The World of Numbers - Anshul Academy

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1From Counting to Zero and Integers

2Rational Numbers

3Irrational Numbers

4Real Numbers and Decimal Expansions

What you will learn in this chapter

How the number system grew from counting, to zero, to negative numbers, to fractions
How to add, subtract, multiply and divide rational numbers, and place them on a number line
How to prove that √2 is irrational, and how to construct an irrational length
How to predict whether a fraction’s decimal expansion terminates or repeats, and convert between the two forms

natural numbersintegersrational numbersirrational numbersreal numbersdensityterminating decimalrepeating decimal

1From Counting to Zero and Integers

~35,000 years agoThe Lebombo Bone: 29 tally notches, among the earliest evidence of counting

Vedic eraSanskrit texts name powers of 10 up to

10^{12}

; the Ṛigveda uses powers of 10 for large quantities

628 CEBrahmagupta’s Brāhmasphuṭasiddhānta formally defines zero and gives rules for negative numbers (“debts”)

Learn by heartDefinition 1

Natural numbers $\mathbb{N} = \{1, 2, 3, 4, \ldots\}$ are the counting numbers. They are closed under addition (the sum of two natural numbers is always natural) but not under subtraction ( $3-5$ is not a natural number).

Rules for zero

a + 0 = a

a – 0 = a

a \times 0 = 0

Brahmagupta (628 CE) gave these three. He also wrote

0 \div 0 = 0

, and that one is wrong: division by zero is not defined at all.

Learn by heartDefinition 2

Integers $\mathbb{Z} = \{\ldots, -2, -1, 0, 1, 2, \ldots\}$ combine the natural numbers, zero, and the negative numbers.

Once negative numbers exist, every calculation needs a rule for what sign the answer carries. These five rules cover every case a paper can ask.

Operation	Rule for the sign	Example
Add, both signs the same	Add the two numbers and keep that common sign	$5 + 4 = 9$ $(-5) + (-4) = -9$
Add, signs different	Ignore the signs, subtract the smaller number from the bigger one, then give the answer the bigger number’s sign	$8 + (-3) = 5$ $3 + (-8) = -5$
Subtract	Change the sign of the number being subtracted, then use the addition rules above	$5 – (-3) = 5 + 3 = 8$ $(-5) – 3 = (-5) + (-3) = -8$
Multiply or divide, both signs the same	The answer is positive	$3 \times 4 = 12$ $(-3) \times (-4) = 12$ $(-12) \div (-4) = 3$
Multiply or divide, signs different	The answer is negative	$(-3) \times 4 = -12$ $12 \div (-4) = -3$

Memory Trick

Same signs give +, different signs give −

That shortcut is for multiplying and dividing only. In addition the sign comes from the bigger number, which is why $(-5) + 4 = -1$ but $(-5) \times 4 = -20$ .

Common Mistake

Writing $(-5) + (-4) = -1$ . Both numbers are negative, so they are added and the answer keeps the minus sign: $-9$ . You only subtract when the two signs are different.

2Rational Numbers

Learn by heartDefinition 3

A rational number is any number expressible as $\dfrac{p}{q}$ , where $p, q$ are integers and $q \ne 0$ . The set of rational numbers is denoted $\mathbb{Q}$ .

Real Numbers (R)

Rational (Q)p/q form; includes N, Z, fractions

IrrationalCannot be written as p/q, e.g. √2, π

Every rational number has infinitely many equivalent forms: $-\dfrac{1}{3} = -\dfrac{2}{6} = -\dfrac{3}{9}$ . Its standard form is the one where $p$ and $q$ share no common factor other than 1 (they are co-prime).

Operation	Rule	Example
Equality	$\frac{a}{b}=\frac{c}{d}$ exactly when $ad=bc$	$\frac{2}{3}=\frac{6}{9}$ since $2\times9=3\times6$
Add or subtract, same denominator	$\frac{a}{b}\pm\frac{c}{b}=\frac{a\pm c}{b}$	$\frac{2}{7}+\frac{3}{7}=\frac{5}{7}$
Add or subtract, different denominators	Take the LCM of the denominators first, or use $\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}$	$\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}=\frac{5}{6}$
Multiply	$\frac{a}{b}\times\frac{c}{d}=\frac{ac}{bd}$	$\frac{2}{3}\times\frac{5}{7}=\frac{10}{21}$
Divide	$\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}$ , with $c\ne0$	$\frac{3}{4}\div\frac{2}{5}=\frac{3}{4}\times\frac{5}{2}=\frac{15}{8}$

Closure Property

Rational numbers are closed under addition, subtraction and multiplication, and under division too, as long as we never divide by zero. The result of any of these operations on two rationals is always another rational.

To place $\dfrac{p}{q}$ on the number line, divide the unit interval into $q$ equal parts and count $p$ parts from 0 (right if positive, left if negative).

1/2

3/2

Learn by heartDefinition 4

The absolute value $|x|$ of a rational number is its distance from 0, so $|x| \ge 0$ always. The distance between two rationals $a$ and $b$ on the number line is $|a-b|$ .

Solved Example

Q. Find a rational number between $\dfrac{1}{2}$ and $\dfrac{2}{3}$ .

Take the average: $\dfrac{\frac{1}{2}+\frac{2}{3}}{2} = \dfrac{\frac{3+4}{6}}{2} = \dfrac{7}{12}$ . Since $\dfrac{1}{2} < \dfrac{7}{12} < \dfrac{2}{3}$ , this works.

Did you know?

The averaging trick above never runs out: between any two rational numbers, however close, there is always another one. This property is called density, and it means there are infinitely many rational numbers between any two rationals.

3Irrational Numbers

Learn by heartDefinition 5

An irrational number cannot be written as $\dfrac{p}{q}$ for any integers $p, q$ . Its decimal expansion never terminates and never repeats. Examples: $\sqrt{2}$ , $\sqrt{3}$ , $\pi$ .

A unit square’s diagonal has length $\sqrt{1^2+1^2}=\sqrt{2}$ by the Baudhāyana-Pythagoras theorem. Is $\sqrt2$ rational? We prove it is not, using proof by contradiction: assume the opposite of what we want to prove, and show it leads to a logical impossibility.

Assume

\sqrt{2}=\dfrac{p}{q}

, in lowest terms (

p,q

co-prime)Step 1: the assumption to be contradicted

2 = \dfrac{p^2}{q^2} \Rightarrow 2q^2 = p^2

Step 2-3: square both sides, clear the fraction

p^2

is even

\Rightarrow

p

is even. Let

p=2k

Step 4: a perfect square is even only if its root is even

2q^2=4k^2 \Rightarrow q^2=2k^2

Step 5-6: substitute and simplify

q^2

is even

\Rightarrow

q

is evenStep 7: same reasoning as for p

p

and

q

are both even, contradicting “co-prime”Step 8: the contradiction

Conclusion

Since the assumption leads to a contradiction, it must be false. Therefore $\sqrt2$ cannot be written as $\dfrac{p}{q}$ ; it is irrational.

Irrational lengths can be constructed exactly with ruler and compass, using the same right-triangle idea repeatedly. Each new leg has length 1, perpendicular to the previous hypotenuse:

Figure 1: The square root spiral. Each new unit-length leg, drawn perpendicular to the last hypotenuse, produces the next integer's square root. — **Figure 1**: The square root spiral. Each new unit-length leg, drawn perpendicular to the last hypotenuse, produces the next integer’s square root.

Exam Tip

To mark $\sqrt2$ on a number line itself (not just build the triangle): after constructing $OB=\sqrt2$ as above, open a compass to length $OB$ , place the point at O, and draw an arc that cuts the number line. That crossing point is exactly $\sqrt2$ .

4Real Numbers and Decimal Expansions

Learn by heartDefinition 6

Real numbers $\mathbb{R}$ are the union of all rational and irrational numbers, filling the number line completely with no gaps.

Number set	Symbol	Decimal expansion
Rational	$\mathbb{Q}$	Always terminating or repeating
Irrational	$\mathbb{R}\setminus\mathbb{Q}$	Never terminates, never repeats

Predicting the type without dividing

Write $\dfrac{p}{q}$ in lowest terms. The decimal terminates exactly when the only prime factors of $q$ are 2, 5, or both. Any other prime factor (3, 7, 11, …) in $q$ means the decimal repeats.

Solved Example

Q. Will $\dfrac{7}{20}$ terminate?

$20 = 2^2\times 5$ , only 2s and 5s. So it terminates: $\dfrac{7}{20}=\dfrac{7\times5}{20\times5}=\dfrac{35}{100}=\;$ $0.35$

Converting a repeating decimal back to $\dfrac{p}{q}$ uses the same trick every time: multiply to shift the repeating block, then subtract to cancel it.

Given

$x = 0.4\overline{5} = 0.4555\ldots$ (5 repeats, 4 does not)

Find

$x$ as $\dfrac{p}{q}$

Solution

Non-repeating digit: 1, so multiply by 10: $10x = 4.5\overline{5}$

Repeating block: 1 digit, so multiply again by 10: $100x = 45.5\overline{5}$

Subtract: $100x – 10x = 45.5\overline5 – 4.5\overline5 \Rightarrow 90x = 41 \Rightarrow x = \;$ $\dfrac{41}{90}$

Did you know? Cyclic numbers

$\dfrac{1}{7}=0.\overline{142857}$ . Multiplying $142857$ by $1,2,3,4,5,6$ just cycles the same six digits in the same order: $142857, 285714, 428571, \ldots$ This is called a cyclic number.

Common Mistake

$0.\overline{9} = 0.999\ldots$ is exactly equal to $1$ , not “just below” it. Proof: let $x=0.\overline9$ . Then $10x=9.\overline9$ , so $10x-x=9.\overline9-0.\overline9=9$ , giving $9x=9$ , so $x=1$ .

Every number is either rational (its decimal terminates or repeats) or irrational (its decimal never settles into a pattern). There is no third option.

the single idea to carry out of this chapter

Check yourself before the exam

Can I give the sign rule for adding, subtracting, multiplying and dividing signed numbers without hesitating?
Can I add, subtract, multiply and divide two rational numbers?
Can I reproduce the full proof that √2 is irrational, step by step?
Can I predict terminating vs repeating from the denominator alone, and convert a repeating decimal to p/q?

Quick Revision: read this the night before the exam

N ⊂ Z ⊂ Q ⊂ R; irrational numbers are in R but not in Q
Zero: a+0=a, a-0=a, a×0=0. Signs: same signs multiply to +, different signs to −
Rational = p/q, q≠0; closed under +, −, ×, ÷ (except ÷0)
Density: the average of two rationals always lies strictly between them
√2 is irrational: proof by contradiction, assume p/q in lowest terms, reach p and q both even
Terminates iff denominator’s only prime factors are 2 and/or 5; otherwise repeats
0.999… = 1 exactly, proved algebraically

Practice Questions

2 marksConvert $\dfrac{3}{50}$ and $\dfrac{2}{9}$ to decimals by long division. State which terminates.
3 marksProve that $\sqrt{5}$ is an irrational number.
1 markConvert $0.0125$ to the form $\dfrac{p}{q}$ .
2 marksConvert $2.4\overline{18}$ to the form $\dfrac{p}{q}$ .
2 marksFind 4 rational numbers between $\dfrac{1}{4}$ and $\dfrac{2}{5}$ .
2 marksIf $\dfrac{x}{4}+\dfrac{x}{6}=\dfrac{25}{12}$ , find the rational number $x$ .
2 marksWithout dividing, state whether $\dfrac{17}{40}$ terminates. If it does, how many decimal places?
3 marksA rational number in lowest form has denominator $2^3\times5^2$ . How many decimal places will it have? Explain.
2 marksLet $a$ and $b$ be non-zero rationals with $a+\dfrac{1}{b}=0$ . Is the product $ab$ positive or negative? Justify.
3 marksShow that $\dfrac{a+b}{2}$ always lies between $a$ and $b$ , for any two distinct rationals $a < b$ .

Multiple Choice

1 markWhich of these is irrational?
(a) $\sqrt{16}$ (b) $\sqrt{18}$ (c) $0.3\overline{3}$ (d) $\frac{22}{7}$
1 markThe decimal expansion of $\dfrac{7}{16}$ is:
(a) terminating(b) repeating(c) neither(d) cannot say
1 mark $(-7) \times (-8)$ equals:
(a) –56(b) 56(c) –15(d) 15

Assertion (A): There are infinitely many rational numbers between 0 and 1.
Reason (R): Rational numbers are dense: the average of any two rationals is another rational strictly between them.

Case Study

A student draws a unit square and its diagonal, then uses the diagonal as one leg of the next right triangle, repeating the construction to build a square root spiral, exactly as in Figure 1.

1 markWhat is the length of the diagonal of the first unit square?
1 markIs this length rational or irrational? How do you know?
2 marksWhat length will the hypotenuse of the 4th triangle in the spiral be?

Answer Key

3/50=0.06

(terminates);

2/9=0.\overline{2}

(repeats)

Q2Proof by contradiction, same structure as √2’s proof, using

p^2=5q^2

0.0125 = \dfrac{125}{10000}=\dfrac{1}{80}

x=2.41818\ldots \Rightarrow \dfrac{133}{55}

(using the two-stage multiply-subtract method)

Q5e.g.

0.26, 0.28, 0.30, 0.32

(any 4 values strictly between 0.25 and 0.4)

\dfrac{3x}{12}+\dfrac{2x}{12}=\dfrac{25}{12} \Rightarrow 5x=25 \Rightarrow x=5

40=2^3\times5

, only 2s and 5s, so it terminates, in 3 decimal places

Q83 decimal places (denominator’s higher power is

10^3

after balancing 2s and 5s)

a=-1/b \Rightarrow ab=-1

, so

ab

is negative

Q10Since

a \lt b

, adding

a

to both sides gives

2a \lt a+b

, so

a \lt \frac{a+b}{2}

; similarly

\frac{a+b}{2} \lt b

MCQ 1–3(b) · (a) · (b)

A/RBoth A and R are true, and R is the correct explanation of A

Case Study√2 · irrational, since 2 is not a perfect square (proved by contradiction) · √5