Real Numbers - Class 10 Maths

1. Lemma and Algorithm

Lemma

A lemma is a proven statement used as a stepping stone to prove another major statement or theorem.

Role: Acts as a foundation or auxiliary proposition in formal proofs.

Algorithm

An algorithm is a well-defined sequence of computational steps that gives a procedure for solving a specific class of problems.

Origin: Derived from the name of the 9th-century Persian mathematician Al-Khwarizmi.

2. Euclid's Division Lemma

Statement: Given two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ satisfying:

Euclid's Division Lemma $$a = bq + r, \qquad 0 \le r < b$$

$a$ is the dividend.
$b$ is the divisor.
$q$ is the quotient ($q \ge 0$).
$r$ is the remainder, which is non-negative and strictly smaller than the divisor $b$.
If $r = 0$, then $b$ divides $a$ exactly ($b \mid a$).

3. Euclid's Division Algorithm

This algorithm is a systematic technique used to compute the Highest Common Factor (HCF) of two positive integers $a$ and $b$ (with $a > b$):

Algorithmic Steps to Find HCF(a, b)

Step 1: Apply Euclid's Division Lemma to find $q$ and $r$ such that: $$a = bq + r, \quad 0 \le r < b$$

Step 2:
• If $r = 0$, then $b$ is the $\text{HCF}(a, b)$.
• If $r \neq 0$, apply the division lemma again to divisor $b$ and remainder $r$.

Step 3: Continue the process until the remainder becomes zero. The divisor at this final stage is $\text{HCF}(a, b)$.

4. Fundamental Theorem of Arithmetic

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

Canonical Prime Factorization $$x = p_1^{a_1} \cdot p_2^{a_2} \dots p_k^{a_k}$$ $$\text{where } p_1 < p_2 < \dots < p_k \text{ are distinct primes and } a_i \in \mathbb{N}$$

Factor Tree illustrating unique prime decomposition.

5. Product Relation Between HCF and LCM

For any two positive integers $a$ and $b$:

Fundamental Formula $$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$$

Critical Note for Three Numbers: This relation holds only for two numbers. For three positive integers $p, q, r$: $$\text{HCF}(p, q, r) \times \text{LCM}(p, q, r) \neq p \times q \times r$$

6. Number Systems Classification

Nested hierarchy of the Real Number System.

Set ℕ

Natural Numbers

Counting numbers starting from $1$:

$$\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}$$

Set W

Whole Numbers

Natural numbers along with zero:

$$\mathbb{W} = \{0, 1, 2, 3, 4, \dots\}$$

Set ℤ

Integers

Whole numbers and negatives of naturals:

$$\mathbb{Z} = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$$

Set ℚ

Rational Numbers

Numbers of the form $\frac{p}{q}$ ($p, q \in \mathbb{Z}, q \neq 0$):

$$\mathbb{Q} = \left\{\frac{p}{q} : p, q \in \mathbb{Z}, q \neq 0, \gcd(p, q) = 1\right\}$$

7. Rational Numbers in Decimal Form

Every rational number $\frac{p}{q}$ can be expressed in decimal form, which is either terminating or non-terminating repeating (recurring).

8. Testing Whether Terminating or Non-Terminating

Let $x = \frac{p}{q}$ be a rational number in simplest form (where $p$ and $q$ are co-prime integers, $\gcd(p, q) = 1$):

Condition for Terminating Decimal The prime factorization of $q$ is of the form: $$q = 2^m \cdot 5^n$$ where $m$ and $n$ are non-negative integers ($m, n \in \{0, 1, 2, \dots\}$).

Condition for Non-Terminating Repeating Decimal If the prime factorization of $q$ contains any prime factor other than 2 or 5 (such as 3, 7, 11, etc.), the decimal expansion is non-terminating recurring.

9. Irrational Numbers

A number is called irrational if it cannot be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Decimal Characterization: The decimal expansion of an irrational number is always non-terminating and non-repeating (e.g., $\sqrt{2} = 1.41421356\dots$, $\pi = 3.14159265\dots$, $0.1010010001\dots$).

10. Theorem: If $p$ is prime and $p \mid a^2$, then $p \mid a$

Proof

Given: $p$ is a prime number and $a$ is a positive integer such that $p$ divides $a^2$ ($p \mid a^2$).

To Prove: $p$ divides $a$ ($p \mid a$).

Proof:
1. Let the prime factorization of $a$ be: $$a = p_1 p_2 p_3 \dots p_n$$ where $p_1, p_2, \dots, p_n$ are primes (not necessarily distinct).
2. Squaring both sides: $$a^2 = (p_1 p_2 \dots p_n)(p_1 p_2 \dots p_n) = p_1^2 p_2^2 \dots p_n^2$$ 3. We are given that $p$ divides $a^2$.
4. By the Fundamental Theorem of Arithmetic, the prime factorization of $a^2$ is unique. The only prime factors of $a^2$ are $p_1, p_2, \dots, p_n$.
5. Therefore, $p$ must be one of the prime factors $p_1, p_2, \dots, p_n$.
6. Since $a = p_1 p_2 \dots p_n$ and $p$ is one of these factors, it follows directly that: $$p \text{ divides } a$$ Hence proved.

11. Proof: $\sqrt{2}$ is Irrational

Proof (By Contradiction)

1. Assume, to the contrary, that $\sqrt{2}$ is rational.

2. Then there exist co-prime positive integers $r$ and $s$ ($s \neq 0$) such that: $$\sqrt{2} = \frac{r}{s} \qquad (\gcd(r, s) = 1)$$

3. Squaring both sides: $$2 = \frac{r^2}{s^2} \implies r^2 = 2s^2 \quad \text{--- (1)}$$

4. This shows that $2$ divides $r^2$.
By Theorem 10, since $2$ is prime: $$2 \text{ divides } r \quad \text{--- (2)}$$

5. Since $2 \mid r$, we can write $r = 2k$ for some integer $k$.
Substitute $r = 2k$ into equation (1): $$(2k)^2 = 2s^2 \implies 4k^2 = 2s^2 \implies s^2 = 2k^2$$

6. This means that $2$ divides $s^2$.
Again by Theorem 10: $$2 \text{ divides } s \quad \text{--- (3)}$$

7. From statements (2) and (3), $2$ is a common factor of both $r$ and $s$.
This contradicts our initial fact that $r$ and $s$ are co-prime ($\gcd(r, s) = 1$).

8. Hence, our assumption that $\sqrt{2}$ is rational was false. $$\therefore \sqrt{2} \text{ is irrational.}$$ Hence proved.

12. Proof: $\sqrt{p}$ is Irrational (where $p$ is Prime)

Proof (By Contradiction)

1. Let $p$ be a prime. Assume, to the contrary, that $\sqrt{p}$ is rational.

2. Then $\sqrt{p} = \dfrac{a}{b}$, where $a, b \in \mathbb{Z}^+$ and $\gcd(a, b) = 1$.

3. Squaring both sides: $$p = \frac{a^2}{b^2} \implies a^2 = pb^2 \quad \text{--- (1)}$$

4. Equation (1) implies $p \mid a^2$.
Since $p$ is prime, $p \mid a$.

5. Let $a = pc$ for some integer $c$. Substituting into (1): $$(pc)^2 = pb^2 \implies p^2c^2 = pb^2 \implies b^2 = pc^2$$

6. This implies $p \mid b^2$, which in turn implies $p \mid b$.

7. Thus, $p$ is a common factor of both $a$ and $b$, contradicting $\gcd(a, b) = 1$.

8. Hence, the assumption is false. $$\therefore \sqrt{p} \text{ is irrational for every prime } p.$$ Hence proved.

13. Proof: If $a$ is Rational and $\sqrt{b}$ is Irrational, then $a + \sqrt{b}$ is Irrational

Proof (By Contradiction)

1. Given $a \in \mathbb{Q}$ and $\sqrt{b} \notin \mathbb{Q}$.
Assume, to the contrary, that $x = a + \sqrt{b}$ is rational ($x \in \mathbb{Q}$).

2. Isolating the radical term $\sqrt{b}$: $$\sqrt{b} = x - a$$

3. Since $x$ is rational (by assumption) and $a$ is rational (given), their difference $(x - a)$ must also be rational (since $\mathbb{Q}$ is closed under subtraction).

4. This implies that $\sqrt{b}$ is rational: $$\sqrt{b} \in \mathbb{Q}$$

5. But this directly contradicts the given statement that $\sqrt{b}$ is irrational.

6. Therefore, our assumption that $a + \sqrt{b}$ is rational is incorrect. $$\therefore a + \sqrt{b} \text{ is irrational.}$$ Hence proved.

14. Proof: If $a$ is Non-Zero Rational and $\sqrt{b}$ is Irrational, then $a\sqrt{b}$ is Irrational

Proof (By Contradiction)

1. Given $a \in \mathbb{Q}$ ($a \neq 0$) and $\sqrt{b} \notin \mathbb{Q}$.
Assume, to the contrary, that $y = a\sqrt{b}$ is rational ($y \in \mathbb{Q}$).

2. Since $a \neq 0$, we can divide both sides by $a$: $$\sqrt{b} = \frac{y}{a}$$

3. Because $y$ is rational and $a$ is a non-zero rational, the quotient $\dfrac{y}{a}$ must be rational (since non-zero rationals are closed under division).

4. This requires $\sqrt{b}$ to be rational, which contradicts the given fact that $\sqrt{b}$ is irrational.

5. Hence, our assumption that $a\sqrt{b}$ is rational is false. $$\therefore a\sqrt{b} \text{ is irrational.}$$ Hence proved.