Real Numbers: A Comprehensive Study Guide

This chapter serves as the foundation for understanding various types of numbers and their properties. We'll explore the classification of numbers into rational and irrational categories, delve into the crucial Fundamental Theorem of Arithmetic which underpins many number theory concepts, and learn how to use it to find HCF and LCM. We'll also tackle proofs of irrationality for numbers like $\sqrt{2}$ and understand the conditions for rational numbers to have terminating or non-terminating repeating decimal expansions. Mastering these concepts is vital for your future mathematical studies.

1. Introduction to Real Numbers

Before diving deep, let's quickly review the number system.

• Natural Numbers ($\boldsymbol{N}$): The counting numbers: $\{1, 2, 3, ...\}$
• Whole Numbers ($\boldsymbol{W}$): Natural numbers including zero: $\{0, 1, 2, 3, ...\}$
• Integers ($\boldsymbol{Z}$): Whole numbers and their negatives: $\{..., -2, -1, 0, 1, 2, ...\}$
• Rational Numbers ($\boldsymbol{Q}$): Numbers that can be expressed in the form $p/q$, where $p$ and $q$ are integers and $q \neq 0$. Examples: $1/2, -3/4, 5, 0$.
• Irrational Numbers: Numbers that cannot be expressed in the form $p/q$. Their decimal expansions are non-terminating and non-repeating. Examples: $\sqrt{2}, \sqrt{3}, \pi$.
• Real Numbers ($\boldsymbol{R}$): The collection of all rational and irrational numbers. Every point on the number line represents a unique real number.

2. The Fundamental Theorem of Arithmetic

This theorem is one of the most important concepts in number theory.

Statement: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

• A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself (e.g., 2, 3, 5, 7, 11).
• A composite number is a natural number greater than 1 that is not prime (e.g., 4, 6, 8, 9, 10).

Example: The number 12 can be written as $2 \times 2 \times 3$ or $2^2 \times 3$. No matter how you start factorising 12, you will always end up with two 2s and one 3. The order might change ($2 \times 3 \times 2$), but the prime factors themselves and their count remain unique. This process is called prime factorisation.

3. HCF and LCM by Prime Factorisation

The Fundamental Theorem of Arithmetic provides a systematic way to find the Highest Common Factor (HCF) and Least Common Multiple (LCM) of two or more positive integers.

• HCF (Highest Common Factor): The largest positive integer that divides each of the given integers without leaving a remainder. Also known as GCD (Greatest Common Divisor).
• LCM (Least Common Multiple): The smallest positive integer that is a multiple of two or more given integers.

Method for finding HCF and LCM using Prime Factorisation:

1. Prime Factorise each of the given numbers.
2. For HCF: Identify the common prime factors. For each common prime factor, take the smallest power (lowest exponent) it has in the factorisations. Multiply these together.
3. For LCM: Identify all prime factors (common and non-common) that appear in the factorisations. For each prime factor, take the largest power (highest exponent) it has. Multiply these together.

Important Formula: For any two positive integers $a$ and $b$, $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$

Example 1: Find the HCF and LCM of 96 and 404 by the prime factorisation method. Also, verify that HCF $\times$ LCM $=$ product of the two numbers.

Solution: Step 1: Prime factorise 96 and 404. $96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3^1$ $404 = 2 \times 2 \times 101 = 2^2 \times 101^1$

Step 2: Find HCF. Common prime factor is 2. Smallest power of 2 is $2^2$. So, HCF$(96, 404) = 2^2 = 4$.

Step 3: Find LCM. All prime factors involved are 2, 3, and 101. Highest power of 2 is $2^5$. Highest power of 3 is $3^1$. Highest power of 101 is $101^1$. So, LCM$(96, 404) = 2^5 \times 3^1 \times 101^1 = 32 \times 3 \times 101 = 96 \times 101 = 9696$.

Step 4: Verify the formula. Product of the two numbers $= 96 \times 404 = 38784$. HCF $\times$ LCM $= 4 \times 9696 = 38784$. Since $38784 = 38784$, the formula is verified.

4. Proving Irrationality

One of the key topics in this chapter is proving that certain numbers are irrational. This often involves using a proof by contradiction. The most common proof is for $\sqrt{2}$.

Theorem: If $p$ is a prime number and $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer. (This theorem is crucial for the proof of irrationality).

Proof that $\sqrt{2}$ is irrational:

Step 1: Assume the contrary. Let's assume that $\sqrt{2}$ is rational.

Step 2: Express it as a fraction in simplest form. If $\sqrt{2}$ is rational, we can write it as $\sqrt{2} = a/b$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are coprime (i.e., they have no common factors other than 1).

Step 3: Manipulate the equation. Squaring both sides: $(\sqrt{2})^2 = (a/b)^2$ $2 = a^2/b^2$ $2b^2 = a^2$ (Equation 1)

Step 4: Draw a conclusion based on divisibility. From Equation 1, $a^2$ is an even number (since it's $2$ times some integer $b^2$). If $a^2$ is even, then $a$ must also be even (using the theorem: if 2 divides $a^2$, then 2 divides $a$).

Step 5: Substitute and find another conclusion. Since $a$ is even, we can write $a = 2c$ for some integer $c$. Substitute $a = 2c$ into Equation 1: $2b^2 = (2c)^2$ $2b^2 = 4c^2$ Divide both sides by 2: $b^2 = 2c^2$

Step 6: Draw a further conclusion. From this, $b^2$ is an even number (since it's $2$ times some integer $c^2$). If $b^2$ is even, then $b$ must also be even (using the theorem: if 2 divides $b^2$, then 2 divides $b$).

Step 7: Arrive at a contradiction. We have shown that both $a$ and $b$ are even. This means that $a$ and $b$ have a common factor of 2. However, in Step 2, we assumed that $a$ and $b$ are coprime (have no common factors other than 1). This is a contradiction!

Step 8: Conclude. Our initial assumption that $\sqrt{2}$ is rational must be false. Therefore, $\sqrt{2}$ is irrational.

Example 2: Prove that $3 + 2\sqrt{5}$ is irrational.

Solution: Step 1: Assume the contrary. Let's assume that $3 + 2\sqrt{5}$ is rational.

Step 2: Express it as a fraction. If $3 + 2\sqrt{5}$ is rational, we can write it as $3 + 2\sqrt{5} = a/b$, where $a$ and $b$ are coprime integers and $b \neq 0$.

Step 3: Isolate the irrational part. $2\sqrt{5} = a/b - 3$ $2\sqrt{5} = (a - 3b)/b$ $\sqrt{5} = (a - 3b)/(2b)$

Step 4: Arrive at a contradiction. Since $a$ and $b$ are integers, $(a - 3b)$ is an integer, and $2b$ is a non-zero integer. Therefore, $(a - 3b)/(2b)$ is a rational number. This implies that $\sqrt{5}$ is rational. However, we know that $\sqrt{5}$ is irrational (this is a standard result, similar to $\sqrt{2}$, and can be assumed or proven separately if required). This is a contradiction.

Step 5: Conclude. Our initial assumption that $3 + 2\sqrt{5}$ is rational must be false. Therefore, $3 + 2\sqrt{5}$ is irrational.

5. Decimal Expansion of Rational Numbers

Rational numbers can have two types of decimal expansions:

1. Terminating Decimal Expansion: The decimal expansion ends after a finite number of digits. Example: $1/4 = 0.25$, $1/8 = 0.125$.
2. Non-terminating Repeating (or Recurring) Decimal Expansion: The decimal expansion continues infinitely, but a block of digits repeats periodically. Example: $1/3 = 0.333...$, $1/7 = 0.142857142857...$.

Theorem on Terminating Decimal Expansion: A rational number $x = p/q$, where $p$ and $q$ are coprime integers and $q \neq 0$, has a terminating decimal expansion if and only if the prime factorisation of the denominator $q$ is of the form $2^n 5^m$, where $n$ and $m$ are non-negative integers.

Theorem on Non-terminating Repeating Decimal Expansion: A rational number $x = p/q$, where $p$ and $q$ are coprime integers and $q \neq 0$, has a non-terminating repeating decimal expansion if and only if the prime factorisation of the denominator $q$ is not of the form $2^n 5^m$, where $n$ and $m$ are non-negative integers. This means $q$ has at least one prime factor other than 2 or 5.

Key point: Always simplify the rational number $p/q$ to its simplest form before checking the denominator $q$.

Example: Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: (i) $13/3125$ $(ii) \frac{64}{455}$ (iii) $15/1600$

Solution: (i) $13/3125$ The fraction is already in its simplest form (13 is prime, 3125 is not a multiple of 13). Denominator $q = 3125$. Prime factorisation of $3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$. This is of the form $2^n 5^m$, where $n=0$ and $m=5$. Therefore, $13/3125$ has a terminating decimal expansion.

$**(ii) \frac{64}{455}**$ The fraction is already in its simplest form (64 is $2^6$, 455 is not a multiple of 2). Denominator $q = 455$. Prime factorisation of $455 = 5 \times 7 \times 13$. This is not of the form $2^n 5^m$, as it contains prime factors 7 and 13, which are neither 2 nor 5. Therefore, $64/455$ has a non-terminating repeating decimal expansion.

(iii) $15/1600$ First, simplify the fraction: $\frac{15}{1600} = (3 \times 5) / (320 \times 5) = \frac{3}{320}.$ Now the fraction is in its simplest form. Denominator $q = 320$. Prime factorisation of $320 = 32 \times 10 = 2^5 \times 2 \times 5 = 2^6 \times 5^1$. This is of the form $2^n 5^m$, where $n=6$ and $m=1$. Therefore, $15/1600$ has a terminating decimal expansion.

Common Mistakes

1. Confusing HCF and LCM: Students often mix up the rules for choosing powers of prime factors. Remember: HCF takes lowest powers of common factors, LCM takes highest powers of all factors.
2. Incorrect Prime Factorisation: Errors in breaking down numbers into their prime factors will lead to incorrect HCF/LCM.
3. Applying HCF $\times$ LCM $=$ Product of Numbers for Three Numbers: This formula is only valid for two positive integers. Do not use it for three or more numbers.
4. Flaws in Irrationality Proofs:

• Not stating that $a$ and $b$ are coprime at the beginning of the proof. This is crucial for the contradiction.
• Failing to correctly use the theorem: "If $p$ divides $a^2$, then $p$ divides $a$."
• Not clearly showing the contradiction at the end.

5. Not Simplifying Fractions for Decimal Expansion: For decimal expansion problems, if the fraction $p/q$ is not in its simplest form, you might incorrectly conclude it's non-terminating. For example, $6/15 = 2/5$. The denominator of $2/5$ is $5$, which is $2^0 5^1$, so it terminates. If you looked at $15$ (which is $3 \times 5$), you might wrongly say it's non-terminating.
6. Misidentifying Prime Factors in Denominator: When checking for terminating decimals, ensure all prime factors in the denominator are only 2s or 5s. If there's any other prime factor (like 3, 7, 11, etc.), it's non-terminating repeating.

Quick Revision

• Real Numbers include all rational and irrational numbers.
• Rational Numbers can be written as $p/q$. Irrational Numbers cannot.
• Fundamental Theorem of Arithmetic: Every composite number has a unique prime factorisation (ignoring order).
• HCF by Prime Factorisation: Product of the smallest powers of common prime factors.
• LCM by Prime Factorisation: Product of the largest powers of all prime factors.
• HCF$(a, b) \times$ LCM$(a, b) = a \times b$ (for two numbers $a, b$).
• Proof of Irrationality: Uses proof by contradiction. Assume rational, derive a contradiction (e.g., common factors when none should exist).
• Terminating Decimal Expansion: A rational number $p/q$ (in simplest form) has a terminating decimal if the prime factorisation of $q$ is of the form $2^n 5^m$.
• Non-terminating Repeating Decimal Expansion: A rational number $p/q$ (in simplest form) has a non-terminating repeating decimal if the prime factorisation of $q$ has prime factors other than 2 or 5.