Number Systems

1. The number hierarchy

Numbers are built up in nested layers, each layer adding something new to the previous one.

• Natural numbers N = {1, 2, 3, 4, …} — the counting numbers. The smallest natural number is 1.
• Whole numbers W = {0, 1, 2, 3, …} — naturals plus zero. The only whole number that is not a natural number is 0. Every natural number is a whole number.
• Integers Z = {…, −3, −2, −1, 0, 1, 2, 3, …} — whole numbers plus negatives. Every whole number is an integer. The negative integers are new additions not found in W.
• Rational numbers Q — all numbers of the form p/q where p and q are integers and q ≠ 0. Examples: 3/4, −7/2, 0, −3, 0.25, 0.̅3. Every integer n is rational (write it as n/1). Their decimal expansions either terminate (like 3/4 = 0.75) or eventually repeat (like 1/3 = 0.333… = 0.̄3).
• Irrational numbers — real numbers that are NOT rational. Cannot be written as p/q for any integers p, q. Examples: √2, √3, √5, π. Their decimals are non-terminating and non-recurring. Rationals and irrationals together make up all of R.
• Real numbers R = rationals + irrationals. Every point on the number line corresponds to exactly one real number, and vice versa. Together, rationals and irrationals leave no gaps on the line.

Think of the sets as nested circles (a Venn diagram): N is the innermost circle, then W, then Z, then Q, then R as the outermost. Irrationals fill the gap between Q and R.

Key consequence: if a number is rational it is NOT irrational, and if it is irrational it is NOT rational. There is no overlap. This "no overlap" fact is exactly what powers every irrationality proof later in the chapter (we use it to get a contradiction).

2. Irrational numbers — definition, examples, non-examples

A real number is called irrational if it cannot be expressed in the form p/q where p and q are integers and q ≠ 0. Equivalently, its decimal expansion is non-terminating and non-recurring.

Classic examples of irrational numbers:

• √2 ≈ 1.41421356237… — the length of the diagonal of a unit square. Non-terminating, non-repeating.
• √3 ≈ 1.73205080757… — similarly irrational.
• √5 ≈ 2.23606797749… — irrational.
• π ≈ 3.14159265358… — proved irrational in 1768 by Lambert. Note carefully: 22/7 is only a rational approximation, not the exact value of π. 22/7 ≠ π.
• e ≈ 2.71828182845… — Euler's number, irrational.
• 0.101001000100001… — a constructed non-terminating non-recurring decimal; irrational by definition.
• √2 + √3, √2 − 1, 3√2 — combinations of irrationals and rationals can also be irrational.

Non-examples (things that look irrational but are NOT):

• √4 = 2 — rational (2 is a perfect square)
• √9 = 3 — rational
• √(1/4) = 1/2 — rational
• √0.36 = 0.6 — rational
• √a is irrational only if a is a positive integer that is not a perfect square.

Quick test: Is √n irrational? Check if n is a perfect square (1, 4, 9, 16, 25, …). If yes, √n is rational. If no, √n is irrational.

3. Real numbers and the real number line

Every real number (rational or irrational) corresponds to exactly one point on the number line. Conversely, every point corresponds to exactly one real number. This is called the completeness of the real number system.

The real numbers are also dense: between any two distinct real numbers, there are infinitely many rational numbers and infinitely many irrational numbers. So you can always find more numbers between any two given ones.

Visualising on the number line:

• Integers sit at the tick marks: …, −2, −1, 0, 1, 2, …
• Rationals like 1/2, 3/4, −5/3 sit between the tick marks at exact fractional positions.
• Irrationals like √2 ≈ 1.414… and π ≈ 3.141… sit at points that no fraction can name exactly, but they are still definite, precise positions on the line.

NCERT Theorem 1.1 (stated): For every real number there is a unique point on the number line, and for every point on the number line there is a unique real number. This one-to-one correspondence makes real numbers and geometry deeply connected.

4. Locating irrational numbers on the number line

Rational numbers are easy to locate (divide intervals into equal parts). But can we locate irrationals geometrically? Yes, using the Pythagorean theorem and a compass.

Locating √2 on the number line:

1. Draw a number line. Mark O at 0 and A at 1 so that OA = 1 unit.
2. At A, draw AB perpendicular to the number line with AB = 1 unit.
3. By the Pythagorean theorem: OB² = OA² + AB² = 1 + 1 = 2, so OB = √2.
4. With centre O and radius OB (= √2), draw an arc. Where it meets the number line (to the right of O) is the point P. Then OP = √2.
5. P is the exact location of √2 on the number line.

Locating √3:

1. We have the segment OB = √2 from the previous construction.
2. At B, erect a perpendicular BC = 1 unit.
3. Then OC² = OB² + BC² = 2 + 1 = 3, so OC = √3.
4. Transfer OC to the number line with a compass: this locates √3.

Successive magnification (zooming in): We can also locate irrationals by using their decimal expansion to zoom in on the number line. For √2 = 1.41421…:

• First zoom: it lies between 1.4 and 1.5
• Second zoom: between 1.41 and 1.42
• Third zoom: between 1.414 and 1.415
• Each magnification pins down the location more precisely. The point is unique and definite.

5. Decimal expansions — the full classification

The type of decimal expansion a number has tells you exactly what kind of number it is.

NCERT Theorem: The decimal expansion of a rational number is either terminating or non-terminating recurring, and conversely every such decimal represents a rational number.

NCERT Theorem: The decimal expansion of an irrational number is non-terminating and non-recurring, and conversely every such decimal represents an irrational number.

When does p/q terminate? A rational p/q in lowest terms (HCF of p and q is 1) gives a terminating decimal if and only if the prime factorisation of q has no prime factors other than 2 and 5.

• 7/8 — denominator 8 = 2³, only factor is 2, so terminates: 0.875
• 3/40 — denominator 40 = 2³ × 5, only factors 2 and 5, so terminates: 0.075
• 1/6 — denominator 6 = 2 × 3, factor 3 is present, so non-terminating recurring: 0.1̄6̄
• 1/7 — denominator 7 is prime and not 2 or 5, so non-terminating recurring: 0.̄142857̄

6. Operations on real numbers

Real numbers are closed under +, −, ×, and ÷ (by nonzero). When you mix rationals and irrationals the results follow predictable rules.

Useful surd identities (memorise these):

Adding like surds: c√a + d√a = (c + d)√a. Example: 3√5 + 7√5 = 10√5. But √3 + √5 cannot be simplified further (different surds).

7. Rationalisation of denominators

A fraction with an irrational denominator (like 1/√3 or 5/(2+√3)) is valid but awkward. Rationalisation rewrites it so the denominator is rational, without changing the value. This is always required in board answers.

Case 1: Single surd √a in denominator

Multiply top and bottom by √a:

Case 2: Binomial a ± √b or √a ± √b in denominator

Multiply top and bottom by the conjugate (same expression, opposite sign in the middle).

Step-by-step method for any rationalisation problem:

1. Identify whether the denominator is a single surd or a binomial.
2. Write the conjugate (for binomials) or the surd itself (for single surds).
3. Multiply both numerator and denominator by it.
4. Simplify: the denominator becomes rational; expand the numerator.
5. Collect like terms and write the final answer with rational denominator.

8. Laws of exponents for real numbers

The laws of exponents you know for positive integers extend to ALL real exponents (fractions, decimals, even irrationals), provided the base is a positive real number.

Connecting radicals and fractional exponents:

The seven laws (a, b > 0; m, n are real numbers):

Common mistakes with exponent laws:

• a^m × a^n ≠ a^(mn). It should be a^(m+n). The exponents ADD when you multiply (same base).
• a^m + a^n ≠ a^(m+n). Laws of exponents apply only to products and quotients, NOT sums.
• (ab)^m = a^m × b^m — correct. But (a+b)^m ≠ a^m + b^m in general.
• 2^3 × 3^3 = (2×3)^3 = 6^3 — correct (same exponent 3). Do NOT write it as 6^6.

9. All NCERT exercises — fully solved

Exercise 1.1

Q1. Is zero a rational number? Can you write it in the form p/q where p and q are integers and q ≠ 0?

Yes. 0 = 0/1 = 0/2 = 0/5. We can write 0 as p/q with q = any nonzero integer and p = 0. So 0 is rational.

Q2. Find six rational numbers between 3 and 4.

Write 3 = 21/7 and 4 = 28/7. Six rationals between them: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7.
Alternatively in decimal: 3.1, 3.2, 3.3, 3.4, 3.5, 3.6.

Q3. Find five rational numbers between 3/5 and 4/5.

Convert to denominator 30: 3/5 = 18/30 and 4/5 = 24/30. Five rationals: 19/30, 20/30, 21/30, 22/30, 23/30 (= 2/3).

Q4. State true or false:
(i) Every natural number is a whole number. TRUE (N ⊂ W)
(ii) Every integer is a whole number. FALSE (−1 is integer but not whole)
(iii) Every rational number is a whole number. FALSE (1/2 is rational but not whole)

Exercise 1.2

Q1. State whether rational or irrational:
(i) √(23) — 23 not a perfect square → irrational
(ii) √(225) = 15 → rational
(iii) 0.3796 (terminating) → rational
(iv) 7.478478… = 7.̄478̄ (recurring) → rational
(v) 1.101001000100001… (non-terminating, non-recurring) → irrational

Q2. Simplify:
(i) 5√3 + √3 = 6√3
(ii) 3√2 − 2√2 = √2
(iii) (3 + √3)(2 + √3) = 6 + 3√3 + 2√3 + 3 = 9 + 5√3
(iv) (√5 + √2)² = 7 + 2√10
(v) (√5 − √2)(√5 + √2) = 5 − 2 = 3

Q3. Recall π ≈ 22/7. Is 22/7 = π? No. 22/7 is a rational approximation. π is irrational; 22/7 terminates and is therefore rational. They are different numbers.

Q4. Represent √9.3 on the number line. See the geometric mean construction in Section 4 above.

Exercise 1.3

Q1. Write decimal form and classify:
(i) 36/100 = 0.36 → terminating
(ii) 1/11 = 0.̄09̄ → non-terminating recurring
(iii) 4 and 1/8 = 33/8 = 4.125 → terminating
(iv) 3/13 = 0.̄230769̄ → non-terminating recurring
(v) 2/11 = 0.̄18̄ → non-terminating recurring
(vi) 329/400 = 0.8225 → terminating

Q2. Given 1/7 = 0.̄142857̄, deduce 2/7 through 6/7 without long division:
2/7 = 0.̄285714̄; 3/7 = 0.̄428571̄; 4/7 = 0.̄571428̄; 5/7 = 0.̄714285̄; 6/7 = 0.̄857142̄
(Each is the same block "142857" shifted cyclically.)

Q3. Express as p/q:
0.̄6̄ = 6/9 = 2/3;   0.47̄ = 43/90;   0.̄001̄ = 1/999

Q4. 0.9999… = x → 10x = 9.999… = 9 + x → 9x = 9 → x = 1. So 0.̄9̄ = 1 exactly. Surprising but mathematically correct.

Q5. Maximum digits in repeating block of 1/17: when dividing by 17, remainders are 1–16, so the block repeats in at most 16 digits.

Q6. Terminating decimal ⇔ denominator (in lowest terms) has only 2 and 5 as prime factors.

Q7. Three non-terminating non-recurring decimals: 0.101001000100001…; 0.202002000200002…; 0.312311231112…

Q8. Three irrationals between 2/5 = 0.4 and 3/7 ≈ 0.4285…: e.g. 0.41041004100041…; 0.41141114…; 0.42042004…

Exercise 1.4

Q1. Rational or irrational?
(i) 2 − √5 → irrational (rational − irrational = irrational)
(ii) (3 + √23) − √23 = 3 → rational
(iii) 2√7 / (7√7) = 2/7 → rational (the √7 cancels)
(iv) 1/√2 → irrational
(v) 2π → irrational

Q2. Simplify:
(i) (3+√3)(2+√3) = 9 + 5√3
(ii) (√3 + √7)² = 10 + 2√21
(iii) (√5 − √2)(√5 + √2) = 3
(iv) (√5 − √3)² = 8 − 2√15

Exercise 1.5

Q1. Rationalise the denominator:
(i) 1/√7 = √7/7
(ii) 1/(√7 − √6) = √7 + √6
(iii) 1/(√5 + √2) = (√5 − √2)/3
(iv) 1/(√7 − 2) = (√7 + 2)/3

Q2. Simplify:
(i) 7^(1/3) × 7^(1/3) = 7^(2/3)
(ii) 11^(1/2) / 11^(1/4) = 11^(1/4)
(iii) 7^(1/2) × 8^(1/2) = √56 = 2√14