§2 The Set ℚ of Rational Numbers

Subtraction forces us from $\mathbb{N}$ to the integers $\mathbb{Z} = \{0, 1, -1, 2, -2, \dots\}$, and division forces us on to the rational numbers

$$\mathbb{Q} = \left\{ \tfrac{m}{n} : m, n \in \mathbb{Z},\ n \ne 0 \right\}.$$

$\mathbb{Q}$ is a very satisfactory system for addition, subtraction, multiplication and division. It breaks down when we try to solve equations like $x^2 = 2$. A unit square has a diagonal $d$ with $d^2 = 1^2 + 1^2 = 2$, so there “ought” to be a number $\sqrt2$. We will see that no rational number works, even though rationals like $1.4142$ and $1.4143$ come very close: their squares are $1.99996164$ and $2.00024449$.

So $\mathbb{Q}$ has gaps. The graph of $y = x^2 - 2$ crosses the $x$-axis, yet it slips between all the rational points on that axis. Numbers like $\pi$ and $e$ also fall into these gaps.

Why this matters

This section is not really about algebraic numbers for their own sake. Its job is to show, beyond doubt, that $\mathbb{Q}$ has holes, which is why analysis needs a bigger number system with no holes at all (§4). The tool for this is the Rational Zeros Theorem below. It turns an infinite search (“is there any fraction whose square is $2$?”) into a short, finite list of candidates to check.

2.1 Definition.

A number is called an algebraic number if it satisfies a polynomial equation

$$c_n x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0 = 0$$

where the coefficients $c_0, c_1, \dots, c_n$ are integers, $c_n \ne 0$ and $n \ge 1$.

Used in §16

Every rational $r = \frac mn$ is algebraic, since it solves $nx - m = 0$. Expressions built from rationals with roots $\sqrt{\ }, \sqrt[3]{\ }, \dots$ and ordinary arithmetic are algebraic as well.

Example 1.

$\sqrt{3}$, $\tfrac{4}{17}$, $\sqrt[3]{17}$, $\sqrt{2 + \sqrt[3]{5}}$ and $\sqrt{\tfrac{4 - 2\sqrt3}{7}}$ are algebraic. The first three solve $x^2 - 3 = 0$, $17x - 4 = 0$ and $x^3 - 17 = 0$.

If $a = \sqrt{2 + \sqrt[3]5}$, then $a^2 - 2 = \sqrt[3]5$, so $(a^2 - 2)^3 = 5$. Expanding, $a$ satisfies

$$x^6 - 6x^4 + 12x^2 - 13 = 0.$$

If $b = \sqrt{(4 - 2\sqrt3)/7}$, then $7b^2 = 4 - 2\sqrt3$, so $2\sqrt3 = 4 - 7b^2$ and $12 = (4 - 7b^2)^2$. So $b$ satisfies $49x^4 - 56x^2 + 4 = 0$.

The next theorem tells us exactly which rationals could possibly solve such an equation. It lets us prove that algebraic numbers that look irrational usually are. If the statement seems complicated, read Corollary 2.3 and Examples 2–5 first.

2.2 Rational Zeros Theorem.

Suppose $c_0, c_1, \dots, c_n$ are integers and $r$ is a rational number satisfying

$$c_n x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0 = 0 \tag{1}$$

where $n \ge 1$, $c_n \ne 0$ and $c_0 \ne 0$. Write $r = \frac cd$ where $c, d$ are integers with no common factors and $d \ne 0$. Then $c$ divides $c_0$ and $d$ divides $c_n$.

Used in Corollary 2.3, §4, Exercise 6.5

In other words, the only rational candidates for solutions of (1) have the form $\frac cd$ where $c$ divides $c_0$ and $d$ divides $c_n$.

Proof

We are given $c_n \left(\frac cd\right)^n + c_{n-1} \left(\frac cd\right)^{n-1} + \cdots + c_1 \frac cd + c_0 = 0$. Multiplying through by $d^n$,

$$c_n c^n + c_{n-1} c^{n-1} d + c_{n-2} c^{n-2} d^2 + \cdots + c_1 c\, d^{n-1} + c_0 d^n = 0. \tag{2}$$

Solving for $c_0 d^n$,

$$c_0 d^n = -c\,\bigl[c_n c^{n-1} + c_{n-1} c^{n-2} d + \cdots + c_1 d^{n-1}\bigr],$$

so $c$ divides $c_0 d^n$. Since $c$ and $d^n$ have no common factors, $c$ divides $c_0$. (Here we use the basic fact that if $c$ divides $ab$ and has no common factor with $b$, then $c$ divides $a$.)

Solving (2) for $c_n c^n$ instead,

$$c_n c^n = -d\,\bigl[c_{n-1} c^{n-1} + c_{n-2} c^{n-2} d + \cdots + c_0 d^{n-1}\bigr],$$

so $d$ divides $c_n c^n$. Since $d$ and $c^n$ have no common factors, $d$ divides $c_n$.

∎

Intuitionwhy the proof works

After clearing denominators in (2), every term except $c_0 d^n$ visibly contains a factor of $c$. Since the terms add up to $0$, which $c$ divides, the leftover term $c_0 d^n$ must be divisible by $c$ too, and $c$ shares no factor with $d$. The same balancing trick, run with $d$ instead of $c$, handles $c_n c^n$. This is the familiar “even/odd” proof that $\sqrt2$ is irrational, with divisibility by $c$ and $d$ in place of divisibility by $2$.

2.3 Corollary.

Consider the polynomial equation $x^n + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0 = 0$, where $c_0, \dots, c_{n-1}$ are integers and $c_0 \ne 0$. Any rational solution of this equation must be an integer that divides $c_0$.

Used in Exercise 2.1, Exercise 2.2, Exercise 2.8, Exercise 4.12

Proof

In Theorem 2.2 the denominator $d$ must divide the leading coefficient, which is $1$. So $r$ is an integer, and it divides $c_0$.

∎

Polynomials whose leading coefficient is $1$ are called monic.

Try it: rational root candidates

Enter integer coefficients from the highest power down. The widget lists every candidate $\pm c/d$ allowed by Theorem 2.2 and evaluates the polynomial at each one exactly, using integer arithmetic.

Coefficients cₙ, …, c₀

6x³ − 11x² + 6x − 1 = 0

Divisors of c₀ = -1: 1. Divisors of cₙ = 6: 1, 2, 3, 6. So the only possible rational roots are the 8 numbers ±c/d below.

-1P = -24

-1/2P = -15/2

-1/3P = -40/9

-1/6P = -7/3

1/6P = -5/18

1/3root ✓

1/2root ✓

1root ✓

Rational roots: 1/3, 1/2, 1.

Example 2.

$\sqrt2$ is not a rational number.

Proof

By Corollary 2.3 the only possible rational solutions of $x^2 - 2 = 0$ are $\pm1, \pm2$. None of these four numbers works. Since $\sqrt2$ is a solution, it is not rational.

∎

Example 3.

$\sqrt{17}$ is not a rational number.

Proof

The only possible rational solutions of $x^2 - 17 = 0$ are $\pm1, \pm17$, and none of them is a solution.

∎

Example 4.

$\sqrt[3]{6}$ is not a rational number.

Proof

The only possible rational solutions of $x^3 - 6 = 0$ are $\pm1, \pm2, \pm3, \pm6$, and none of these eight numbers works.

∎

Example 5.

$a = \sqrt{2 + \sqrt[3]{5}}$ is not a rational number.

Proof

By Example 1, $a$ solves $x^6 - 6x^4 + 12x^2 - 13 = 0$, whose only possible rational solutions are $\pm1, \pm13$. At $x = \pm1$ the left side is $-6$. At $x = \pm13$ we can avoid the arithmetic: $a$ is obviously between $1$ and $13$. Alternatively, $13^6 - 6\cdot13^4 + 12\cdot13^2 - 13 = 13\,(13^5 - 6\cdot13^3 + 12\cdot13 - 1) \ne 0$, since the term in parentheses is one less than a multiple of $13$.

∎

Example 6.

$b = \sqrt{(4 - 2\sqrt3)/7}$ is not a rational number.

Proof

By Theorem 2.2 the possible rational roots of $49x^4 - 56x^2 + 4 = 0$ are $\pm1, \pm\frac17, \pm\frac1{49}, \pm2, \pm\frac27, \pm\frac2{49}, \pm4, \pm\frac47, \pm\frac4{49}$. Checking all 18 is discouraging, so here is a cleverer route. Example 1 showed $12 = (4 - 7b^2)^2$. If $b$ were rational, so would be $4 - 7b^2$ (Exercise 2.6), and then $x^2 = 12$ would have a rational solution. But the only candidates $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$ all fail. So $b$ is irrational.

∎

In practice, estimating the number in question eliminates most candidates at once. The numbers in Examples 2–5 are clearly not integers. In Example 6, $b \approx 0.2767$, and the nearest candidate is $\frac27 \approx 0.2857$. Be careful, though: not every irrational-looking expression is irrational. See Exercise 2.7.

Common pitfall

• The theorem produces candidates. It never says that any of them is actually a root; you still have to check each one.
• It needs $c_0 \ne 0$. If the constant term is $0$, factor out a power of $x$ first: $x^3 - 2x = x(x^2 - 2)$, and apply the theorem to $x^2 - 2$.
• The fraction must be in lowest terms. The number $\frac24$ is a root of $2x - 1 = 0$, yet $2$ does not divide $c_0 = -1$. Writing it as $\frac12$ fixes this.
• To prove a number is irrational this way, you first need an integer polynomial that it satisfies. Finding that polynomial, as in Example 1, is often the real work.

2.4 Remark.

The Rational Zeros Theorem finds the rational zeros of integer polynomials. To approximate other zeros, of these or other functions, one uses Newton’s method and the secant method, discussed in §31.

Reading alongside Abbott

Abbott opens with the classical proof that $\sqrt2$ is irrational in Section 1.1: assume $\sqrt2 = \frac pq$ in lowest terms and show that $p$ and $q$ would both be even. Theorem 2.2 is that argument made to work for every integer polynomial at once. In Section 1.3, Abbott uses a set of rationals with no rational least upper bound to motivate completeness. Here that idea appears in §4, right after the Completeness Axiom.

Check your understanding

Take the quiz (3 questions)

By Corollary 2.3, which numbers are the only possible rational solutions of $x^2 - 6 = 0$?

$\pm 1, \pm 2, \pm 3, \pm 6$$\pm 1, \pm 6$Every fraction $\frac cd$ with $c$ dividing $6$

Suppose $r = \frac cd$ in lowest terms solves $c_n x^n + \cdots + c_1 x + c_0 = 0$ with integer coefficients and $c_0 \ne 0$. What does the Rational Zeros Theorem say?

$c$ divides $c_0$ and $d$ divides $c_n$.$c$ divides $c_n$ and $d$ divides $c_0$.Both $c$ and $d$ divide $c_0$.

Is every algebraic number rational?

Yes, by definition.No: $\sqrt 2$ is algebraic, since it solves $x^2 - 2 = 0$, but it is irrational.No: algebraic numbers are never rational.

Exercises

Exercise 2.1

Show $\sqrt3$, $\sqrt5$, $\sqrt7$, $\sqrt{24}$ and $\sqrt{31}$ are not rational numbers.

Hint

Apply Corollary 2.3 to $x^2 - k = 0$. In each case the candidates are the divisors of $k$, and none of them squares to $k$.

Solution

Suppose $\sqrt k$ were rational. It would be a rational solution of $x^2 - k = 0$, so by Corollary 2.3 it would be an integer dividing $k$. We check that no divisor $d$ of $k$ (positive or negative) has $d^2 = k$.

• $k = 3$: divisors $\pm1, \pm3$ have squares $1, 9$.
• $k = 5$: divisors $\pm1, \pm5$ have squares $1, 25$.
• $k = 7$: divisors $\pm1, \pm7$ have squares $1, 49$.
• $k = 24$: divisors $\pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24$ have squares $1, 4, 9, 16, 36, 64, 144, 576$.
• $k = 31$: divisors $\pm1, \pm31$ have squares $1, 961$.

None equals $k$, so none of these square roots is rational.

Tags: StandardContradiction

Exercise 2.2

Show $\sqrt[3]{2}$, $\sqrt[7]{5}$ and $\sqrt[4]{13}$ are not rational numbers.

Hint

The candidates for $x^3 - 2$ are $\pm1, \pm2$; for $x^7 - 5$ they are $\pm1, \pm5$; for $x^4 - 13$ they are $\pm1, \pm13$.

Solution

Each number is a root of $x^n - k = 0$ for the indicated $n, k$, so a rational one would be an integer dividing $k$ (Corollary 2.3).

• $\sqrt[3]2$: $x^3 - 2$; candidates $\pm1, \pm2$ have cubes $\pm1, \pm8 \neq 2$.
• $\sqrt[7]5$: $x^7 - 5$; candidates $\pm1, \pm5$ have seventh powers $\pm1, \pm78125 \neq 5$.
• $\sqrt[4]{13}$: $x^4 - 13$; candidates $\pm1, \pm13$ have fourth powers $1, 28561 \neq 13$.

So none of them is rational.

Tags: StandardContradiction

Exercise 2.3

Show $\sqrt{2 + \sqrt2}$ is not a rational number.

Solution

If $a = \sqrt{2 + \sqrt2}$ then $(a^2 - 2)^2 = 2$, so $a$ solves $x^4 - 4x^2 + 2 = 0$. The candidates are $\pm1, \pm2$, giving values $-1$ and $2$. None is zero.

Tags: StandardContradiction

Exercise 2.4

Show $\sqrt[3]{5 - \sqrt3}$ is not a rational number.

Solution

If $a = \sqrt[3]{5 - \sqrt3}$ then $(5 - a^3)^2 = 3$, so $a$ solves $x^6 - 10x^3 + 22 = 0$. The candidates are $\pm1, \pm2, \pm11, \pm22$. Since $1 < a < 2$, none of them is $a$.

Tags: StandardContradiction

Exercise 2.5

Show $\left[3 + \sqrt2\right]^{2/3}$ is not a rational number.

Solution

If $a = (3 + \sqrt2)^{2/3}$ then $a^3 = (3 + \sqrt2)^2 = 11 + 6\sqrt2$, so $(a^3 - 11)^2 = 72$ and $a$ solves $x^6 - 22x^3 + 49 = 0$. The candidates are $\pm1, \pm7, \pm49$, but $a^3 \approx 19.5$ gives $a \approx 2.69$.

Tags: StandardContradiction

Exercise 2.6

In connection with Example 6, discuss why $4 - 7b^2$ is rational if $b$ is rational.

Solution

$\mathbb{Q}$ is closed under multiplication and subtraction: if $b = \frac mn$ then $4 - 7b^2 = \frac{4n^2 - 7m^2}{n^2}$.

Tags: Warm-upExplain whyComputation

Exercise 2.7

Show the following irrational-looking expressions are actually rational numbers.

1. $\sqrt{4 + 2\sqrt3} - \sqrt3$
2. $\sqrt{6 + 4\sqrt2} - \sqrt2$

Solution

(a) $4 + 2\sqrt3 = (1 + \sqrt3)^2$, so the expression is $1$. (b) $6 + 4\sqrt2 = (2 + \sqrt2)^2$, so the expression is $2$.

Tags: Warm-upComputation

Exercise 2.8

Find all rational solutions of the equation $x^8 - 4x^5 + 13x^3 - 7x + 1 = 0$.

Solution

By Corollary 2.3 the only candidates are $\pm1$. At $x = 1$: $1 - 4 + 13 - 7 + 1 = 4$. At $x = -1$: $1 + 4 - 13 + 7 + 1 = 0$. So $x = -1$ is the only rational solution. Try it in the widget above.

Tags: StandardComputationContradiction