Real Numbers | CBSE Class 10 Maths Notes

This Mathematics note covers real numbers, rational and irrational numbers, unique prime factorisation, the Fundamental Theorem of Arithmetic, highest common factors, lowest common multiples, divisibility, and proofs of irrationality. It includes factor trees, calculations with two and three positive integers, repeated journeys, and expressions involving square roots.

How are rational and irrational numbers related to real numbers?

Real numbers include both rational numbers and irrational numbers. These two types belong to the same number system, but they differ in whether they can be expressed as a ratio of integers. That distinction is the starting point for proving irrationality.

Definition: A rational number can be written as $\frac{a}{b}$, where $a$ and $b$ are integers and $b\ne0$. An irrational number cannot be written in this form.

The restriction on the denominator matters. A fraction with a zero denominator is not an allowed representation of a rational number. When a proof begins by assuming a number is rational, it must include a non-zero denominator.

Why do proofs use coprime integers?

Two integers are coprime when their only common positive factor is $1$. If the numerator and denominator of a fraction share another factor, divide both by that factor. Continue cancelling common factors to obtain a fraction in lowest terms.

This does not change the value of the fraction. It gives a useful starting condition: the numerator and denominator have no common prime factor. An irrationality proof will show that this condition cannot hold for the number under investigation.

For example, the proof for $\sqrt{2}$ starts by supposing that it has a rational representation in lowest terms. It eventually forces both the numerator and denominator to be divisible by $2$. That conflicts with the starting condition.

What is a proof by contradiction?

In a proof by contradiction, assume the opposite of the required conclusion and follow its consequences. If those consequences contradict an established fact or the starting conditions, reject the assumption. State the contradiction precisely instead of merely announcing that something is impossible.

For square roots, the contradiction usually concerns a common prime factor. For an expression such as $5-\sqrt{3}$, it concerns the already established irrationality of the square root. Both arguments require an explicit chain of reasons.

What does unique prime factorisation tell us?

Theorem: Fundamental Theorem of Arithmetic

Definition: Every composite number has a prime factorisation. Its prime factors and their repetitions are fixed; changing their order does not produce a different factorisation.

The theorem contains two claims. First, a composite number can be broken into prime factors. Second, the resulting collection of prime factors is unique, apart from order. A factorisation containing a composite factor is not yet a complete prime factorisation.

Repeated occurrences of the same prime can be collected using powers. Writing the primes in ascending order makes the result easier to compare with other factorisations. The powers record how many times each prime occurs, so repetitions must not be discarded.

How does a factor tree reach prime factors?

The top box contains $32760$. Its branches lead to $2$ and $16380$. Successive branches split the right-hand composite factors, ending with boxes containing $7$ and $13$. The terminal prime boxes contain three copies of $2$, two of $3$, and one each of $5$, $7$, and $13$.

Reference: NCERT Class 10, page 2, unnumbered figure

Worked example 1. Basic factorisation: express $32760$ as a product of powers of primes.

1. Begin with the first split: $$32760=2\times16380.$$
2. Split the remaining even factor: $$16380=2\times8190.$$
3. Continue until the remaining factor is odd: $$8190=2\times4095.$$
4. Divide the remaining composite factors by primes: $$4095=3\times1365,\qquad1365=3\times455.$$
5. Complete the branches: $$455=5\times91,\qquad91=7\times13.$$
6. Collect the terminal primes: $$32760=2\times2\times2\times3\times3\times5\times7\times13=2^3\times3^2\times5\times7\times13.$$
7. Check the multiplication: $$8\times9=72,\quad72\times5=360,\quad360\times7=2520,\quad2520\times13=32760.$$

Answer: $32760=2^3\times3^2\times5\times7\times13$.

At every branch, the product of the two child numbers equals their parent number. Stopping at $91$ would leave a composite factor unresolved. Stopping only at primes gives the factorisation required by the theorem.

A tree is a way to organise the calculation; uniqueness concerns its final prime factors. Changing the sequence of splits cannot change the final collection of primes. This is why a missing prime cannot appear merely by choosing a different factorisation route.

How do prime factors give the HCF and LCM of two integers?

The highest common factor, or HCF, is the greatest positive integer that divides each given integer. The lowest common multiple, or LCM, is the least positive integer that is a multiple of each given integer.

Both calculations start with complete prime factorisations, but the selection rules differ. A common factor must fit inside both numbers. A common multiple must contain enough copies of every prime to include both numbers as factors.

How are the two selection rules applied?

Worked example 2. Basic calculation: find the HCF and LCM of $6$ and $20$.

1. Factorise the first integer: $$6=2\times3.$$
2. Factorise the second integer: $$20=2\times10=2\times2\times5=2^2\times5.$$
3. The only common prime is $2$. Choose its smaller power: $$\operatorname{HCF}(6,20)=2.$$
4. For the LCM, include the greatest powers of all primes involved: $$\operatorname{LCM}(6,20)=2^2\times3\times5=4\times3\times5=60.$$
5. Check the common factor and common multiple: $$6\div2=3,\quad20\div2=10,\quad60\div6=10,\quad60\div20=3.$$

Answer: HCF $=2$; LCM $=60$.

The HCF excludes $3$ because it is absent from the factorisation of $20$. It excludes $5$ because that prime is absent from the factorisation of $6$. In contrast, the LCM needs both of these primes.

For the LCM, one copy of $2$ would not be enough to include $20$ as a factor. Two copies are necessary. For the HCF, two copies would not divide $6$. These observations explain the power rules rather than leaving them as instructions to memorise.

Keep the complete factorisations visible until both answers have been found. This makes it easier to distinguish a shared prime from a prime needed only for the common multiple.

How can the HCF help us calculate an LCM?

Result: The product relation for two positive integers

For two positive integers, multiplying their HCF and LCM gives the product of the original numbers. Once the HCF is known, division therefore gives the LCM. The relation provides a useful check on separate prime-factor calculations.

1. Write the relation for positive integers $a$ and $b$: $$\operatorname{HCF}(a,b)\times\operatorname{LCM}(a,b)=a\times b.$$
2. Divide both sides by the non-zero HCF: $$\operatorname{LCM}(a,b)=\frac{a\times b}{\operatorname{HCF}(a,b)}.$$

The two-number condition is essential. Do not insert a third integer into this formula. First identify the pair of numbers, then substitute their HCF and their product into the correct positions.

Worked example 3. Linked calculation: find the HCF of $96$ and $404$, then their LCM.

1. Factorise by repeated division: $$96=2\times48=2^2\times24=2^3\times12=2^4\times6=2^5\times3.$$
2. Factorise the other integer: $$404=2\times202=2^2\times101.$$
3. The smaller power of the only common prime gives $$\operatorname{HCF}(96,404)=2^2=4.$$
4. Substitute into the product relation: $$\operatorname{LCM}(96,404)=\frac{96\times404}{4}=24\times404=9696.$$
5. Verify the result as a common multiple: $$9696\div96=101,\qquad9696\div404=24.$$

Answer: HCF $=4$; LCM $=9696$.

What if the HCF is supplied?

Worked example 4. Direct application: given $\operatorname{HCF}(306,657)=9$, find their LCM.

1. Use the relation for two positive integers: $$\operatorname{LCM}(306,657)=\frac{306\times657}{9}.$$
2. Cancel the given HCF before multiplying: $$306\div9=34.$$
3. Complete the product: $$34\times657=30\times657+4\times657=19710+2628=22338.$$
4. Check divisibility by both integers: $$22338\div306=73,\qquad22338\div657=34.$$

Answer: $\operatorname{LCM}(306,657)=22338$.

In this second calculation, refactorising both integers is unnecessary because the HCF is given. Cancelling before multiplication also reduces the size of the arithmetic. The final divisibility checks confirm that the calculated answer is a common multiple of the required pair.

How do HCF and LCM calculations change for three integers?

The prime factorisation method works for three integers as well as for two. For the HCF, a selected prime must occur in every integer. For the LCM, collect every prime occurring anywhere among the integers and select its greatest power.

A prime shared by only two of the three integers cannot contribute to the HCF of all three. However, it still contributes to their LCM. Check all three factorisations before choosing powers.

How can all three factorisations be compared?

Worked example 5. Multi-step calculation: find the HCF and LCM of $6$, $72$, and $120$.

1. Factorise the first integer: $$6=2\times3.$$
2. Factorise the second: $$72=8\times9=2^3\times3^2.$$
3. Factorise the third: $$120=8\times15=2^3\times3\times5.$$
4. The primes common to all three are $2$ and $3$. Select their smallest powers: $$\operatorname{HCF}(6,72,120)=2\times3=6.$$
5. Select the greatest powers of all primes present: $$\operatorname{LCM}(6,72,120)=2^3\times3^2\times5=8\times9\times5=360.$$
6. Check the common multiple: $$360\div6=60,\qquad360\div72=5,\qquad360\div120=3.$$

Answer: HCF $=6$; LCM $=360$.

Here, $5$ is present only in the third integer. That is enough to require it in the LCM, but not enough to include it in the HCF. The exponent of $3$ is greater in $72$ than in the other two integers.

Why must the two-number product shortcut be avoided?

1. Multiply the three original integers: $$6\times72\times120=432\times120=51840.$$
2. Multiply their HCF and LCM: $$6\times360=2160.$$
3. Compare the results: $$51840\ne2160.$$

This calculation shows why the two-number identity cannot be used as a general rule for three numbers. It does not mean that prime factorisation has failed. Both the HCF and the LCM were obtained correctly; only the proposed extension of the product rule fails.

How do prime factors solve divisibility and timing problems?

Prime factors help answer questions without listing many powers or multiples. A question about a final zero asks whether certain prime factors must be present. A question about simultaneous returns to a starting point asks for a common multiple of the journey times.

Why can a missing prime rule out a final zero?

Worked example 6. Reasoning problem: can $4^n$ end in zero for any natural number $n$?

1. Rewrite the base as a prime power: $$4=2^2.$$
2. Express the whole power using that prime: $$4^n=(2^2)^n=2^{2n}.$$
3. A final zero would require divisibility by $5$, so the prime factorisation would have to contain $5$.
4. The displayed factorisation contains only $2$. By uniqueness, it cannot also contain the different prime $5$.

Answer: No natural number $n$ makes $4^n$ end in zero. The required prime factor $5$ is absent.

The argument concerns every natural exponent at once. Calculating a few initial powers would suggest a pattern, but it would not explain why the pattern must continue. Unique factorisation supplies that reason.

How does an LCM describe a simultaneous return?

Worked example 7. Application: Sonia takes $18$ minutes and Ravi takes $12$ minutes to complete a round of a circular path. They start together at the same point and travel in the same direction. When will they next meet at the starting point?

1. A return to the starting point takes a whole number of rounds. The required elapsed time is therefore a common multiple of $18$ and $12$.
2. Factorise the two times: $$18=2\times9=2\times3^2,\qquad12=4\times3=2^2\times3.$$
3. Select the greatest powers for the least positive common time: $$\operatorname{LCM}(18,12)=2^2\times3^2=4\times9=36.$$
4. Check the completed rounds: $$36\div18=2,\qquad36\div12=3.$$

Answer: They next meet at the starting point after $36$ minutes. Sonia completes $2$ rounds and Ravi completes $3$ rounds.

The phrase at the starting point determines the calculation. Both travellers must have completed whole rounds. An HCF would describe a common divisor of the times, whereas this situation requires a time divisible by each round time.

How can a common factor establish that a number is composite?

Worked example 8. Factor-based reasoning: explain why $7\times11\times13+13$ is composite.

1. Take out the common factor: $$7\times11\times13+13=13(7\times11+1).$$
2. Evaluate the remaining factor: $$7\times11+1=77+1=78.$$
3. Complete the calculation: $$13\times78=1014.$$
4. Both factors $13$ and $78$ exceed $1$, so this is a non-trivial factorisation.

Answer: $1014$ is composite because $1014=13\times78$.

It is enough to exhibit factors greater than one; a complete prime factorisation is not necessary merely to prove that a number is composite.

Why does a prime dividing a square also divide its base?

Theorem: A prime divisor of a square divides the original integer

Let $p$ be prime and $a$ be a positive integer. If $p$ divides $a^2$, then $p$ divides $a$. The condition that $p$ is prime is part of the theorem and must remain in its statement.

This is the link between prime factorisation and the irrationality proofs that follow. Squaring repeats the prime factors already present. It does not introduce a new prime into the factorisation.

How does uniqueness prove the result?

1. Write the prime factors of the positive integer, allowing repeated primes: $$a=p_1p_2\cdots p_n,$$ where $p_1,p_2,\ldots,p_n$ are the prime factors and $n$ is their total number, counting repetitions.
2. Square the product: $$a^2=(p_1p_2\cdots p_n)(p_1p_2\cdots p_n)=p_1^2p_2^2\cdots p_n^2.$$
3. Because $p$ is prime and divides $a^2$, it must be one of the primes in the prime factorisation of $a^2$.
4. Uniqueness identifies those primes as the primes already listed in $a$. Thus $p$ is one of $p_1,p_2,\ldots,p_n$.
5. Therefore the original product contains $p$ as a factor: $$p\mid a.$$

The result allows a divisibility statement about a square to become a divisibility statement about its base. The vertical bar in $p\mid a$ means that $p$ divides $a$ exactly.

In applying this result, identify the prime first. In the proof for $\sqrt{2}$, the prime is $2$. In the proof for $\sqrt{3}$, it is $3$. Each proof uses the result twice, once for the numerator and once for the denominator.

Note: Do not confuse divisibility of $a^2$ by a prime with divisibility of $a$ by the square of that prime. The theorem concludes divisibility by the prime itself.

The proof also explains why uniqueness matters. Merely knowing that a square has some factorisation would be insufficient. We need to know that its prime factors are exactly those obtained by repeating the factors of the original integer.

How are square roots proved to be irrational?

The central strategy is to assume a fraction in lowest terms, remove the square root by squaring, and establish a common prime factor in both numerator and denominator. The final conclusion must refer back to the assumption that these integers were coprime.

Theorem: Irrationality of $\sqrt{2}$

1. Assume that $\sqrt{2}$ is rational. Choose positive coprime integers $a$ and $b$, with $b\ne0$, such that $$\sqrt{2}=\frac{a}{b}.$$
2. Multiply by the denominator: $$b\sqrt{2}=a.$$
3. Square both sides: $$2b^2=a^2.$$
4. Thus $2$ divides $a^2$. Since $2$ is prime, it divides $a$. Write $$a=2c$$ for an integer $c$.
5. Substitute into the squared equation: $$2b^2=(2c)^2=4c^2.$$
6. Divide by $2$: $$b^2=2c^2.$$
7. Now $2$ divides $b^2$, so it also divides $b$. Both $a$ and $b$ therefore have the common factor $2$.
8. This contradicts their being coprime. Reject the rationality assumption and conclude that $\sqrt{2}$ is irrational.

The contradiction is not that even numbers cannot form a fraction. They can. It is that a fraction deliberately chosen in lowest terms cannot have an even numerator and an even denominator. This distinction makes the logical conclusion precise.

How does the proof work with another prime?

Worked example 9. Proof application: prove that $\sqrt{3}$ is irrational.

1. Assume a representation with positive coprime integers: $$\sqrt{3}=\frac{a}{b},\qquad b\ne0.$$
2. Multiply and square: $$b\sqrt{3}=a,\qquad3b^2=a^2.$$
3. Since $3$ divides $a^2$ and is prime, it divides $a$. Hence $$a=3c$$ for an integer $c$.
4. Substitute and simplify: $$3b^2=(3c)^2=9c^2,\qquad b^2=3c^2.$$
5. Thus $3$ divides $b^2$, and the prime-divisor theorem gives that $3$ divides $b$.
6. The common factor $3$ contradicts coprimality. Therefore the assumed rational representation cannot exist.

Answer: $\sqrt{3}$ is irrational.

The substitutions must be squared correctly. Replacing $a$ by $3c$ means replacing $a^2$ by $9c^2$. Both the coefficient and the variable are squared; losing the square on the coefficient breaks the argument.

How can the same reasoning prove irrationality of $\sqrt{5}$?

Worked example 10. Independent proof: prove that $\sqrt{5}$ is irrational.

1. Suppose $\sqrt{5}$ is rational, with positive coprime integers satisfying $$\sqrt{5}=\frac{a}{b},\qquad b\ne0.$$
2. Multiply by the denominator and square: $$b\sqrt{5}=a,\qquad5b^2=a^2.$$
3. The prime $5$ divides $a^2$, so it divides $a$. Write $$a=5c.$$
4. Substitute this into the squared equation: $$5b^2=(5c)^2=25c^2.$$
5. Divide by $5$: $$b^2=5c^2.$$
6. Therefore $5$ divides $b^2$, and hence $b$. The common factor $5$ contradicts the choice of coprime integers.

Answer: $\sqrt{5}$ is irrational.

The structure is unchanged because each chosen number under the square root is prime. At each stage, name the reason for moving from divisibility of a square to divisibility of the original integer. Do not leave that crucial inference unexplained.

How can irrationality be proved for expressions containing square roots?

Once the irrationality of a square root has been established, it can support further proofs. Assume the whole expression is rational, then rearrange to isolate the known irrational number. The contradiction is that the rearrangement would give it a ratio of integers.

The sum or difference of a rational and an irrational number is irrational. Multiplication or division involving a non-zero rational number and an irrational number also gives an irrational result. The non-zero restriction matters when applying these multiplication and division properties.

How does subtraction lead to a contradiction?

Worked example 11. Rearrangement proof: show that $5-\sqrt{3}$ is irrational.

1. Assume the expression is rational. For integers $a$ and $b$, write $$5-\sqrt{3}=\frac{a}{b},\qquad b\ne0.$$
2. Move the square root and the fraction to isolate the root: $$\sqrt{3}=5-\frac{a}{b}.$$
3. Use a common denominator: $$\sqrt{3}=\frac{5b-a}{b}.$$
4. The numerator $5b-a$ is an integer, and the integer denominator $b$ is non-zero. The fraction is therefore rational.
5. This contradicts the established irrationality of $\sqrt{3}$. Reject the assumption about the original expression.

Answer: $5-\sqrt{3}$ is irrational.

Notice the sign in the numerator. Subtracting the fraction from $5$ produces $5b-a$, not a sum. Writing the common denominator explicitly makes the reasoning easy to check.

How does a non-zero multiplier affect the proof?

Worked example 12. Multiplication proof: show that $3\sqrt{2}$ is irrational.

1. Suppose the expression is rational: $$3\sqrt{2}=\frac{a}{b},\qquad b\ne0,$$ where $a$ and $b$ are integers.
2. Divide by the non-zero integer $3$: $$\sqrt{2}=\frac{a}{3b}.$$
3. The numerator is an integer. The denominator is an integer and remains non-zero, so this fraction is rational.
4. This would make $\sqrt{2}$ rational, contradicting its proved irrationality.

Answer: $3\sqrt{2}$ is irrational.

No new square-root proof is needed here. The argument depends on a result already established. What must be shown is that the rationality of the whole expression would force the rationality of that square root.

How can addition and multiplication be handled together?

Worked example 13. Combined proof: show that $3+2\sqrt{5}$ is irrational.

1. Assume a rational representation: $$3+2\sqrt{5}=\frac{a}{b},\qquad b\ne0,$$ with integer numerator and denominator.
2. Subtract the rational constant: $$2\sqrt{5}=\frac{a}{b}-3=\frac{a-3b}{b}.$$
3. Divide by the non-zero multiplier: $$\sqrt{5}=\frac{a-3b}{2b}.$$
4. The numerator is an integer and the integer denominator is non-zero. Thus the right-hand side is rational.
5. This contradicts the irrationality of $\sqrt{5}$, so the initial assumption is false.

Answer: $3+2\sqrt{5}$ is irrational.

These proofs follow a shared sequence: assume rationality, isolate the root, justify the rationality of the resulting fraction, and state the contradiction. Each line must preserve equality. The final sentence then returns to the original expression and classifies it.

Glossary

• Real numbers — The number system formed by rational numbers together with irrational numbers.
• Rational number — A number expressible as a ratio of integers with a non-zero denominator.
• Irrational number — A number that cannot be expressed as a ratio of integers with a non-zero denominator.
• Prime number — A positive integer greater than one whose only positive factors are one and itself.
• Composite number — A positive integer greater than one that has a positive factor other than one and itself.
• Prime factorisation — An expression of an integer as a product of primes, with repeated primes retained.
• Factor tree — A branching arrangement that splits composite factors until its terminal factors are all prime.
• Highest common factor — The greatest positive integer that divides every integer in a given collection.
• Lowest common multiple — The least positive integer that is a multiple of every given integer.
• Coprime integers — Integers whose only common positive factor is one, as required in a lowest-terms fraction.
• Unique factorisation — The property that prime factors and their repetitions are fixed, apart from their order.
• Proof by contradiction — A proof that rejects an assumption after its consequences conflict with an established condition or fact.

Common errors and misconceptions

• Misconception: A change in the order of prime factors creates a new prime factorisation. Correct: Uniqueness allows a change of order; it fixes the primes and the number of times each occurs.
• Misconception: Use the greatest common prime power for the HCF, giving the wrong line $\operatorname{HCF}(6,20)=2^2$. Correct: Use the smallest common power: $\operatorname{HCF}(6,20)=2$.
• Misconception: Use only shared primes for the LCM, giving the wrong line $\operatorname{LCM}(6,20)=2^2$. Correct: Include all primes involved: $\operatorname{LCM}(6,20)=2^2\times3\times5=60$.
• Misconception: The wrong line $6\times72\times120=\operatorname{HCF}(6,72,120)\times\operatorname{LCM}(6,72,120)$ extends the product rule to three numbers. Correct: The two sides are $51840$ and $2160$, respectively; the two-number identity is not a general three-number identity.
• Misconception: Showing that the numerator is even completes the irrationality proof for $\sqrt{2}$. Correct: Show that the denominator is also even, then explain why their common factor contradicts the assumed lowest-terms representation.
• Misconception: Substituting $a=3c$ gives the wrong line $a^2=3c^2$. Correct: Square the entire product: $a^2=(3c)^2=9c^2$.
• Misconception: The prime-divisor theorem needs no restriction on the divisor. Correct: Its hypothesis specifies a prime divisor. State and use that condition when passing from a squared integer to its base.
• Misconception: An HCF gives the next common return time in the circular-path problem. Correct: The time must be a multiple of each round time, so use their LCM.

Exam-style questions with model answers

Q1. State the Fundamental Theorem of Arithmetic and explain uniqueness. [2 marks]

1. Every composite number can be expressed as a product of prime numbers.
2. The primes and their repetitions are fixed. Reordering those factors does not count as a different prime factorisation.

Q2. Can $4^n$ end in zero for a natural number $n$? Give a reason. [2 marks]

1. Its prime factorisation is $4^n=2^{2n}$, so its only prime factor is $2$.
2. A final zero would require the prime factor $5$. Uniqueness excludes that factor, so no such natural number exists.

Q3. Find the HCF of $96$ and $404$, and hence their LCM. [3 marks]

1. Factorise both integers: $96=2\times48=2^2\times24=2^3\times12=2^4\times6=2^5\times3$ and $404=2\times202=2^2\times101$. The only common prime is $2$; neither remaining prime occurs in both of the given integers.
2. Choose its smaller power to obtain $\operatorname{HCF}(96,404)=2^2=4$.
3. For two positive integers, their HCF times their LCM equals their product. Thus $\operatorname{LCM}(96,404)=\frac{96\times404}{4}=24\times404=9696$. As a check, $9696\div96=101$ and $9696\div404=24$, so both original integers divide the answer.

Q4. Find the HCF and LCM of $6$, $72$, and $120$. Check whether their product equals the product of the three numbers. [4 marks]

1. Prime factorisation gives $6=2\times3$, $72=8\times9=2^3\times3^2$, and $120=8\times15=2^3\times3\times5$.
2. Take the smallest powers of primes common to all three: $\operatorname{HCF}=2\times3=6$. The prime $5$ is excluded because it is not common to every number.
3. Take the greatest powers of every prime involved: $\operatorname{LCM}=2^3\times3^2\times5=360$.
4. Now $\operatorname{HCF}\times\operatorname{LCM}=6\times360=2160$, whereas $6\times72\times120=51840$. These results differ, so the identity for two integers does not extend to this group of three.

Q5. Prove that $\sqrt{2}$ is irrational. [5 marks]

1. Assume the contrary: $\sqrt{2}=\frac{a}{b}$, where $a$ and $b$ are positive coprime integers and $b\ne0$. Choosing lowest terms ensures that they have no common prime factor.
2. Multiplying gives $b\sqrt{2}=a$. Squaring gives $2b^2=a^2$. Thus $2$ divides $a^2$. Since $2$ is prime, the prime-divisor theorem implies that it divides $a$.
3. Write $a=2c$ for an integer $c$. Substitution gives $2b^2=4c^2$, and division gives $b^2=2c^2$.
4. Consequently $2$ divides $b^2$, so it divides $b$ as well. Both integers now have the common factor $2$.
5. This contradicts their assumed coprimality. The assumption of rationality is false, and therefore $\sqrt{2}$ is irrational.

Q6. Prove that $\sqrt{3}$ is irrational and hence that $5-\sqrt{3}$ is irrational. [5 marks]

1. Suppose $\sqrt{3}=\frac{a}{b}$ for positive coprime integers with $b\ne0$. Multiplying and squaring gives $b\sqrt{3}=a$ and $3b^2=a^2$.
2. The prime $3$ divides $a^2$, so it divides $a$. Put $a=3c$. Then $3b^2=9c^2$, giving $b^2=3c^2$.
3. It follows that $3$ divides $b$ too. This contradicts coprimality and proves that $\sqrt{3}$ is irrational.
4. Now assume $5-\sqrt{3}=\frac{r}{s}$, where $r$ and $s$ are integers and $s\ne0$. Rearranging gives $\sqrt{3}=5-\frac{r}{s}=\frac{5s-r}{s}$.
5. This is a ratio of integers with non-zero denominator, so it would make $\sqrt{3}$ rational. That contradicts the first part. Therefore $5-\sqrt{3}$ is irrational as well. The contradiction rules out the assumed rational representation of the entire expression.

Q7. Sonia takes $18$ minutes and Ravi takes $12$ minutes per round. Starting together, when do they next meet at the starting point? [3 marks]

1. Each return to the starting point occurs after a whole number of rounds. The next common return therefore requires the least positive common multiple of the two round times.
2. Factorise the times: $18=2\times3^2$ and $12=2^2\times3$. Select the greatest powers: $\operatorname{LCM}(18,12)=2^2\times3^2=36$.
3. They next meet at the starting point after $36$ minutes. The checks $36\div18=2$ and $36\div12=3$ show that each has completed a whole number of rounds.

Q8. Show that $3\sqrt{2}$ is irrational. [3 marks]

1. Assume that $3\sqrt{2}$ is rational. Write $3\sqrt{2}=\frac{a}{b}$, where $a$ and $b$ are integers and $b\ne0$.
2. Divide by the non-zero integer $3$ to obtain $\sqrt{2}=\frac{a}{3b}$. The denominator is still a non-zero integer, so the right-hand side is rational.
3. This contradicts the proved irrationality of $\sqrt{2}$. Hence the initial assumption must be rejected, and $3\sqrt{2}$ is irrational.

Key takeaways

• Real numbers include rational and irrational numbers; rational numbers have an integer-ratio representation with a non-zero denominator.
• Prime factorisation fixes the primes and their repetitions, while allowing those factors to appear in any order.
• Find the HCF by selecting the smallest powers of primes common to every given integer.
• Find the LCM by selecting the greatest powers of all primes appearing in the given integers.
• The product of the HCF and LCM equals the original product for two positive integers.
• A prime that divides the square of a positive integer also divides that integer itself.
• Square-root irrationality proofs begin with a lowest-terms fraction and end by contradicting the coprimality assumption.
• For expressions containing known irrational roots, assume rationality and rearrange to force a rational representation of the root.

Test yourself

What remains unchanged when prime factors are reordered?

The primes and their repetitions remain unchanged; the order of multiplication does not affect uniqueness.

What is the prime factorisation of $32760$?

It is $2^3\times3^2\times5\times7\times13$, retaining all the terminal primes and their repetitions in the factor tree.

Why does $5$ belong to the LCM but not the HCF of $6$ and $20$?

It occurs in the factorisation of $20$ but not in that of $6$, so it is not common.

Which relation finds an LCM from the HCF of two positive integers?

Use $\operatorname{LCM}(a,b)=\frac{ab}{\operatorname{HCF}(a,b)}$. The condition that this formula concerns two positive integers must be retained.

What prime factor is missing from $4^n$ if a final zero is required?

The missing prime is $5$; the factorisation $4^n=2^{2n}$ contains only the prime $2$.

Why must the numerator and denominator be coprime at the start of a square-root proof?

This condition makes the later discovery of a common prime factor a contradiction, ruling out the assumed rational representation.

Which step lets divisibility of a square imply divisibility of its base?

The prime-divisor theorem supplies that inference, provided the divisor is prime and the base is a positive integer.

What is contradicted when $5-\sqrt{3}$ is assumed rational?

Rearrangement would make $\sqrt{3}$ a ratio of integers, contradicting the established irrationality of that square root.