**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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