Rational and Irrational Numbers | ICSE Class 9 Maths Notes

This note covers rational and irrational numbers, their place among real numbers, decimal expansions, fractions on the number line, constructions of irrational lengths, operations with surds, rationalisation of denominators, and proofs of the irrationality of the square roots of two, three and five.

How do rational and irrational numbers fit into the number system?

Natural numbers are the counting numbers 1, 2, 3 and so on. Whole numbers include these numbers and zero. Integers include the positive counting numbers, zero and the corresponding negative numbers. A collection of numbers is called a set.

Definition: A rational number can be expressed as p/q, where p and q are integers and q ≠ 0. Here p is the numerator, q is the denominator, the slash means division, and ≠ means “is not equal to”.

An irrational number cannot be expressed as such a ratio of integers. Rational and irrational numbers together form the real numbers. Every rational number is real, and every irrational number is real, but no number belongs to both of these two groups.

Why are integers rational?

An integer can be written with denominator 1. For example, 5 = 5/1 and −10 = −10/1. The symbol = means “is equal to”, and the sign − denotes a negative number or subtraction, according to its position.

The test is whether a suitable integer ratio exists, rather than whether a number is already displayed as a fraction. Changing how a number is written does not change the number itself or its classification.

How do equivalent fractions and rational arithmetic work?

Equivalent fractions represent the same number. Multiplying or dividing their numerator and denominator by the same non-zero number preserves their value. A factor of an integer divides it exactly; a common factor divides each of the integers being considered.

Two integers are coprime when their only common positive factor is 1. A fraction is in lowest terms when its numerator and denominator are coprime. This condition becomes essential when proving that a square root is irrational.

Worked example 1. Express 12/30 in lowest terms.

Answer: Both 12 and 30 have the common factor 6. Dividing numerator and denominator by 6 gives 12/30 = 2/5. The integers 2 and 5 have no common positive factor other than 1, so the resulting fraction is in lowest terms.

Property: Rational numbers are closed under arithmetic, with a division condition

Closure means that an operation on members of a set produces a member of the same set. Adding, subtracting or multiplying rational numbers gives a rational number. Dividing one rational number by another also gives a rational number, provided the divisor is non-zero.

For addition or subtraction, first use a common denominator, meaning the same denominator in both fractions. Add or subtract the numerators and retain that denominator. The signs + and × mean addition and multiplication; ÷ means division.

Worked example 2. Find the sum 2/5 + 3/10.

Answer: Rewrite 2/5 as 4/10 by multiplying its numerator and denominator by 2. Then 2/5 + 3/10 = 4/10 + 3/10 = 7/10. The result is rational because both 7 and 10 are integers and the denominator is non-zero.

How are rational numbers represented on the number line?

A number line is a straight line on which numbers are placed in order using a fixed unit length. The point labelled zero is the origin. Positive numbers lie to its right, while negative numbers lie to its left.

A unit interval is the distance between consecutive integers. To represent a positive fraction with positive denominator, divide each unit into as many equal parts as the denominator specifies. Count the number of parts specified by the numerator from zero.

How do quarter units locate a fraction?

1. Choose zero and one on the number line, fixing a unit length.
2. Divide the interval from zero to one into four equal parts.
3. Count three of these parts to the right of zero.
4. Label the resulting point 3/4, keeping the quarter-unit spacing uniform.

The line has labelled points 0, 1/4, 1/2, 3/4 and 1. Consecutive labelled points are equally spaced, showing the division of one unit into four equal parts.

See Fig. 3.5 in your NCERT textbook

For −3/4, count three quarter-unit parts to the left of zero. For 9/4, recognise that nine quarters equal two whole units and one quarter. The point is therefore one quarter of a unit to the right of 2, between 2 and 3.

Property: There are infinitely many rational numbers between distinct rational numbers

Let a and b denote two distinct rational numbers. Their average, calculated by adding them and dividing by 2, is (a + b)/2. It is rational and lies between them. Applying the same process repeatedly produces further intermediate rational numbers.

For example, 3/2 lies between 1 and 2. The average of 1 and 3/2 is 5/4, which lies between those two numbers. Having infinitely many rational points between rational endpoints does not mean that every point on the line is rational.

How do decimal expansions distinguish rational and irrational numbers?

A decimal expansion expresses a number using decimal place values. A terminating decimal ends after finitely many decimal places. A non-terminating recurring decimal continues indefinitely, with a fixed digit or block of digits repeating from some point onwards.

Result: Rational decimals terminate or recur

The decimal expansion of a rational number is terminating or non-terminating recurring. The decimal expansion of an irrational number is non-terminating and non-recurring. “Non-recurring” means that no fixed block eventually repeats indefinitely; it does not mean that individual digits cannot reappear.

The notation … means that the displayed sequence continues. Whenever it follows a recurring decimal here, the repeating block is stated explicitly. A finite list of digits without a continuation rule cannot establish that a number is irrational.

Why does rational long division repeat?

A remainder is the amount left after a division step. In dividing by a fixed positive integer, there are finitely many possible remainders. If a remainder becomes zero, the decimal terminates. If a non-zero remainder repeats, the subsequent division steps repeat too.

For example, during the non-terminating division for 1/7, the possible remainders are 1, 2, 3, 4, 5 and 6. A remainder must recur, producing the repeating block 142857. Thus an infinite decimal need not be irrational.

Note: In lowest terms with a positive denominator greater than 1, a rational fraction terminates precisely when its denominator has no prime factors other than 2 and 5. A prime number is an integer greater than 1 whose only positive factors are 1 and itself.

How can terminating and recurring decimals be converted into fractions?

For a terminating decimal, use its place value to write an integer numerator over a power of ten. A power represents repeated multiplication: 10² means 10 × 10. The raised 2 is the exponent, indicating two factors of 10.

Worked example 3. Convert 0.35 into a fraction in lowest terms.

Answer: There are two decimal places, so 0.35 = 35/100 = 7/20. Dividing both numerator and denominator by 5 reduces the fraction without changing its value. The result is an integer ratio with a non-zero denominator.

How is a pure recurring decimal converted?

A pure recurring decimal begins repeating immediately after the decimal point. Introduce a letter for its value, multiply by a power of ten matching the length of the repeating block, and subtract the original equation. The identical recurring tails then cancel.

Worked example 4. Convert 0.454545…, with 45 repeating indefinitely, into a fraction.

Answer: Let x denote the decimal's value. Then x = 0.454545… and 100x = 45.454545…, where 100x means 100 × x. Subtracting gives 99x = 45. Therefore x = 45/99 = 5/11.

The multiplier is 100 because the repeating block has two digits. Multiplication moves one complete block to the left of the decimal point. Align the equations by decimal place before subtracting so that the identical infinite tails are paired correctly.

How is a mixed recurring decimal converted?

A mixed recurring decimal has a non-repeating beginning after the decimal point, followed by a recurring block. First move the non-repeating digits to the left of the decimal point. Then move one full repeating block and subtract these two shifted equations.

Worked example 5. Convert 0.16666…, in which only 6 repeats, into a fraction.

Answer: Let x denote this decimal. Then 10x = 1.6666… and 100x = 16.6666…. Subtraction gives 100x − 10x = 15, so 90x = 15. Hence x = 15/90 = 1/6.

The non-repeating digit 1 must be treated separately from the recurring digit 6. In both recurring examples, subtraction eliminates an identical continuing decimal part. The resulting equation has integer quantities and can be solved to give the required rational fraction.

How can irrational square roots be located exactly on the number line?

The symbol √ denotes the principal square root: the non-negative number whose square equals the number inside the sign. A square means a number multiplied by itself. Thus √2 is positive and (√2)² = 2, where brackets group an expression.

An irrational length can be constructed geometrically without replacing it by a terminating decimal. A right-angled triangle has one right angle, measuring 90 degrees. Its hypotenuse is the side opposite that right angle.

How does the construction of √2 work?

Pythagoras' theorem states that, in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. Two perpendicular sides of length 1 unit therefore give a hypotenuse whose square is 2 square units.

1. Mark O as the origin and A as the point one unit to its right. The notation OA denotes the segment from O to A or its length, as appropriate.
2. At A, draw a perpendicular, meaning a line meeting OA at a right angle.
3. Mark B on this perpendicular so that AB is one unit, then join O to B. Pythagoras' theorem gives OB² = 1² + 1² = 2, so OB = √2 units.
4. With centre O and radius OB, draw a compass arc meeting the positive number line at P. A radius is the distance from the centre to the arc; thus OP = OB = √2 units.

The drawing shows a unit square with vertices O, A, B and C, diagonal OB labelled √2, and a dashed arc from B to P on the number line. P lies between 1 and 2.

See Fig. 3.11 in your NCERT textbook

For √3, use perpendicular sides of lengths √2 and 1; for √5, use lengths 2 and 1. Their squared hypotenuses are respectively 3 and 5. Transfer each hypotenuse from the origin to the positive number line using a compass arc.

These constructions locate exact lengths. Writing only a few decimal digits gives a rational approximation to an irrational value. The point defined by the full irrational number is not replaced by that approximation.

What are surds, and how are they simplified?

A surd is an irrational root of a positive rational number. In square-root calculations, the number inside the root sign is the radicand. A perfect square is the square of an integer, so its principal square root is an integer.

The numbers √2, √3 and √5 are surds. By contrast, √81 = 9 is rational, so the presence of a root sign alone does not make an expression a surd. Simplify the root before deciding the number's classification.

Identity: Products and quotients of positive square roots

An identity is an equality valid for every permitted value of its letters. Let a and b denote positive real numbers. Then √a × √b = √(ab) and √a/√b = √(a/b). Juxtaposition, as in ab, means multiplication.

To simplify a square root, separate a perfect-square factor from the radicand. Taking its square root moves an integer factor outside the radical. A number multiplying a surd is its coefficient; for instance, the coefficient in 2√3 is 2.

Worked example 6. Simplify √10 × √15.

Answer: Using the product identity gives √10 × √15 = √150. Since 150 = 25 × 6, this becomes √25 × √6. Therefore √10 × √15 = 5√6. The perfect-square factor 25 has been removed from inside the root.

How are like surds combined?

Like surds have the same root part after simplification. Add or subtract their coefficients, retaining that common root part. This follows the distributive law: multiplying a sum gives the sum of the separate products.

Worked example 7. Simplify 2√3 + √3.

Answer: The root part √3 is common to both terms. Their coefficients are 2 and 1, so 2√3 + √3 = (2 + 1)√3 = 3√3. Adding coefficients preserves the root part; it does not add the radicands.

What happens when rational and irrational numbers are combined?

Let r denote a rational number and s an irrational number. The sum r + s and the difference r − s are irrational. If their result were rational, rearranging the equality would express s as a difference of rational numbers, contradicting its irrationality.

Result: Multiplication requires a non-zero rational factor

If r is a non-zero rational number, the product rs and quotient r/s are irrational. Keep the non-zero condition: multiplying an irrational number by zero gives zero, which is rational. The irrational denominator s cannot be zero because zero is rational.

Two irrational numbers can behave differently. Their sum or product may be rational. For example, the irrational numbers 3 + √2 and 3 − √2 have sum 6 and product 7. The irrational parts cancel in the sum and through the difference-of-squares identity in the product.

Identity: Conjugate expressions give a difference of squares

For real numbers u and v, (u + v)(u − v) = u² − v². The letters u and v represent the two terms. Expressions with the same two terms but opposite signs between them are called conjugates.

Multiplying out explains the identity: the two cross-products, uv and −uv, cancel. This cancellation makes conjugates useful when simplifying surd products and when removing square roots from a denominator.

Worked example 8. Simplify (3√5 − 5√2)(4√5 + 3√2).

Answer: Multiply every term in the first bracket by every term in the second. The result is 60 + 9√10 − 20√10 − 30. Combine the rational terms and then the like surds to obtain 30 − 11√10.

The expression in this example is not a conjugate pair, so its root terms do not cancel completely. Expanding all four products before collecting terms avoids losing a cross-product or changing the sign of a subtracted term.

How is a denominator rationalised?

Rationalisation rewrites a fraction so that its denominator is rational. Multiply the numerator and denominator by the same suitable non-zero expression. This preserves the fraction's value because the multiplier divided by itself equals 1.

A rationalising factor is a non-zero expression which, when multiplied by the given denominator, makes the product rational. With a single square-root term, multiplication by the same root is sufficient. With two terms, use a suitable conjugate.

How is a single surd removed?

Worked example 9. Rationalise the denominator of 2/(3√3).

Answer: Multiply numerator and denominator by √3. This gives 2√3/(3 × √3 × √3) = 2√3/(3 × 3). Thus 2/(3√3) = 2√3/9. The denominator is now the rational integer 9.

Only the denominator needs to become rational. A root may remain in the numerator, and the value of the entire fraction may still be irrational. Rationalisation changes the form of the expression, rather than changing the number represented.

How does a conjugate remove a two-term denominator?

Worked example 10. Rationalise the denominator of 1/(√7 − 2).

Answer: The conjugate of √7 − 2 is √7 + 2. Multiply both numerator and denominator by it. The denominator becomes (√7 − 2)(√7 + 2) = 7 − 4 = 3. Hence 1/(√7 − 2) = (√7 + 2)/3.

The minus sign in the denominator determines the plus sign in the rationalising factor. Preserve the entire original numerator during multiplication. The new denominator is a difference of squares, so the square applies to each whole term, including any coefficient.

Worked example 11. Rationalise and simplify 6/(3√2 − 2√3).

Answer: Multiply numerator and denominator by 3√2 + 2√3. The denominator is (3√2)² − (2√3)² = 18 − 12 = 6. The fraction becomes 6(3√2 + 2√3)/6, giving 3√2 + 2√3.

In this last example, cancellation of the common factor 6 completes the simplification. Cancelling a factor is valid because it multiplies the whole numerator and denominator. Individual terms joined by addition or subtraction cannot be cancelled in this way.

What reasoning is needed before proving a square root irrational?

A proof by contradiction starts by assuming the opposite of the desired conclusion. Valid steps then lead to an impossibility or a statement conflicting with the assumption. The assumption must therefore be false, establishing the desired result.

Theorem: A prime dividing a square divides its positive integer base

Let p denote a prime number and a a positive integer. If p divides a², then p divides a. Here divides means “is an exact factor of”. The letter a is the base of the square a².

This result follows from unique prime factorisation: a composite positive integer, meaning one greater than 1 that is not prime, has a unique expression as a product of primes, apart from their order. Squaring repeats its prime factors rather than introducing a new prime.

Thus, if a prime occurs as a factor of a², it must already occur in a. The condition that p is prime matters. In the coming proofs, the relevant primes are 2, 3 and 5.

Why must the assumed fraction be in lowest terms?

Suppose the positive square root under discussion were rational. It could then be written a/b with a and b positive coprime integers. Positivity is possible because the root is positive; b is necessarily non-zero.

The aim is to force a common prime factor into both a and b. That contradicts coprimality. Without the initial lowest-terms condition, merely finding a common factor would not be a contradiction, because an unreduced rational fraction may have common factors.

A complete argument therefore needs the assumption, the squared equation, divisibility of the numerator, substitution, divisibility of the denominator, and the final contradiction. Giving a long decimal approximation supplies none of these logical steps.

How do we prove that √2 is irrational?

Theorem: The square root of two is irrational

Assume, for contradiction, that √2 is rational. Let a and b be positive coprime integers such that √2 = a/b. This notation names the numerator and denominator of a fraction already reduced to lowest terms.

1. Multiply by b to obtain b√2 = a. Squaring both sides gives 2b² = a².
2. The equation shows that 2 divides a². Since 2 is prime, it also divides a. Therefore a is even, meaning divisible by 2.
3. Write a = 2c, where c is a positive integer. Substitution gives 2b² = 4c².
4. Divide by 2 to obtain b² = 2c². Hence 2 divides b² and, by the same prime-divisibility theorem, 2 divides b.
5. Both a and b are divisible by 2. This contradicts their being coprime. Therefore the initial assumption is false, and √2 is irrational.

Where does the contradiction occur?

The contradiction is not simply that an even number appears. An even integer has 2 as a factor, and that alone is entirely possible. The problem is that the reasoning forces both integers in the supposed lowest-terms fraction to have that factor.

Keep the roles of the equations clear. The first squared equation forces divisibility of a. Substituting a = 2c then creates the new equation that forces divisibility of b. Each step depends on the preceding one.

The result explains why the constructed unit-square diagonal cannot be expressed exactly as a ratio of integers. It has a definite length and a definite number-line position even though its decimal expansion neither terminates nor recurs.

How do the proofs for √3 and √5 follow the same method?

The same contradiction method applies because 3 and 5 are primes. Change the prime consistently throughout the algebra. The assumed numerator and denominator must again be positive coprime integers, rather than unspecified quantities with no lowest-terms condition.

Theorem: The square root of three is irrational

1. Assume √3 = a/b, where a and b are positive coprime integers.
2. Squaring gives a² = 3b². Therefore 3 divides a² and hence a, since 3 is prime.
3. Write a = 3c, where c is a positive integer. Then 9c² = 3b², giving b² = 3c².
4. Thus 3 divides b² and hence b. Both a and b have the common factor 3, contradicting their being coprime. Therefore √3 is irrational.

The expression (3c)² is 9c² because both factors in 3c are squared. Dividing the resulting equation by 3 is what leaves b² as three times an integer square. This step transfers the divisibility argument from numerator to denominator.

Theorem: The square root of five is irrational

1. Assume √5 = a/b, where a and b are positive coprime integers.
2. Squaring gives a² = 5b². Therefore 5 divides a² and hence a, since 5 is prime.
3. Write a = 5c, where c is a positive integer. Substitution gives 25c² = 5b², so b² = 5c².
4. Thus 5 divides b² and hence b. Both a and b share the factor 5, contradicting coprimality. Therefore √5 is irrational.

In each proof, the square-root assumption leads to a shared prime factor. The conclusion rejects the existence of any integer fraction for that root. It does not merely reject one particular pair of trial values for the numerator and denominator.

A reliable final check is to read the two divisibility deductions separately. Explain why the prime divides the numerator, and then why it divides the denominator. Finish by connecting their common factor explicitly to the lowest-terms assumption.

Glossary

• Rational number — A number expressible as a ratio of integers with a non-zero denominator.
• Irrational number — A real number that cannot be expressed as an integer ratio with a non-zero denominator.
• Real numbers — The collection formed by all rational numbers together with all irrational numbers.
• Coprime integers — Integers whose only common positive factor is the number one.
• Terminating decimal — A decimal representation that ends after a finite number of decimal places.
• Recurring decimal — A decimal in which a fixed digit or block repeats indefinitely from some point onwards.
• Principal square root — The non-negative number whose square equals the number under the square-root sign.
• Surd — An irrational root of a positive rational number, retained in exact root form.
• Like surds — Surd terms with the same root part after the expressions have been simplified.
• Conjugate expressions — Two expressions with the same two terms but opposite signs between those terms.
• Rationalisation — Rewriting a fraction with a rational denominator while preserving its original value.
• Prime number — An integer greater than one with exactly two positive factors, one and itself.
• Proof by contradiction — An argument that disproves an assumption by deriving a conclusion incompatible with that assumption.
• Hypotenuse — The side opposite the right angle in a right-angled triangle.

Common errors and misconceptions

• Misconception: Every rational number is an integer. Correct: Integers are rational, but fractions such as 2/5 are rational without being integers.
• Misconception: Every non-terminating decimal is irrational. Correct: Recurring decimals are rational; irrational decimals are both non-terminating and non-recurring.
• Misconception: Every expression containing a square-root sign is irrational. Correct: Simplify it first. For example, √81 = 9 is rational.
• Misconception: Adding like surds means adding their radicands. Correct: Add their coefficients and retain the common root part, as in 2√3 + √3 = 3√3.
• Misconception: The product of a rational and an irrational number must be irrational. Correct: The rational factor must be non-zero; multiplication by zero gives zero.
• Misconception: Rationalisation permits multiplication of the denominator alone. Correct: Multiply both numerator and denominator by the same suitable non-zero factor to preserve the value.
• Misconception: Showing a common factor completes an irrationality proof without any initial restriction. Correct: First assume a lowest-terms integer fraction; the common factor then contradicts coprimality.

Exam-style questions with model answers

Q1. Classify 0.375 and 0.454545… as rational or irrational, giving a reason for each. In the second decimal, the block 45 repeats indefinitely. [2 marks]

1. 0.375 is rational because its decimal expansion terminates. It equals the integer fraction 3/8.
2. 0.454545… is rational because the fixed block 45 recurs indefinitely. It equals 5/11.

Q2. Describe how to locate 3/4 and −3/4 on a number line, using a fixed unit length. [3 marks]

1. Mark the origin, labelled zero, and mark equal unit intervals on both sides. Divide each unit interval needed for the construction into four equal parts, so that each small part represents one quarter.
2. Count three quarter-unit parts to the right of zero. Label the resulting point 3/4.
3. Count three quarter-unit parts to the left of zero. Label this point −3/4, between −1 and zero.

Q3. Convert 0.16666… into a fraction in lowest terms, where the digit 1 does not repeat and the digit 6 repeats indefinitely. [3 marks]

1. Let x be the value of the given decimal. Multiplying by 10 shifts the single non-repeating digit before the decimal point, giving 10x = 1.6666….
2. Multiply again by 10 to shift one full repeating block: 100x = 16.6666…. Subtract the equations with identical decimal tails to obtain 90x = 15.
3. Divide by 90 and reduce the fraction: x = 15/90 = 1/6.

Q4. Rationalise the denominator of 1/(√7 − 2), showing the factor used and the simplified result. [3 marks]

1. The conjugate of the denominator √7 − 2 is √7 + 2. Multiply both the numerator and denominator by this same non-zero expression so that the fraction keeps its value.
2. The denominator becomes (√7 − 2)(√7 + 2) = (√7)² − 2² = 7 − 4 = 3.
3. The numerator becomes √7 + 2. Hence the rationalised form is (√7 + 2)/3, which has a rational denominator.

Q5. Prove that √2 is irrational. You may use the result that a prime dividing the square of a positive integer also divides that integer. [5 marks]

1. Assume that √2 is rational. Then √2 = a/b for positive coprime integers a and b, with b non-zero. Coprime means that their only common positive factor is 1.
2. Squaring and multiplying by b² gives a² = 2b². Hence 2 divides a²; because 2 is prime, it also divides a.
3. Write a = 2c for a positive integer c. Substituting this expression into the equation gives 4c² = 2b².
4. Dividing by 2 gives b² = 2c². Thus 2 divides b² and, by the given result, it also divides b.
5. Both a and b have the common factor 2, contradicting coprimality. The assumption is false, so √2 is irrational.

Q6. Prove that √5 is irrational. You may use the result that a prime dividing the square of a positive integer also divides that integer. [5 marks]

1. Suppose √5 were rational. Write √5 = a/b, where a and b are positive coprime integers and b is non-zero. This chooses a fraction already reduced to lowest terms.
2. Squaring gives a² = 5b². Therefore 5 divides a² and, since 5 is prime, it follows from the given result that 5 divides a.
3. Put a = 5c, where c is a positive integer. Substitution gives 25c² = 5b², which simplifies to b² = 5c².
4. Consequently 5 divides b². Applying the prime-divisibility result once more shows that 5 also divides b.
5. The common factor 5 contradicts the coprimality of a and b. Therefore the rationality assumption is false and √5 is irrational.

Q7. Explain a ruler-and-compass construction locating √2 on the positive number line. Use two perpendicular sides each of length 1 unit and justify the constructed length. [4 marks]

1. Let O be the origin. Mark A one unit to its right on the number line, so OA = 1 unit.
2. Draw a perpendicular at A and mark B on it with AB = 1 unit. Join O to B.
3. By Pythagoras' theorem, OB² = OA² + AB² = 1² + 1² = 2. Hence the positive length OB is √2 units.
4. Draw an arc with centre O and radius OB to meet the positive number line at P. Since OP = OB, point P represents √2.

Q8. Simplify (3√5 − 5√2)(4√5 + 3√2), showing the products and collection of like terms. [3 marks]

1. Distribute each term in the first bracket over both terms in the second bracket. The four products are 60, 9√10, −20√10 and −30; the last two carry the negative sign from −5√2.
2. Add the rational terms: 60 − 30 = 30. Combine the like surds separately: 9√10 − 20√10 = −11√10.
3. Therefore the complete simplified expression is 30 − 11√10, with its rational and irrational parts retained.

Key takeaways

• Rational numbers are integer ratios with non-zero denominators; irrational numbers cannot be written in that form.
• Natural numbers, whole numbers and integers are included among rational numbers, and all are real numbers.
• A rational decimal terminates or recurs, whereas an irrational decimal continues indefinitely without an eventually repeating block.
• Equal subdivisions locate rational fractions, while right-angled triangles and compass arcs can locate exact irrational square-root lengths.
• Simplify roots before classifying numbers or combining like surds; the presence of a root sign alone proves nothing.
• Rationalisation preserves value by multiplying numerator and denominator by the same suitable non-zero expression.
• Conjugates remove cross-products through the difference-of-squares identity, often producing a rational denominator from a surd expression.
• Irrationality proofs begin with a lowest-terms fraction and end by contradicting its coprime numerator and denominator.

Test yourself

Why is −10 a rational number?

It can be written −10/1, a ratio of two integers with a non-zero denominator.

What rational number is halfway between 1 and 3/2?

Their average is (1 + 3/2)/2 = 5/4, which lies between the two given numbers.

Is 0.454545…, with 45 repeating indefinitely, irrational?

No. Its fixed repeating block makes it rational; its exact fractional value is 5/11.

Why is √81 not a surd?

Its principal square root is 9, an integer and therefore a rational number.

What is the conjugate of √7 − 2, and what is their product?

The conjugate is √7 + 2. Their product is 7 − 4 = 3, a rational number.

What is the rationalised form of 2/(3√3)?

Multiply numerator and denominator by √3 to obtain 2√3/9, which has a rational denominator.

Why does the proof for √3 start with coprime integers?

The proof forces both integers to share the factor 3, contradicting their assumed coprimality.

Must the product of two irrational numbers be irrational?

No. The irrational numbers 3 + √2 and 3 − √2 have the rational product 7.