REAL NUMBERS

Real numbers, denoted by \( \mathbb{R} \), mein rational aur irrational numbers dono shamil hote hain.

Number System Hierarchy

• Natural Numbers (\( \mathbb{N} \)): Counting numbers. \({1, 2, 3, ...}\)
• Whole Numbers (\( \mathbb{W} \)): Natural numbers + zero. \({0, 1, 2, 3, ...}\)
• Integers (\( \mathbb{Z} \)): Whole numbers + negative natural numbers. \({..., -3, -2, -1, 0, 1, 2, 3, ...}\)
• Rational Numbers (\( \mathbb{Q} \)): Numbers jo \(p/q\) form mein likhe ja sakte hain, jahan \(p, q\) integers hain aur \(q \neq 0\). Decimal expansion ya toh terminating hota hai ya non-terminating repeating.
• Examples: \(1/2 = 0.5\) (terminating), \(1/3 = 0.333...\) (non-terminating repeating), \(5 = 5/1\).
• Irrational Numbers: Numbers jo \(p/q\) form mein nahi likhe ja sakte. Decimal expansion non-terminating non-repeating hota hai.
• Examples: \( \sqrt{2}, \sqrt{3}, \pi, 0.101101110... \).

Key Properties of Real Numbers

• Closure Property: Addition, subtraction, multiplication, division (by non-zero) ke under closed hain.
• Commutative Property: \(a+b = b+a\) aur \(a \times b = b \times a\).
• Associative Property: \((a+b)+c = a+(b+c)\) aur \((a \times b) \times c = a \times (b \times c)\).
• Distributive Property: \(a \times (b+c) = a \times b + a \times c\).
• Identity Elements: \(0\) for addition (additive identity), \(1\) for multiplication (multiplicative identity).
• Inverse Elements: \(-a\) for addition (additive inverse), \(1/a\) for multiplication (multiplicative inverse, \(a \neq 0\)).

Euclid's Division Lemma (EDL)

Given positive integers \(a\) and \(b\), there exist unique integers \(q\) and \(r\) satisfying \(a = bq + r\), where \(0 \le r < b\).

• Yahan, \(a\) hai dividend, \(b\) hai divisor, \(q\) hai quotient, aur \(r\) hai remainder.
• EDL ek statement hai jo prove ho chuka hai aur dusre statements ko prove karne ke liye use hota hai.

Euclid's Division Algorithm (EDA)

Yeh ek technique hai do positive integers ke HCF (Highest Common Factor) ko compute karne ke liye.

Steps to find HCF of two positive integers \(c\) and \(d\) (where \(c > d\)) using EDA:

1. Step 1: Apply Euclid's Division Lemma to \(c\) and \(d\). Find whole numbers \(q\) and \(r\) such that \(c = dq + r\), where \(0 \le r < d\).
2. Step 2:

• Agar \(r = 0\) hai, toh \(d\) hi HCF hai.
• Agar \(r \neq 0\) hai, toh divisor \(d\) aur remainder \(r\) par EDL apply karo.

1. Step 3: Process ko tab tak continue karo jab tak remainder zero na ho jaye. Jis stage par remainder zero hota hai, us stage ka divisor hi HCF hota hai.

Applications of EDA

• HCF nikalna.
• Integers ki properties ko prove karna, jaise ki 'every positive even integer is of the form \(2q\)' ya 'every positive odd integer is of the form \(2q+1\)'.

Statement

Har composite number ko primes ke product ke roop mein uniquely express kiya ja sakta hai, irrespective of the order of the prime factors.

• Composite Number: Ek positive integer jiske do se zyada factors hote hain (e.g., 4, 6, 8, 9, 10...). Prime numbers composite nahi hote.
• Prime Number: Ek positive integer jiske exactly do factors hote hain: 1 aur woh number khud (e.g., 2, 3, 5, 7, 11...).

Example:

• \(12 = 2 \times 2 \times 3 = 2^2 \times 3\)
• \(90 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5\)

Applications of FTA

1. HCF aur LCM nikalna (Prime Factorisation Method)

• HCF (Highest Common Factor): Involved prime factors ki sabse chhoti power ka product.
• LCM (Least Common Multiple): Involved prime factors ki sabse badi power ka product.

Formula for two positive integers \(a\) and \(b\):

\(HCF(a, b) \times LCM(a, b) = a \times b\)

• Note: Yeh formula sirf do numbers ke liye valid hai, teen ya zyada numbers ke liye nahi.

2. Irrational numbers ko prove karna

• FTA ka use karke \( \sqrt{2}, \sqrt{3}, \sqrt{5} \) jaise numbers ko irrational prove kiya ja sakta hai (Proof by Contradiction).

3. Rational numbers ke decimal expansions ko determine karna

• Agar ek rational number \(p/q\) (jahan \(p\) aur \(q\) coprime hain) ka denominator \(q\) ke prime factors sirf \(2\) aur \(5\) hain (i.e., \(q = 2^m 5^n\) form ka hai), toh uska decimal expansion terminating hoga.
• Agar \(q\) ke prime factors mein \(2\) aur \(5\) ke alawa koi aur prime factor hai, toh uska decimal expansion non-terminating repeating hoga.

Definition

Ek number \(s\) ko irrational kehte hain agar usse \(p/q\) ke form mein express nahi kiya ja sakta, jahan \(p\) aur \(q\) integers hain aur \(q \neq 0\).

• Inke decimal expansions non-terminating aur non-repeating hote hain.

Proving Irrationality (Proof by Contradiction)

Yeh ek standard method hai jo board exams mein frequently pucha jaata hai.

General Steps to prove \( \sqrt{p} \) is irrational (where \(p\) is a prime number):

1. Assume the opposite: Maan lo \( \sqrt{p} \) rational hai. Toh isse \(a/b\) ke form mein likha ja sakta hai, jahan \(a\) aur \(b\) coprime integers hain (i.e., unka HCF 1 hai) aur \(b \neq 0\).
2. Square both sides: \( \sqrt{p} = a/b \implies p = a^2/b^2 \implies pb^2 = a^2 \).
3. Deduction 1: Iska matlab \(p\) divides \(a^2\). Fundamental Theorem of Arithmetic ke according, agar \(p\) ek prime number hai aur \(p\) divides \(a^2\), toh \(p\) must divide \(a\). So, \(a = pk\) for some integer \(k\).
4. Substitute and Deduce 2: \(pb^2 = (pk)^2 \implies pb^2 = p^2k^2 \implies b^2 = pk^2 \).

Iska matlab \(p\) divides \(b^2\). Again, FTA ke according, \(p\) must divide \(b\).

1. Contradiction: Humne shuru mein assume kiya tha ki \(a\) aur \(b\) coprime hain (unka koi common factor nahi hai except 1). Lekin humne prove kar diya ki \(p\) dono \(a\) aur \(b\) ka common factor hai. Yeh ek contradiction hai.
2. Conclusion: Hamari initial assumption galat thi. Isliye, \( \sqrt{p} \) irrational hai.

Proving \(a + b\sqrt{p}\) is irrational:

• Isme bhi contradiction method use hota hai. Assume karo ki \(a + b\sqrt{p}\) rational hai, toh isse \(x\) ke equal rakho (jahan \(x\) rational hai).
• \(a + b\sqrt{p} = x \implies b\sqrt{p} = x - a \implies \sqrt{p} = (x-a)/b\).
• Right hand side ek rational number hai (since \(x, a, b\) rational hain aur \(b \neq 0\)).
• Lekin hum jaante hain ki \( \sqrt{p} \) irrational hai. Irrational = Rational, jo ki impossible hai.
• Hence, hamari assumption galat thi, aur \(a + b\sqrt{p}\) irrational hai.

Terminating Decimal Expansions

Ek rational number \(x = p/q\) (jahan \(p\) aur \(q\) coprime hain) ka decimal expansion terminating hoga agar \(q\) ka prime factorisation \(2^m 5^n\) ke form ka hai, jahan \(m\) aur \(n\) non-negative integers hain.

• Example: \(3/8 = 3/2^3 = 3/(2^3 \times 5^0)\). Denominator \(2^m 5^n\) form ka hai, toh terminating decimal \(0.375\) hoga.
• Example: \(7/20 = 7/(2^2 \times 5^1)\). Denominator \(2^m 5^n\) form ka hai, toh terminating decimal \(0.35\) hoga.

Non-terminating Repeating Decimal Expansions

Ek rational number \(x = p/q\) (jahan \(p\) aur \(q\) coprime hain) ka decimal expansion non-terminating repeating hoga agar \(q\) ka prime factorisation \(2^m 5^n\) ke form ka nahi hai, matlab \(q\) ke prime factors mein \(2\) aur \(5\) ke alawa koi aur prime number bhi hai.

• Example: \(1/3\). Denominator \(3\) hai, jo \(2^m 5^n\) form ka nahi hai. Decimal expansion \(0.333...\) (non-terminating repeating).
• Example: \(1/7\). Denominator \(7\) hai. Decimal expansion \(0.142857142857...\) (non-terminating repeating).
• Example: \(1/12 = 1/(2^2 \times 3)\). Denominator mein \(3\) bhi hai. Decimal expansion \(0.08333...\) (non-terminating repeating).

Without Actual Division

Board exams mein aksar bina actual division ke decimal expansion ka type pucha jaata hai. Iske liye sirf denominator ke prime factors ko check karna hota hai.

Steps:

1. Rational number ko \(p/q\) form mein likho, jahan \(p\) aur \(q\) coprime hon (agar common factors hain toh simplify karo).
2. Denominator \(q\) ka prime factorisation karo.
3. Agar \(q\) sirf \(2\) aur \(5\) ke powers mein hai (i.e., \(2^m 5^n\) form), toh terminating.
4. Agar \(q\) mein \(2\) aur \(5\) ke alawa koi aur prime factor bhi hai, toh non-terminating repeating.

Irrational numbers jaise \( \sqrt{2}, \sqrt{3}, \sqrt{5} \) ko number line par represent karne ke liye Pythagoras Theorem aur compass ka use kiya jaata hai.

Representation of \( \sqrt{2} \) on the Number Line

1. Number line par point O ko 0 mark karo aur point A ko 1 mark karo. Toh OA = 1 unit.
2. Point A par OA ke perpendicular ek line segment AB draw karo, jahan AB = 1 unit ho.
3. O aur B ko join karo. Triangle OAB ek right-angled triangle hai.
4. Pythagoras Theorem se, \(OB^2 = OA^2 + AB^2 = 1^2 + 1^2 = 1 + 1 = 2\). So, \(OB = \sqrt{2}\) units.
5. Compass ko O par rakho aur pencil ko B par. Compass ko rotate karo taaki woh number line ko intersect kare. Jis point par intersect karega, woh point \( \sqrt{2} \) ko represent karta hai.

Representation of \( \sqrt{3} \) on the Number Line

1. Pehle \( \sqrt{2} \) ko number line par represent karo (jaise upar bataya gaya hai). Let this point be B.
2. Point B par OB ke perpendicular ek line segment BC draw karo, jahan BC = 1 unit ho.
3. O aur C ko join karo. Triangle OBC ek right-angled triangle hai.
4. Pythagoras Theorem se, \(OC^2 = OB^2 + BC^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3\). So, \(OC = \sqrt{3}\) units.
5. Compass ko O par rakho aur pencil ko C par. Compass ko rotate karo taaki woh number line ko intersect kare. Jis point par intersect karega, woh point \( \sqrt{3} \) ko represent karta hai.

Representation of \( \sqrt{x} \) for any positive real number \(x\) (Spiral Method)

• Yeh method \( \sqrt{4}, \sqrt{5}, \sqrt{6} \) etc. ko represent karne ke liye use hota hai, jahan har naye square root ko pichle wale se construct kiya jaata hai.
• General Idea: Agar humne \( \sqrt{n} \) ko number line par represent kar liya hai, toh us point par 1 unit ka perpendicular bana kar aur hypotenuse draw karke \( \sqrt{n+1} \) ko represent kar sakte hain.
• Hypotenuse \( = \sqrt{(\sqrt{n})^2 + 1^2} = \sqrt{n+1} \).

Successive Magnification

Yeh process real numbers ke decimal representation ko visualize karne ke liye use hota hai. Isse hum number line par numbers ko zoom in karke unki exact position dekh sakte hain.

• Example: \(3.765\) ko number line par represent karna.

1. Pehle \(3\) aur \(4\) ke beech dekho.
2. Phir \(3.7\) aur \(3.8\) ke beech zoom karo.
3. Phir \(3.76\) aur \(3.77\) ke beech zoom karo.
4. Ant mein, \(3.765\) ko locate karo.

• Yeh method terminating decimals ke liye useful hai. Non-terminating decimals ke liye, hum sirf unke approximate values ko hi represent kar sakte hain.