Rational and Irrational Numbers

Definition

Rational numbers are numbers that can be expressed as the ratio of two integers. Rational numbers follow the rules of arithmetic and all rational numbers can be reduced to the form ab\frac{a}{b}, where b0b\neq0 and gcd(a,b)=1gcd(a,b)=1.

Rational numbers are often denoted by Q\mathbb{Q}. These numbers are a subset of the real numbers, which comprise the complete number line and are often denoted by R\mathbb{R}. Real numbers that cannot be expressed as the ratio of two integers are called irrational numbers.

Worked Examples

1. Determine the value of 0.238095 0.\overline{238095}.

The line over 238095 238095 denotes that it is a repeating decimal, of the form 0.238095238095238095 0.238095238095238095\ldots.

Solution. Let S=0.238095 S = 0.\overline{238095}. Then 1000000S=238095.2380951000000S = 238095.\overline{238095}, and taking the sum, we obtain

S=000000.2380951000000S=238095.238095999999S=238095\begin{aligned} - S & = -000000.\overline{238095} \\ 1000000S & = 238095.\overline{238095}\\ \hline \\ 999999S & = 238095& \\ \end{aligned}

Hence, S=238095999999=521S = \frac {238095}{999999} = \frac {5}{21}.

 

2. Show that 2+3 \sqrt{2} + \sqrt{3} is not rational.

Solution: We give a proof by contradiction. If 2+3 \sqrt{2}+\sqrt{3} is rational, then (32)×1(2+3)=32 (3-2) \times \frac {1}{(\sqrt{2} + \sqrt{3})} = \sqrt{3}-\sqrt{2}, implying 32 \sqrt{3} - \sqrt{2} is also rational. Since (3+2)(32)=22 (\sqrt{3} + \sqrt{2}) - (\sqrt{3}-\sqrt{2}) = 2 \sqrt{2}, we obtain 22 2 \sqrt{2} is rational. Thus, 22×12=2 2 \sqrt{2} \times \frac {1}{2} = \sqrt{2} is also rational, which is a contradiction.

We generalize the result that D \sqrt{D} is rational if and only if D D is a perfect square.

Theorem. Given integers n n and m m, if n1m n^{\frac {1}{m}} is rational, then n1m n^{\frac {1}{m}} is an integer. In particular, the only rational mth m^{th} roots of integers n n are the integers.

Proof: Let n1m=ab n^{\frac {1}{m}}= \frac {a}{b} , where a a and b b are coprime integers. Then taking powers and clearing denominators gives bmn=am b^m n = a^m . If p p is a prime that divides b b, then p p divides bmn b^m n, so p p divides am a^m and thus p p divides a a. Since a a and b b are coprime, there is no prime that divides both a a and b b. Hence, no prime divides b b, implying b=1 b=1. Therefore, n1m=a n^{\frac {1}{m}} = a is an integer. _\square

 

Rational Root Theorem. If f(x)=pnxn+pn1xn1++p1x+p0 f(x) = p_n x^n + p_{n-1} x^{n-1} + \ldots + p_1 x + p_0 has a rational root of the form r=ab r = \frac {a}{b} with gcd(a,b)=1 \gcd (a,b)=1, then ap0 a \vert p_0 and bpn b \vert p_n.

Proof: Suppose ab \frac {a}{b} is a root of f(x) f(x). Then
pn(ab)n+pn1(ab)n1++p1ab+p0=0. p_n \left(\frac {a}{b} \right)^n + p_{n-1} \left(\frac {a}{b} \right)^{n-1} + \ldots + p_1 \frac {a}{b} + p_0 = 0. By shifting the p0 p_0 term to the right hand side, and multiplying throughout by bn b^n, we obtain pnan+pn1an1b++p1abn1=p0bn p_n a^n + p_{n-1} a^{n-1} b + \ldots + p_1 ab^{n-1} = -p_0 b^n. Notice that the left hand side is a multiple of a a, thus ap0bn a| p_0 b^n. Since gcd(a,b)=1 \gcd(a, b)=1, Euclid's Lemma implies ap0 a | p_0.

Similarly, if we shift the pn p_n term to the right hand side and multiply throughout by bn b^n, we obtain pn1an1b+pn2an2b2++p1abn1+p0bn=pnan. p_{n-1} a^{n-1} b + p_{n-2} a ^{n-2} b^2 + \ldots + p_1 a b^{n-1} + p_0 b^n = -p_n a^n. Note that the left hand side is a multiple of b b, thus bpnan b | p_n a^n. Since gcd(a,b)=1 \gcd(a,b)=1, Euclid's Lemma implies bpn b | p_n. _\square

In particular, this tells us that if we want to check for 'nice' rational roots of a polynomial f(x) f(x), we only need to check finitely many numbers of the form ±ab \pm \frac {a}{b}, where ap0 a | p_0 and bpn b | p_n. This is a great tool for factorizing polynomials.

 

Integer Root Theorem. If f(x) f(x) is a monic polynomial (leading coefficient of 1), then the rational roots of f(x) f(x) must be integers.

Proof: By the rational root theorem, if r=ab r = \frac {a}{b} is a root of f(x) f(x), then bpn b | p_n. But since pn=1 p_n = 1 by assumption, hence b=1 b=1 and thus r=a r=a is an integer. _\square

Worked Examples

1. Factorize the cubic polynomial f(x)=2x3+7x2+5x+1 f(x) = 2x^3 + 7x^2 + 5x + 1 .

By the rational root theorem, any rational root of f(x) f(x) has the form r=ab r = \frac {a}{b} where a1 a | 1 and b2 b | 2. Thus, we only need to try the numbers ±11,±12 \pm \frac {1}{1}, \pm \frac {1}{2}. We see that f(1)>0, f(1) >0, f(12)>0,f(\frac {1}{2}) > 0, f(12)=28+7452+1=0, f(-\frac {1}{2}) = -\frac {2}{8} + \frac {7}{4} - \frac {5}{2} + 1 = 0,
f(1)=2+75+1=10. f(-1) = -2 + 7 - 5 + 1 = 1 \neq 0. Hence, by the remainder-factor theorem, (2x+1) (2x+1) is a factor of f(x) f(x), implying f(x)=(2x+1)(x2+3x+1) f(x) = (2x+1) (x^2 + 3x + 1). We can then use the quadratic formula to factorize the quadratic if irrational roots are desired.

 

2. Using the rational root theorem, show that 2 \sqrt{2} is not rational.

Since 2 \sqrt{2} is a root of the polynomial f(x)=x22 f(x) = x^2-2, the rational root theorem tells us that the rational roots of f(x) f(x) are of the form ±1,21 \pm \frac { 1, 2}{ 1}. It is easy to check that none of these are roots of f(x) f(x), hence f(x) f(x) has no rational roots. Thus, 2 \sqrt{2} is not rational.

 

3. Show that if x x is a positive rational such that x2+x x^2 + x is an integer, then x x must be an integer.

Let x2+x=n x^2 + x = n, where n n is an integer. This is equivalent to finding the roots of f(x)=x2+xn f(x) = x^2+x-n. Since f(x) f(x) is a monic polynomial, by the integer root theorem, if x x is a rational root of f(x) f(x), then it is an integer root.

 

4. Prove Theorem 1 using the Rational Root Theorem.

Consider the polynomial f(x)=xmn f(x) = x^m -n. By the rational root theorem, if r=ab r = \frac {a}{b} is a rational root of f(x) f(x), then we must have b1 b| 1, and hence b=1 b=1 (or -1). Thus, r=a r = a is an integer. [Pop Quiz: Give a 1 line proof of Theorem 1 using the Integer Root Theorem].

Note by Calvin Lin
5 years, 7 months ago

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but , tell me who was the first mathematician to discover rational numbers ? Please answer this question as soon as possible because this is very urgent ...........

Gauravi Saxena - 5 years, 6 months ago

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