The Rational Root Theorem and Some Consequences

LAUNCH

State whether the following are rational or irrational and justify your answer.

1 √

How sure do you feel about your answers to the launch questions? After reading this section, you will want to revisit your responses to see if you were indeed correct, or if in fact, anyone really knows the answers.

As we alluded to earlier, the study of polynomial equations allows us to investigate some very interesting mathematical questions. For example, we have seen in chapter 1 that√

2 is irrational.

A similar proof shows that√

3 is irrational and in fact,√

N is irrational when N is a positive integer which is not a perfect square. What about numbers like√

2 +√

Are these also irrational? You might think, “Sure. √ 2 +√

3 is irrational because the sum of two irrational numbers is irrational.” Well, that is false, as the following example shows: 1 +√

2 is irrational and so is 1−√

2. Yet their sum is 2 which is rational. In fact, the third number we presented,

3− 2√ 2 +

3 + 2√

2, looks pretty irrational to most people, but in fact, it is rational!

Certain numbers “look” like they should be irrational, like 2^π andπ^π but no one knows if these are rational or not. Today’s best mathematicians, with all the computer technology available, have not determined the nature of these numbers. And what about numbers like sin 1^◦ or log23? Are these irrational?

The purpose of this section is to treat a large number of these expressions from a single and rather elegant point of view which uses a theorem about the roots of polynomials: the Rational Root Theorem. This is taught in many secondary school precalculus courses. Our goal in this section is to present the theoretical background for some sharp mathematical observations that have been used to solve some very difficult problems in mathematics and give us a powerful arsenal of useful information as well.

Theorem 3.12 (Rational Root Theorem) If a polynomial with integer coefficients p(x) = anxⁿ+ an−1xⁿ⁻¹+ an−2xⁿ⁻²+.... + a0

has a rational root a

b where a

b is in lowest terms, then the numerator a must divide the constant term a0and the denominator b must divide the lead coefficient an.

Proof. We give the proof for the specific polynomial p(x) = 3x³+ 2x + 5 since it will make it easier for you to follow. Afterwards, we give the general proof. Now, saying that a/b is a root of p(x) means that p(a/b) = 0. Substituting x = a/b into p(x) = 0 yields

3(a/b)³+ 2(a/b) + 5 = 0 or just

3a³ b³ + 2a

b+ 5 = 0.

Multiplying both sides by b³we get

3a³+ 2ab²+ 5b³= 0. (3.4)

Now, if we subtract 5b³from both sides of the equation we get 3a³+ 2ab²=−5b³

or just

a(3a²+ 2b²) =−5b³. (3.5)

Thus a is a divisor of the left side of equation (3.5). So it must divide the right side of equation (3.5) also. That is, it must divide −5b³. Now, a/b is in lowest terms. So, a and b have no common factors. Therefore, since a divides −5b³, it must be that a divides 5, since it can’t divide b³ by Theorem 2.22 of Chapter 2. In summary, a divides the constant term, 5, of p(x).

Now we use a similar method to make the conclusion we want about b, namely, that it divides 3. We subtract from both sides all the terms of equation (3.4) that have b in them. This yields

3a³=−2ab²− 5b³. (3.6)

This shows that b is a divisor of the right side of equation (3.6) and hence must divide the left side of equation (3.6, ) which is 3a³. Since b has no common factor with a, b must divide 3, which is the lead coefficient of p(x).

In summary, we have shown that any rational roots a

b of this polynomial p(x) = 3x³+ 2x + 5 have the property that a divides 5 and b divides 3. (Thus, the only possible rational roots of this equation are±5 Multiplying both sides of equation (3.7) by bⁿand simplifying, we get

anaⁿ+ an−1aⁿ⁻¹b +. . . + a0bⁿ= 0.

We subtract the last term a0bⁿfrom both sides and we get

anaⁿ+ an−1aⁿ⁻¹+. . . + a1a =−a0bⁿ (3.8)

and, since a can be factored out of the left side of equation (3.8), the left side is divisible by a. Thus the right side,−a0bⁿ is also divisible by a. Since a and b have no common factor, the only way a can divide the right side is if a divides a0, the constant term. You can finish the proof mimicking what we did earlier to show that b divides an. 

Let us now illustrate how this theorem can help us find the rational roots of a polynomial equation.

Example 3.13 What are the possible rational roots of p(x) = 2x³+ 3x− 5 and which, if any, are actual roots of p(x)?

Solution. Any rational root a/b of p(x) has the property that a must divide the constant term 5 and that b must divide the lead coefficient, 2. Thus, a = ±1, ±5, and b = ±1, ±2. It follows that possible rational roots is not zero. So the only rational root of p(x) is x = 1.

Now we give an example which is more in line with what we have set out to do, which is, to discover whether certain numbers are rational or irrational.

Example 3.14 (a) Show that the only rational roots the equation p(x) = x²− 2 = 0 can have are x =±1 and x = ±2. (b) Show that none of these are roots. (c) Show that√

2 is a root of this equation.

(d) Give another proof using (a), (b) and (c) that√

2 is irrational.

Solution. (a) By the rational root theorem, if a/b is any root of the equation x²− 2 = 0, then a must divide 2 and so must be either±2 or ±1. Also b must divide 1, meaning b must be ±1. Thus, a/b must be either ±2 or ±1.

(b) Substituting each of these values into p(x), we see that p(x) is not zero for any of these values. Thus p(x) = 0 has no rational roots.

(c). It is clear that p(√

2) = 0. (d) Since there are no rational roots, √

2 , which is a root of p(x), cannot be rational.

Can you see how powerful this technique is in proving that a number is irrational? Let us try another example and show that√³

7 is irrational. Since√³

7 satisfies the polynomial p(x) = x³− 7 = 0, and the only rational roots possible for this equation are±7 and ±1, none of which work,√³

7 is irrational. Similarly, we can show√⁴

15 is irrational, or even√ⁿ

A where A is an integer that is not a perfect nth power. Even more elaborate numbers like³

1 +√

2 can be shown to be irrational in a similar manner. For example, if we let x =³

1 +√

2, and cube both sides, we get x³= 1 +√ 2 and subtracting 1 from both sides and squaring, we get that (x³− 1)²= 2 or that x⁶− 2x³− 1 = 0. The only rational roots of this are ±1 by the rational root theorem and none of them work. So, this equation has no rational roots. But ³

1 +√

2 is a root of this equation, so it must be irrational.

What a nice tool the rational root theorem is!

We have just seen that several irrational numbers can be obtained as roots of polynomials.

For example,√

2 is a root of the polynomial p(x) = x²− 2, and ³ 1 +√

2 is a root of p(x) = x⁶− 2x³− 1. It is a natural question to ask if all irrational numbers are roots of polynomials with integer coefficients. For a while, many people believed that. But, to allow for the possibility that this was not true, mathematicians defined the term algebraic number.

An algebraic number is a number that is the root of a polynomial with integral coefficients.

Thus,√

2 and³ 1 +√

2 and√⁴

15 are algebraic as we saw in the last two paragraphs. A number which is not algebraic is called transcendental. Thus, a transcendental number is a number that is not a root of any polynomial with integral coefficients (though it can be a root of a polynomial whose coefficients are not integers).

As we have pointed out, the prevailing thought was that all irrational numbers were algebraic, and thus transcendental numbers did not exist! This was wrong. It took many years for the discovery of the first transcendental number. One number was discovered by the mathematician Louiseville in around 1851 and it is the number 0.1100010000000000000000010000000000000 . . . where the number 1 occurs in only the factorial positions. That is, in the 1!, 2!, 3! and so on positions (in the 1st, 2nd, 6th, 24th, etc. position).

Proving that this number is transcendental is quite involved. We refer the reader to the Internet for several variations on proofs of this or to the book, Numbers, Rational and Irrational (1961) by Ivan Niven. It took until 1873 until the mathematician Hermite proved that the number e so prevalent in the study of calculus, was transcendental, and then another 9 years before the mathematician Lindemann proved thatπ was transcendental. Thus e and π, though irrational, are not roots of polynomials with integral coefficients. Historically, finding numbers that are transcendental was slow. This might lead you to believe that very few exist. But in mathematics, things are not always what they seem.

In 1887, George Cantor surprised the mathematical world when he proved that there were infinitely many transcendental numbers, and in fact, there were more of them than rational numbers! Indeed, “almost all” irrational numbers are transcendental. What a surprise! But even though there are so many transcendental numbers, proving that a number is transcendental seems to be extremely difficult. In fact, it took until 1999 just to prove that numbers like e^π√2 and e^π√3 and so on are transcendental. For more information about transcendental numbers, see Chapter 6 Page 304.

Defining and finding algebraic and transcendental numbers seemed to just be a game intel-lectuals played. But it turned out that these notions held the key to problems that had baffled mathematicians for thousands of years. Some of the problems that were solved by studying algebraic numbers were the problem of squaring the circle, duplicating the cube, and trisecting an angle, using only an unmarked ruler and compass. These problems of antiquity seem to be

irrelevant in today’s world. But they were puzzles that could not be solved by even the best minds for over 2000 years. We discuss these problems in Chapter 14.

We have shown how the rational root theorem could be used to prove that certain numbers are irrational. We can also use this theorem to show that cos n^◦is irrational for all rational values of θ such that 0 < θ < 90^◦, except for cos 60 degrees. Before showing this, we will state the following result will be proven in Chapter 8. (See Example 8.24.) That result is, that for any rational angleθ between 0 and 90 degrees, the quantity 2 cosθ satisfies an equation of the form

1xⁿ+ an−1xⁿ⁻¹+ . . . + a0= 0 (3.9)

where the coefficients are integers. (See the corollary to Example 8.24.) We now use this result.

Example 3.15 Show that cosθ^◦ is irrational for all rational anglesθ where 0^◦< θ < 90^◦except for cos 60 degrees.

Solution. By the rational root theorem, any rational solution of equation (3.9) is of the form a/b, where a divides a0 and b divides 1. Of course, if b divides 1, b is either 1 or −1 and the fraction a/b is an integer. Thus, the only rational roots of equation (3.9) are integers. Now, x = 2 cos θ is a root of equation (3.9) . Thus, if x = 2 cos θ is rational, it must be an integer. Since 2 cos θ is strictly between 0 and 2 (that is, 0< 2 cos θ < 2) when θ is strictly between 0 and 90 degrees, it follows that the only integer x = 2 cosθ can be is 1. And that happens when cos θ = 1/2, which happens whenθ = 60^◦. Thus, the only rational root of this equation occurs when θ = 60^◦.

Student Learning Opportunities

1 Finish the proof of Theorem 3.12.

2 Set up a polynomial that each of the following numbers is a root of, and then use the rational root theorem to show that each of these is irrational.

(a) √⁴

and making a similar observation for 3 +√ 2.

4 (C) A student asks you whether an irrational number raised to an irrational power can be rational? Can it? Explain.

5 Use the rational root theorem to find all the roots of the following equations (a) x³− 4x²+ 3 = 0

(b) 4x³− x²+ 5 = 0 (c) 2x³+ 6x²= 8

(d) 4x³+ 4x²= x + 1 (e) x²(4x + 8)− 11x = 15

(f) x³− 2x²= 1− 2x

6 (C) Your students understand that, since the number 2

3 satisfies the equation 3x− 2 = 0, which has integral coefficients, 2

3 must be algebraic by definition of algebraic. But they have the following questions and need help in answering and then proving their answers. How would you explain the answers to these questions? Use variables in part (a).

(a) Are all rational numbers algebraic?

(b) Are all transcendental numbers irrational?