Rational and Irrational Numbers

A. Formal Definitions:

• Rational Numbers ($\mathbb{Q}$): Any number that can be expressed in the form $\frac{p}{q}$, where $p, q \in \mathbb{Z}$, $q \neq 0$, and $\gcd(p, q) = 1$. Examples: $-\frac{7}{4}, 0 = \frac{0}{1}, 0.35 = \frac{7}{20}, 0.\overline{3} = \frac{1}{3}$.
• Irrational Numbers ($\mathbb{I}$ or $\mathbb{Q}^{\prime}$): A real number that cannot be written in the form $\frac{p}{q}$ ($p, q \in \mathbb{Z}, q \neq 0$). Examples: $\sqrt{2}, \sqrt{3}, \sqrt[3]{5}, \pi, e, 0.1010010001...$
• Real Numbers ($\mathbb{R}$): The union of all rational and irrational numbers: $\mathbb{R} = \mathbb{Q} \cup \mathbb{Q}^{\prime}$. Every point on the continuous number line corresponds uniquely to a real number.

B. Decimal Expansion Criteria:

Step-by-Step Conversion Algorithm:

To convert a recurring decimal $x$ to fractional form $\frac{p}{q}$:

1. Let $x = \text{given decimal}$ (write out several periods of the repeating block).
2. If there are non-repeating digits after the decimal point, multiply by $10^k$ (where $k$ is the number of non-repeating digits) so that only the repeating block remains after the decimal point.
3. Multiply this equation by $10^n$, where $n$ is the period (number of digits in the repeating block).
4. Subtract the earlier equation from the new equation so the infinite repeating decimal tails cancel out completely.
5. Solve the resulting linear algebraic equation for $x$ and reduce $\frac{p}{q}$ to its lowest terms.

Illustrative Example: Express $x = 0.2\overline{35} = 0.2353535...$ in $\frac{p}{q}$ form.

Step 1: Let $x = 0.2353535...$   --- (1)

Step 2: There is 1 non-repeating digit ($2$) after the decimal. Multiply (1) by $10$:

Step 3: The repeating block has $2$ digits ($35$). Multiply (2) by $10^2 = 100$:

Step 4: Subtract (2) from (3):

Since $\gcd(233, 990) = 1$, the canonical fraction is $\mathbf{\frac{233}{990}}$.

Theorem: Prove that $\sqrt{2}$ is an irrational number.

Proof (by Reductio ad Absurdum):

Assume to the contrary that $\sqrt{2}$ is a rational number.

Therefore, there exist two positive integers $a$ and $b$ such that:

Squaring both sides of the equation:

Since $2b^2$ is divisible by $2$, it follows that $a^2$ is divisible by $2$.

Lemma: If a prime number $p$ divides $n^2$, then $p$ divides $n$. Hence, since $2$ divides $a^2$, $2$ divides $a$.

Thus, we can write $a = 2k$ for some integer $k$.

Substitute $a = 2k$ into Equation 1:

This implies that $b^2$ is divisible by $2$, and consequently, by the same lemma, $2$ divides $b$.

Contradiction: We have established that $2$ is a common factor of both $a$ and $b$. But this directly contradicts our initial assumption that $\gcd(a, b) = 1$ (that $a$ and $b$ are coprime).

This contradiction arose solely because of our false assumption that $\sqrt{2}$ is rational.

A. Definition and Terminology:

An irrational number of the form $\sqrt[n]{a}$ is called a surd of order $n$ if $a$ is a positive rational number and $\sqrt[n]{a}$ cannot be evaluated as an exact rational number. Here, $\sqrt{}$ is the radical sign, $n$ is the index/order of the surd, and $a$ is the radicand.

B. Laws of Radicals / Surds:

• $\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{ab}$
• $\frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}$
• $(\sqrt[n]{a})^n = a$
• $\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$
• $(\sqrt[n]{a})^m = \sqrt[n]{a^m}$

C. Conjugate Surds and Rationalisation:

When the product of two surds is a rational number, each surd is called the Rationalising Factor (RF) of the other. The process of converting an irrational denominator into a rational number is called rationalisation.

• For a monomial surd $\sqrt{a}$, the rationalising factor is $\sqrt{a}$ because $\sqrt{a} \cdot \sqrt{a} = a \in \mathbb{Q}$.
• For a binomial quadratic surd $(a + \sqrt{b})$, its conjugate $(a - \sqrt{b})$ is the rationalising factor, because: $$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - (\sqrt{b})^2 = a^2 - b \in \mathbb{Q}$$
• Similarly, the conjugate of $(\sqrt{a} + \sqrt{b})$ is $(\sqrt{a} - \sqrt{b})$, yielding: $$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b \in \mathbb{Q}$$

Problem 1: If $x = 3 + 2\sqrt{2}$, find the value of: (i) $\frac{1}{x}$, (ii) $x + \frac{1}{x}$, (iii) $x^2 + \frac{1}{x^2}$.

Solution:

(i) Find $\frac{1}{x}$ by rationalising the denominator:

(ii) Find $x + \frac{1}{x}$:

(iii) Find $x^2 + \frac{1}{x^2}$:

Using the algebraic expansion $\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$:

Problem 2: Find the values of rational numbers $a$ and $b$ if:

Solution:

Rationalise the left-hand side by multiplying the numerator and denominator by the conjugate of the denominator, which is $(\sqrt{3} + 1)$:

Now equate this to the right-hand side:

Equating rational and irrational parts on both sides: