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Math 0413 Supplement

Binomial Expansions and Roots

October 8, 2010

1 Binomial Expansions

You are familiar with the expansions

(a+b)1 = a+b

(a+b)2 = a2 + 2ab+b2

(a+b)3 = a3 + 3a2b+ 3ab2 +b3

(a+b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 +b4.

These formulas are called binomial expansions. They are obtained by repeatedapplication of the Distributive Law. The Binomial Theorem gives a recipe forthe general binomial expansion of (a + b)n, wheren can be any natural number.

1.1 Binomial CoefficientsThe numbers

n

k

=

n!

k!(n k)!

are called binomial coefficients. Heren and k are non-negative integers withk n. The symbol

nk

is read binomial n k or n choosek .

Theorem 1.1 (Binomial Theorem). Letn N. Then for any real numbersaandb we have

(a+b)n =n

k=0

n

k

ankbk.

The proof of the Binomial Theorem requires

Lemma 1.2. For anyn, k N withk n we have

n 1

k

+

n 1

k 1

=

n

k

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11 1

1 2 11 3 3 1

1 4 6 4 1...

Figure 1: Pascals Triangle

Proof.

n 1

k

+

n 1

k 1

=

(n 1)!

k!(n 1 k)!+

(n 1)!

(k 1)!(n k)!

= (n 1)!(n k) + (n 1)!kk!(n k)!

= n!

k!(n k)!

=

n

k

.

Lemma 1.1 is the basis for Pascals Triangle (Figure 1), which gives a quickway to calculate binomial coefficients for small values of n. The rows in thetriangle are numbered from the top, starting with 0. Row n give the coefficientsin the binomial expansion of (a+ b)n. Notice that any entry in the triangle

which is not on the edge can be calculated as the sum of the entries in thepreceding row to the immediate left and right. This is precisely the content ofLemma 1.1

Proof of the Binomial Theorem. The proof is by induction onn. The casen= 1is trivial, since the assertion in this case is simply a+b = a +b. Assume thatn >1 and that the theorem holds with n replaced byn 1. Then

(a+b)n = (a+b)(a+b)n1

= (a+b)n1k=0

n 1

k

an1kbk

= a

n1k=0

n 1k

an1k

bk

+b

n1k=0

n 1k

an1k

bk

=n1k=0

n 1

k

ankbk +

n1k=0

n 1

k

an1kbk+1

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=n1

k=0

n 1

k

ankbk +

n

k=1

n 1

k 1

ankbk

= an +

n1k=1

n 1

k

+

n 1

k 1

ankbk +bn

= an +n1k=1

n

k

ankbk +bn

=n

k=0

n

k

ankbk

Corollary 1.3.n

k=0

nk

= 2n.

Proof. Takea = b = 1 in the Binomial Theorem.

2 Existence and uniqueness of roots

In this section we will prove existence and uniqueness of positive nth roots ofpositive real numbers. Uniqueness only requires the field and order propertiesof the real numbers. Existence requires the Completeness Axiom.

2.1 Uniqueness

Lemma 2.1. Leta and b be non-negative real numbers and let n N. Thena < b if and only ifan < bn.

Proof. We first prove the if part. Assume 0 a < b. We will show by induc-tion that an < bn for every n N. In the case n = 1, theres nothing to prove.For n > 1, we suppose that the inequality an1 < bn1 holds. Multiplyingthrough by the non-negative numbera gives

an abn1.

Also, multiplying the inequality a < bby the positive number bn1 gives

abn1 < bn.

Transitivity givesan < bn.For the converse, we prove the contrapositive assertion: Ifa b then an

bn. We consider two cases. Ifa = b, then an = bn, and so an bn holds. Ifa > b, then by the part already proved (reversing the roles ofa and b),an > bn,and so again an bn holds. This completes the proof.

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Leta be any real number, and let n N. A real number b will be called annth root ofa ifbn =a.

Lemma 2.2. For anyn N, a real number can have at most one non-negativenth root.

Proof. Suppose a and b are non-negative nth roots for c. Then an and bn areboth equal to c, so

an =bn.

By Lemma 2.1, this is inconsistent with botha < band b < a, so by trichotomywe must have a = b.

2.2 Existence of roots

Lemma 2.3. For anya, b [0, 1] andn N we have

(a+b)n an + 2nb.

Proof. By the Binomial Theorem,

(a+b)n =n

k=0

n

k

ankbk

= an +bn

k=1

ankbk1.

Sincea and b are both less than or equal to 1, we obtain

(a+b)

n

an

+b

n

k=1

nk

an +bn

k=0

n

k

.

But the sum on the right equals 2n, by Corollary 1.3, so the proof is complete.

Theorem 2.4. Leta be any non-negative real number and letn N. There isa unique non-negative real numberb such thatbn =a.

Proof. The uniqueness is established by Lemma 2.2, so the issue here is exis-tence.

We first consider the case 0 a 1. Let

E= {x R :x 0 and xn a}.

Then E=, since 0 E. Also, sincea 1, the number 1 is an upper boundfor E. By completeness, Ehas a least upper bound b which must be less thanor equal to 1. We will show that bn =a.

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Let be any real number satisfying 0 < 1. Then b < b, so b is not an upper bound for E. Therefore, there is anx E with b < x, or

equivalently,b < x+.

By Lemmas 2.1 and 2.3, we have

bn 0 is arbitrary, (1) holds for every (0, 1), and therefore

bn a

2n 0

and sobn a. (2)

It remains to establish the reverse inequality.Again, let be a real number with 0 < 1. Sinceb + > b, and b is an

upper bound for E, it follows thatb +E, and so

a < (b+)n.

By Lemma 2.3, we havea < bn + 2n,

soa bn

2n <

for every (0, 1). It follows that (a bn)/2n 0, and hence

a bn. (3)

Combining (2) and (3) givesbn =a

as required. This establishes the existence when 0 a 1.It remains to consider the case a > 1. In this case, let a = 1/a. Then

0 < a < 1, so by the case already established, there is a positive real numberb such that

bn =a = 1

a.

Lettingb = 1/b gives

bn

=a.The proof is now complete.

For any non-negative real numberaandn N, we denote bya1/n the uniquenon-negative real number b such that bn = a. The numbera1/n is called thenth root ofa.

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2.3 Rational powers

For any rational number r there are integers m and n, with n = 0, such thatr = m/n. Of course, the integers m and n are not uniquely determined. Wewill call m/n the reduced representationof r ifn > 0 and the integers m andn have no common factors. The reduced representation of a rational number isunique.

Leta be a positive real number and let r be a rational number with reducedrepresentationr = m/n. We define

ar = (a1/n)m.

2.4 Irrational powers

Making sense ofab whenb is irrational can be tricky. One approach is to define

ab

by way of the natural exponential and logarithm functions, but its easy endup chasing your own tail here, since you must first give rigorous definitions ofthe log and exponential functions. One way out is todefinethe natural log andexponential functions to be solutions to certain differential equations. Existenceof solutions to these differential equations can be deduced from the FundamentalTheorem of Calculus.

Another approach, which does not rely on calculus, is to base the definitiondirectly on the Completeness Axiom. Whena 1, we could define

ab = sup{ar :r Q and r < b}.

When 0< a < 1, you can then define

ab =1

ab

.

This works, but there are some technical issues that need to be addressed. Youhave to check that this agrees with the previous definition when b is rational,and that the familiar rules for working with exponents hold. We will not pursuethis here.

3 Exercises

1. Prove that ifm is a natural number which is not a perfect square thenm1/2 is irrational.

2. Ifm and n are natural numbers, show that m1/n

is either irrational or aninteger.

3. Prove that for any positive real numbera and any integers m and n withn= 0 we have

am/n = (a1/n)m = (am)1/n.

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4. Prove that for any positivea and any rational r ands we have

(ar)s =ars.

5. Prove that for any positivea and any rational r ands we have

ar+s =aras.

6. Prove that for any positivea and b and any rational r we have

(ab)r =arbr.

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