Number Systems — Class 9 (CBSE)

Mathematics has only one tool: the number. Every chapter of every class, from now till PhD, uses it. Class 9's Number Systems is where you stop treating numbers as a primary-school formality and start understanding what they actually are.

1. A 4000-year story in five acts

Before you learn the rules, see the story. Each act in this play is a real human moment when someone realised the existing numbers weren't enough.

Act I — Counting. A herder watches his goats wander out in the morning and back at sundown. To check none was lost, he matches each goat to a pebble. Natural numbers $1,2,3,\dots$ are born — for counting things.

Act II — Nothing. Indian mathematicians around 5th century CE write the symbol for zero (and call it shoonya). Counting becomes whole numbers $0,1,2,\dots$. The world's bank balances now make sense.

Act III — Owing. Chinese merchants need to record debts. Negative numbers appear. Combined with the whole numbers, you get integers $\dots ,-3,-2,-1,0,1,2,3,\dots$.

Act IV — Sharing. A baker cuts a loaf for three children. Integers can't describe a third of a loaf. Rational numbers appear — anything that can be written $p/q$ with $q\ne 0$. By 500 BCE, the Pythagoreans believe rationals are all the numbers there are.

Act V — The diagonal that broke a cult. Around 530 BCE, a Pythagorean named Hippasus proves that the diagonal of a unit square — which is $\sqrt{2}$ — cannot be written as $p/q$. Legend says he was drowned for blasphemy. Irrational numbers were so threatening they were murderous. Together with rationals they form the real numbers — every point on the number line.

You are now an heir to all five acts. Class 9 makes you fluent in them.

2. The big picture — one diagram

                  ┌─────────────────────────────┐
                  │     REAL NUMBERS  (ℝ)        │
                  │  every point on the number   │
                  │  line                         │
                  └─────────┬───────────────────┘
                            │
              ┌─────────────┴─────────────┐
              │                            │
        RATIONALS (ℚ)                IRRATIONALS
        p/q form                     non-terminating,
        terminating OR               non-repeating decimals
        repeating decimals           e.g. √2, π, e, √3, √5
              │
       ┌──────┴──────┐
       │             │
  INTEGERS (ℤ)   FRACTIONS
  …−2, −1, 0, 1, 2…
       │
       │
  WHOLE NUMBERS (W)
   0, 1, 2, 3, …
       │
       │
  NATURAL NUMBERS (ℕ)
   1, 2, 3, …

Read it bottom-up — each set is a superset of everything beneath it. Real numbers are the biggest set you'll meet in Class 9. (Class 11 adds complex numbers.)

3. Natural numbers, whole numbers, integers, rationals — recap

These four you've seen before but the language matters in the exam.

• Natural numbers $\mathbb{N}=\{1,2,3,\dots \}$ — counting numbers. No zero.
• Whole numbers $W=\{0,1,2,\dots \}$ — naturals plus zero.
• Integers $\mathbb{Z}=\{\dots ,-2,-1,0,1,2,\dots \}$ — whole numbers and their negatives. (From German Zahlen.)
• Rational numbers $\mathbb{Q}=\{p/q:p,q\in \mathbb{Z},q\ne 0\}$ — numbers expressible as a fraction. (From quotient.)

Key fact. Every integer is rational (take $q=1$). So $\mathbb{N}\subset W\subset \mathbb{Z}\subset \mathbb{Q}$.

4. Between any two rationals there are infinitely many — the squeeze trick

A surprising fact: between any two distinct rationals, no matter how close, you can find another rational. And another. And another. Infinitely many.

The averaging trick. Between rationals $r$ and $s$ (with $r<s$), the mean $(r+s)/2$ is a rational that sits strictly between them.

Worked example. Find a rational between $1/3$ and $1/2$.

Average: $\frac{\frac{1}{3}+\frac{1}{2}}{2}=\frac{\frac{2+3}{6}}{2}=\frac{5/6}{2}=\frac{5}{12}$.

Check: $1/3=4/12$, $1/2=6/12$. Indeed $4/12<5/12<6/12$. ✓

Repeat the trick to find a rational between $1/3$ and $5/12$: average $=9/24=3/8$. And so on. The squeeze never stops.

Why this matters. "Between any two there are infinitely many" is called density of the rationals. It's a deep idea you'll use again in calculus when defining limits.

5. Irrational numbers — the gap in the rationals

If the rationals are so dense, surely every point on the number line is a rational? The Pythagoreans thought so. They were wrong.

Claim. $\sqrt{2}$ is irrational. (It cannot be written as $p/q$.)

Proof (by contradiction — a beautiful classic): Suppose $\sqrt{2}=p/q$ where $p,q$ are integers with no common factor (lowest form), $q\ne 0$. Squaring: $2=p^2/q^2\Rightarrow p^2=2q^2$. So $p^2$ is even $\Rightarrow$ $p$ is even (the square of an odd number is odd). Write $p=2k$. Then $(2k{)}^2=2q^2\Rightarrow 4k^2=2q^2\Rightarrow q^2=2k^2$. So $q^2$ is even $\Rightarrow$ $q$ is even. But then both $p$ and $q$ are even — they share a factor 2 — contradicting our "lowest form" assumption. Hence $\sqrt{2}$ is not rational. $■$

Other irrationals you'll see: $\sqrt{3},\sqrt{5},\sqrt[3]{2},\pi ,e,{\log }_{10}2$, $0.101001000100001\dots$ (made-up non-repeating). Any time a "$\sqrt{\text{non-perfect-square}}$" appears, it's irrational.

Together with the rationals, the irrationals fill in every gap on the number line. The result — every conceivable point on the line — is the set of real numbers $\mathbb{R}$.

6. Decimal expansions — the algorithm that classifies any number

You can always classify a number by its decimal expansion:

Theorem (decimal classification).

A real number is rational if and only if its decimal expansion is either terminating or non-terminating-but-repeating. It is irrational if and only if its expansion is non-terminating and non-repeating.

Converting a repeating decimal to $p/q$ — the standard trick:

Express $0.\overline{27}$ as a fraction.

Let $x=0.\overline{27}=0.27272727\dots$ Multiply by $100$ (because the repeat block has 2 digits): $100x=27.272727\dots$ Subtract: $100x-x=27.27\cdots -0.27\cdots \Rightarrow 99x=27\Rightarrow x=27/99=3/11.$

Check: $3/11=0.272727\dots$. ✓

Rule of thumb. If the repeat block is $n$ digits, multiply by $10^n$ and subtract.

7. When does a fraction give a terminating decimal?

Reduce $p/q$ to lowest terms. Then $p/q$ has a terminating decimal expansion if and only if the prime factorisation of $q$ consists only of $2$s and $5$s (no other prime).

Examples.

• $7/8=7/2^3$ — $q$ has only $2$s → terminating. ($7/8=0.875$) ✓
• $3/20=3/(2^2\cdot 5)$ — only $2$s and $5$s → terminating. ($3/20=0.15$) ✓
• $5/12=5/(2^2\cdot 3)$ — $q$ has a $3$ → non-terminating but repeating. ($5/12=0.41\overline{6}$) ✓
• $1/7=1/7$ — $q$ has a $7$ → non-terminating, repeating. ($1/7=0.\overline{142857}$) ✓

Why? A terminating decimal is just a fraction with a power of 10 in the denominator: $10^n=2^n\cdot 5^n$. If $q$ has any other prime factor, you can't possibly turn it into a power of 10.

8. Locating an irrational on the number line — the spiral construction

You can plot $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, … precisely on the number line using a Pythagorean spiral.

Constructing $\sqrt{2}$.

1. Draw a number line. Mark $O$ at $0$ and $A$ at $1$.
2. At $A$, draw a unit perpendicular up to a point $B$.
3. Join $OB$. By Pythagoras, $OB=\sqrt{1^2+1^2}=\sqrt{2}$.
4. With $O$ as centre and radius $OB$, draw an arc cutting the number line at $P$.
5. The point $P$ corresponds to $\sqrt{2}$.

Constructing $\sqrt{3}$. From $B$ (which is $\sqrt{2}$ away from $O$), draw a unit perpendicular to $OB$ up to $C$. Then $OC=\sqrt{(\sqrt{2}{)}^2+1^2}=\sqrt{3}$. Sweep an arc.

Continue: $\sqrt{4},\sqrt{5},\sqrt{6},\dots$ The spiral is called the spiral of Theodorus.

Tip: in exams you'll often be asked to "represent $\sqrt{n}$ on the number line." Always describe the construction step by step — diagram + procedure = full marks.

9. Operations on real numbers

Real numbers obey all the usual rules:

• Closure — sum, difference, product, quotient of two reals is a real (except division by 0).
• Commutative: $a+b=b+a$; $ab=ba$.
• Associative: $(a+b)+c=a+(b+c)$; $(ab)c=a(bc)$.
• Distributive: $a(b+c)=ab+ac$.
• Identity: $a+0=a$; $a\cdot 1=a$.
• Inverse: $a+(-a)=0$; $a\cdot (1/a)=1$ (for $a\ne 0$).

Rational + irrational?

The sum of a rational and an irrational is irrational. The product of a non-zero rational and an irrational is irrational. But the sum or product of two irrationals can be either rational or irrational.

Examples: $\sqrt{2}+(-\sqrt{2})=0$ (rational!), $\sqrt{2}\cdot \sqrt{2}=2$ (rational!), but $\sqrt{2}+\sqrt{3}$ is irrational.

10. Surds and their laws

A surd is a root of a positive rational that comes out irrational — like $\sqrt{2},\sqrt[3]{5},\sqrt[4]{7}$.

Laws of surds (let $a,b>0$):

$\sqrt{ab}=\sqrt{a}\cdot \sqrt{b},\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}.$

$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b.$

$(\sqrt{a}+\sqrt{b}{)}^2=a+b+2\sqrt{ab}.$

$(a+\sqrt{b})(a-\sqrt{b})=a^2-b.$

These four identities account for ~80% of all surd-manipulation marks in Class 9.

11. Rationalisation — moving the irrational out of the denominator

You cannot leave $\sqrt{2}$ in a denominator. The conventional, marker-approved form moves it to the numerator. This is rationalisation.

Type A — Single surd in the denominator

Multiply numerator and denominator by the same surd.

$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}.$

Type B — Binomial denominator

Multiply numerator and denominator by the conjugate (flip the sign between the two terms).

$\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{\sqrt{5}+\sqrt{2}}\cdot \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}.$

Worked example (exam favourite — 3 marks).

Rationalise $\frac{1}{3+2\sqrt{2}}$.

Multiply top and bottom by the conjugate $3-2\sqrt{2}$:

$\frac{1}{3+2\sqrt{2}}\cdot \frac{3-2\sqrt{2}}{3-2\sqrt{2}}=\frac{3-2\sqrt{2}}{3^2-(2\sqrt{2}{)}^2}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}.$

Why does this work? $(a+b)(a-b)=a^2-b^2$ kills the square root.

12. Laws of exponents (revisited for irrationals)

You learned these for rational exponents in Class 8. They extend cleanly to real exponents.

For $a,b>0$ and any real $p,q$:

$a^p\cdot a^q=a^{p+q},\frac{a^p}{a^q}=a^{p-q},(a^p{)}^q=a^{pq},a^p\cdot b^p=(ab{)}^p.$

And: $a^0=1,a^{-p}=1/a^p,a^{1/n}=\sqrt[n]{a}$.

Worked example. Simplify $(81{)}^{3/4}$.

$(81{)}^{3/4}=(3^4{)}^{3/4}=3^{4\cdot 3/4}=3^3=27$.

Worked example 2. Simplify $\sqrt{2}\cdot \sqrt[3]{2}$.

$\sqrt{2}=2^{1/2},\sqrt[3]{2}=2^{1/3}$. Sum the exponents: $2^{1/2+1/3}=2^{5/6}=\sqrt[6]{2^5}=\sqrt[6]{32}$.

13. Seven worked exam examples

Example 1 — Classify (1 mark)

Is $\sqrt{225}$ rational or irrational? $\sqrt{225}=15$, an integer. Rational. (Always check if the radicand is a perfect square first.)

Example 2 — Insert (2 marks)

Insert three rationals between $1/4$ and $1/3$. Convert to common denominator $12$: $1/4=3/12,1/3=4/12$. Now scale further to $60$: $15/60$ and $20/60$. Three rationals between: $16/60=4/15,17/60,18/60=3/10$.

Example 3 — Decimal to fraction (3 marks)

Express $1.\overline{27}$ as $p/q$. Let $x=1.272727\dots$. Then $100x=127.272727\dots$. Subtract: $99x=126\Rightarrow x=126/99=14/11$.

Example 4 — Rationalise (3 marks)

Rationalise $\frac{1}{\sqrt{7}-\sqrt{6}}$. Multiply by conjugate $\sqrt{7}+\sqrt{6}$: $\frac{\sqrt{7}+\sqrt{6}}{7-6}=\sqrt{7}+\sqrt{6}$.

Example 5 — Exponents (2 marks)

Find the value of $(64{)}^{2/3}$. $64=2^6\Rightarrow (2^6{)}^{2/3}=2^4=16$.

Example 6 — Combined operations (4 marks, HOTS)

If $a=2+\sqrt{3}$ and $b=2-\sqrt{3}$, find $a^2+b^2$ and $ab$.

$ab=(2{)}^2-(\sqrt{3}{)}^2=4-3=1$ (conjugate trick). $a+b=4,a-b=2\sqrt{3}$. $a^2+b^2=(a+b{)}^2-2ab=16-2=14$.

Example 7 — Locate $\sqrt{n}$ (3 marks)

Locate $\sqrt{5}$ on the number line. On the number line, mark $O$ at 0 and $A$ at 2. At $A$, draw a unit perpendicular to $B$. Then $OB=\sqrt{2^2+1^2}=\sqrt{5}$. With $O$ as centre and radius $OB$, draw an arc cutting the number line at $P$. Point $P$ is $\sqrt{5}$.

14. Common pitfalls — the exam-killers

1. Calling $\sqrt{4}$ irrational. $\sqrt{4}=2$ is a rational integer. Always check for perfect squares first.
2. Leaving the denominator irrational. Mark-deductions are automatic.
3. Sign-slip in the conjugate. $(a+b)(a+b)=a^2+2ab+b^2$, NOT $a^2-b^2$. The minus comes from the conjugate $(a-b)$.
4. Multiplying surds incorrectly. $\sqrt{2}+\sqrt{3}\ne \sqrt{5}$ (totally wrong). Only $\sqrt{2}\cdot \sqrt{3}=\sqrt{6}$ works.
5. Forgetting "in lowest terms" before applying the terminating-decimal test. $14/20$ looks like it has only $2,5$ in the denominator, but reduce to $7/10$ first.
6. Confusing "non-terminating" with "irrational." Non-terminating but repeating is still rational ($1/3=0.\overline{3}$).
7. Saying "$\pi =22/7$". $22/7$ is a useful approximation, but $\pi$ itself is irrational (genuine non-repeating).

15. Beyond NCERT — stretch problems

Stretch 1 — Olympiad classic

Prove $\sqrt{3}$ is irrational. Adapt the $\sqrt{2}$ proof: assume $\sqrt{3}=p/q$ in lowest terms. Squaring: $3q^2=p^2$ → $p^2$ divisible by 3 → $p$ divisible by 3. Write $p=3k$. Then $3q^2=9k^2\Rightarrow q^2=3k^2$ → $q$ divisible by 3. Both share factor 3 — contradiction.

Stretch 2 — Density meets the irrationals

Find an irrational number between $0.1$ and $0.11$. $0.101001000100001\dots$ — a number with non-repeating zero patterns lies between $0.1$ and $0.11$.

Stretch 3 — Continued fraction (a peek into Class 12 / college)

$\sqrt{2}$ can be written as a never-ending fraction: $\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots }}}$. The deeper you go, the closer you get to the true $\sqrt{2}$. (You don't need this for exams — just appreciate that even infinite expressions can be precise.)

16. Real-world number systems

• Computers store reals approximately. A float is a rational — there's no way to store $\pi$ exactly in any computer. Hence the famous $0.1+0.2\ne 0.3$ bug.
• Music is irrational. The 12-tone equal temperament scale uses the twelfth root of 2 between consecutive notes — an irrational that makes harmony work.
• The golden ratio $\phi =(1+\sqrt{5})/2\approx 1.618$ is irrational and appears in flower petal counts, the Parthenon's proportions, and stock-market technical analysis.
• $\pi$ appears in probability (yes, prime-distribution problems use $\pi$ even though they have nothing to do with circles), in heat flow, and in quantum mechanics.

17. CBSE exam blueprint

Total marks: typically 6–8 / 80. This chapter is a high-value, easy-to-score chapter — if you've practised, expect ≥ 90% of the marks.

Exam-day strategy.

1. Memorise the seven surd identities (Section 10). They appear in every paper.
2. Always reduce $p/q$ to lowest terms before applying the terminating-decimal test.
3. For "find a^2 + b^2 given a" type questions, the magic trick is $(a+b{)}^2-2ab$ or $(a-b{)}^2+2ab$. Use conjugates.

18. 60-second recap

• Sets: $\mathbb{N}\subset W\subset \mathbb{Z}\subset \mathbb{Q}\subset \mathbb{R}$. Irrationals + rationals = reals.
• Density: between any two rationals, infinitely many rationals.
• $\sqrt{2}$ proof is the classic proof-by-contradiction — memorise the structure.
• Decimal classification: terminating or repeating ⇔ rational; non-repeating ⇔ irrational.
• Terminating-decimal test: in lowest terms, denominator must be $2^a{5}^b$.
• Rationalisation: multiply by conjugate.
• Surd laws: $\sqrt{ab}=\sqrt{a}\sqrt{b}$, $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$.
• Exponent laws extend to all real exponents.

Take the practice quiz and the flashcard deck before moving to Polynomials.