k - j ) = ( x - y ) ( ∑ j = 0 k x j y k - j )
(15)

, which shows that the assertion holds for the natural number k+1.k+1. So, by induction, the theorem is proved.

Exercise 3

Let xx and yy be real numbers.

  1. Let nn be an odd natural number; i.e., n=2k+1n=2k+1 for some natural number k.k. Show that
    xn+yn=(x+y)(∑j=0n-1(-1)jxjyn-1-j.xn+yn=(x+y)(∑j=0n-1(-1)jxjyn-1-j.
    (16)
    HINT: Write xn+yn=xn-(-y)n.xn+yn=xn-(-y)n.
  2. Show that x2+y2x2+y2 can not be factored into a product of the form (ax+by)(cx+dy)(ax+by)(cx+dy) for any choices of real numbers a,b,c,a,b,c, and d.d.

Using the Binomial Theorem together with the preceding theorem, we may now investigate the existence of nnth roots of real numbers. This next theorem is definitely not valid in any ordered field, for it again depends on the completeness property.

Theorem 4

Let nn be a natural number and let xx be a positive real number. Then there exists a unique positive real number yy such that yn=x;yn=x; i.e., xx has a unique positive nnth root.

Proof

Note first that if 0≤t<s,0≤t<s, then tn<sn.tn<sn. (To see this, argue by induction, and use part (e) of (Reference).) Using this, we mimic the proof of (Reference). Thus, let SS be the set of all positive real numbers tt for which tn≤x.tn≤x. Then SS is nonempty and bounded above. Indeed, if x≥1,x≥1, then 1∈S,1∈S, while if x<1,x<1, then xx itself is in S.S. Therefore, SS is nonempty. Also, using part (b) of Exercise 2, we see that 1+(x/n)1+(x/n) is an upper bound for S.S. For, if t>1+x/n,t>1+x/n, then

t n > ( 1 + ( x / n ) ) n ≥ 1 + n ( x / n ) > x . t n > ( 1 + ( x / n ) ) n ≥ 1 + n ( x / n ) > x .
(17)

Now let y=supS,y=supS, and let us show that yn=x.yn=x. We rule out the other two possibilities. First, if yn>x,yn>x, let ϵϵ be the positive number yn-x,yn-x, and define ϵ'ϵ' to be the positive number ϵ/(nyn-1).ϵ/(nyn-1). Then, using (Reference), choose t∈St∈S so that y-ϵ'<t≤y.y-ϵ'<t≤y. ((Reference) is where the completeness of the ordered field RR is crucial.) We have

ϵ = y n - x = y n - t n + t n - x ≤ y n - t n = ( y - t ) ( ∑ j = 0 n - 1 y j t n - 1 - j ) ≤ ( y - t ) ( ∑ j = 0 n - 1 y j y n - 1 - j ) = ( y - t ) ( ∑ j = 0 n - 1 y n - 1 < ϵ ' n y n - 1 = ϵ , ϵ = y n - x = y n - t n + t n - x ≤ y n - t n = ( y - t ) ( ∑ j = 0 n - 1 y j t n - 1 - j ) ≤ ( y - t ) ( ∑ j = 0 n - 1 y j y n - 1 - j ) = ( y - t ) ( ∑ j = 0 n - 1 y n - 1 < ϵ ' n y n - 1 = ϵ ,
(18)

and this is a contradiction. Therefore, ynyn is not greater than x.x.

Now, if yn<x,yn<x, let ϵϵ be the positive number x-yn,x-yn, and choose a δ>0δ>0 such that δ<1δ<1 and δ<ϵ/(y+1)n.δ<ϵ/(y+1)n. Then, using the Binomial Theorem, we have that

( y + δ ) n = ∑ k = 0 n n k y k δ n - k = y n + ∑ k = 0 n - 1 n k y k δ n - k = y n + δ ∑ k = 0 n - 1 n k y k δ n - 1 - k < y n + δ ∑ k = 0 n n k y k 1 n - k = y n + δ ( y + 1 ) n = x - ϵ + δ ( y + 1 ) n < x - ϵ + ϵ = x , ( y + δ ) n = ∑ k = 0 n n k y k δ n - k = y n + ∑ k = 0 n - 1 n k y k δ n - k = y n + δ ∑ k = 0 n - 1 n k y k δ n - 1 - k < y n + δ ∑ k = 0 n n k y k 1 n - k = y n + δ ( y + 1 ) n = x - ϵ + δ ( y + 1 ) n < x - ϵ + ϵ = x ,
(19)

implying that y+δ∈S.y+δ∈S. But this is a contradiction, since y=supS.y=supS. Therefore, ynyn is not less than x,x, and so yn=x.yn=x.

We have shown the existence of a positive nnth root of x.x. To see the uniqueness, suppose yy and y'y' are two positive nnth roots of x.x. Then

0 = y n - y ' n = ( y - y ' ) ( ∑ j = 0 n - 1 y j y ' n - j - 1 , 0 = y n - y ' n = ( y - y ' ) ( ∑ j = 0 n - 1 y j y ' n - j - 1 ,
(20)

which implies that either y-y'=0y-y'=0 or ∑j=0n-1yjy'n-j-1=0.∑j=0n-1yjy'n-j-1=0. Since this latter sum consists of positive terms, it cannot be 0, whence y=y'.y=y'. This shows that there is but one positive nnth root of x,x, and the theorem is proved.

Exercise 4

  1. Show that if n=2kn=2k is an even natural number, then every positive real number has exactly two distinct nnth roots.
  2. If n=2k+1n=2k+1 is an odd natural number, show that every real number has exactly one nnth root.
  3. If nn is a natural number greater than 1, prove that there is no rational number whose nnth power equals 2, i.e., the nnth root of 2 is not a rational number.

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