**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 \(\frac{a}{b}\), where \(b\neq0\) and \(gcd(a,b)=1\).

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

## 1. Determine the value of \( 0.\overline{238095}\).

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

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

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

Hence, \(S = \frac {238095}{999999} = \frac {5}{21}\).

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

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

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

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

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

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

Proof: Suppose \( \frac {a}{b}\) is a root of \( f(x)\). Then

\[ p_n \left(\frac {a}{b} \right)^n + p_{n-1} \left(\frac {a}{b} \right)^{n-1} + \ldots + p_1 \frac {a}{b} + p_0 = 0.\] By shifting the \( p_0\) term to the right hand side, and multiplying throughout by \( b^n\), we obtain \( p_n a^n + p_{n-1} a^{n-1} b + \ldots + p_1 ab^{n-1} = -p_0 b^n\). Notice that the left hand side is a multiple of \( a\), thus \( a| p_0 b^n\). Since \( \gcd(a, b)=1\), Euclid's Lemma implies \( a | p_0\).Similarly, if we shift the \( p_n\) term to the right hand side and multiply throughout by \( b^n\), we obtain \[ p_{n-1} a^{n-1} b + p_{n-2} a ^{n-2} b^2 + \ldots + p_1 a b^{n-1} + p_0 b^n = -p_n a^n.\] Note that the left hand side is a multiple of \( b\), thus \( b | p_n a^n\). Since \( \gcd(a,b)=1\), Euclid's Lemma implies \( b | p_n\). \( _\square\)

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

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

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

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

By the rational root theorem, any rational root of \( f(x)\) has the form \( r = \frac {a}{b}\) where \( a | 1\) and \( b | 2\). Thus, we only need to try the numbers \( \pm \frac {1}{1}, \pm \frac {1}{2}\). We see that \[ f(1) >0,\] \[f(\frac {1}{2}) > 0,\] \[ f(-\frac {1}{2}) = -\frac {2}{8} + \frac {7}{4} - \frac {5}{2} + 1 = 0,\]

\[ f(-1) = -2 + 7 - 5 + 1 = 1 \neq 0.\] Hence, by the remainder-factor theorem, \( (2x+1)\) is a factor of \( f(x)\), implying \( f(x) = (2x+1) (x^2 + 3x + 1)\). We can then use the quadratic formula to factorize the quadratic if irrational roots are desired.

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

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

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

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

## 4. Prove Theorem 1 using the Rational Root Theorem.

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

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

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