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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,

  1. If x 1 , then
    j = 0 n x j = 1 - x n + 1 1 - x .
  2. If x = 1 , then
    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

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

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

j = 0 k x j = 1 - x k + 1 1 - x ,

let us show that the assertion holds for the natural number 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 ,

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 × ( n - 1 ) × ( n - 2 ) × ... × 2 × 1 .

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

If k is any integer for which 0 k n , we define the binomial coefficient n k by

n k = n ! k ! ( n - k ) ! = n × ( n - 1 ) × ( n - 2 ) × ... × ( n - k + 1 ) k ! .
  1. Prove that n 0 = 1 , n 1 = n and n n = 1 .
  2. Prove that
    n k 2 n k 2 k
    for all natural numbers n and all integers 0 k n .
  3. Prove that
    n + 1 k = n k + n k - 1
    for all natural numbers n and all integers 1 k n .

If x , y R and n is a natural number, then

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

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

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

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

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

( 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 ,

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.

  1. If x and y are positive real numbers, and if n and k are natural numbers with k n , show that ( x + y ) n n k x k y n - k .
  2. For any positive real number x and natural number n , show that ( 1 + x ) n 1 + n x .
  3. For any real number x > - 1 and natural number n , prove that ( 1 + x ) n 1 + n x . 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 R and n is a natural number, then

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

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

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 )

, 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.

  1. Let n be an odd natural number; i.e., n = 2 k + 1 for some natural number k . Show that
    x n + y n = ( x + y ) ( j = 0 n - 1 ( - 1 ) j x j y n - 1 - j .
    HINT: Write x n + y n = x n - ( - y ) n .
  2. Show that x 2 + y 2 can not be factored into a product of the form ( a x + b y ) ( c x + d y ) 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 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 x . Then S is nonempty and bounded above. Indeed, if x 1 , then 1 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 1 + n ( x / n ) > x .

Now let y = sup S , and let us show that y n = x . We rule out the other two possibilities. First, if y n > x , let ϵ be the positive number y n - x , and define ϵ ' to be the positive number ϵ / ( n y n - 1 ) . Then, using [link] , choose t S so that y - ϵ ' < t y . ( [link] is where the completeness of the ordered field R 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 ,

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

Now, if y n < x , let ϵ be the positive number x - y n , and choose a δ > 0 such that δ < 1 and 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 ,

implying that y + δ S . But this is a contradiction, since y = sup S . 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 ' are two positive n th roots of x . Then

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

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

  1. Show that if n = 2 k is an even natural number, then every positive real number has exactly two distinct n th roots.
  2. If n = 2 k + 1 is an odd natural number, show that every real number has exactly one n th root.
  3. 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.

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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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