Chapter 1: Real Numbers

Solution:
We break each number down by dividing repeatedly by the smallest possible prime until only 1 remains, then write the result in exponential form where applicable.

(i) \(\begin{aligned} 140 &= 2 \times 2 \times 5 \times 7 \\ &= 2^2 \times 5 \times 7 \end{aligned}\)

(ii) \(\begin{aligned} 156 &= 2 \times 2 \times 3 \times 13 \\ &= 2^2 \times 3 \times 13 \end{aligned}\)

(iii) \(\begin{aligned} 3825 &= 3 \times 3 \times 5 \times 5 \times 17 \\ &= 3^2 \times 5^2 \times 17 \end{aligned}\)

(iv) \(5005 = 5 \times 7 \times 11 \times 13\)

(v) \(7429 = 17 \times 19 \times 23\)

Solution:
For each pair, we first write the prime factorisation of both numbers. The HCF is the product of the smallest powers of all common prime factors, and the LCM is the product of the greatest powers of all prime factors that appear.

(i) 26 and 91
\(\begin{aligned} &26 = 2 \times 13 \\ &91 = 7 \times 13 \\ &\text{HCF} = 13 \\ &\text{LCM} = 2 \times 7 \times 13 = 182 \end{aligned}\)

Product of the two numbers \(= 26 \times 91 = 2366\)
\(\text{HCF} \times \text{LCM} = 13 \times 182 = 2366\)
Since both sides are equal, product of two numbers \(=\) HCF \(\times\) LCM is verified.

(ii) 510 and 92
\(\begin{aligned} &510 = 2 \times 3 \times 5 \times 17 \\ &92 = 2 \times 2 \times 23 \\ &\text{HCF} = 2 \\ &\text{LCM} = 2 \times 2 \times 3 \times 5 \times 17 \times 23 = 23460 \end{aligned}\)

Product of the two numbers \(= 510 \times 92 = 46920\)
\(\text{HCF} \times \text{LCM} = 2 \times 23460 = 46920\)
Since both sides are equal, product of two numbers \(=\) HCF \(\times\) LCM is verified.

(iii) 336 and 54
\(\begin{aligned} &336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 = 2^4 \times 3 \times 7 \\ &54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3 \\ &\text{HCF} = 2 \times 3 = 6 \\ &\text{LCM} = 2^4 \times 3^3 \times 7 = 3024 \end{aligned}\)

Product of the two numbers \(= 336 \times 54 = 18144\)
\(\text{HCF} \times \text{LCM} = 6 \times 3024 = 18144\)
Since both sides are equal, product of two numbers \(=\) HCF \(\times\) LCM is verified.

Solution:
We apply prime factorisation to each number in the group, then identify the HCF using the lowest common power of shared primes, and the LCM using the highest power of every prime that appears.

(i) 12, 15 and 21
\(\begin{aligned} &12 = 2^2 \times 3 \\ &15 = 3 \times 5 \\ &21 = 3 \times 7 \\ &\text{HCF} = 3 \\ &\text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420 \end{aligned}\)

(ii) 17, 23 and 29
Since 17, 23, and 29 are all prime numbers, the only common factor among them is 1. Therefore:
\(\begin{aligned} &17 = 1 \times 17 \\ &23 = 1 \times 23 \\ &29 = 1 \times 29 \\ &\text{HCF} = 1 \\ &\text{LCM} = 17 \times 23 \times 29 = 11339 \end{aligned}\)

(iii) 8, 9 and 25
\(\begin{aligned} &8 = 2 \times 2 \times 2 \\ &9 = 3 \times 3 \\ &25 = 5 \times 5 \\ &\text{HCF} = 1 \\ &\text{LCM} = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 = 1800 \end{aligned}\)
Notice that 8, 9, and 25 share no common prime factors at all, which is why their HCF is 1.

Solution:
We use the key relationship between HCF and LCM — that their product always equals the product of the two numbers — to work backwards and find the LCM directly.

\(\begin{aligned} &\text{HCF}(306,657) = 9 \\ &\text{We know that, } \text{LCM} \times \text{HCF} = \text{Product of two numbers} \\ &\therefore \text{LCM} \times 9 = 306 \times 657 \\ &\text{LCM} = \frac{306 \times 657}{9} \\ &\text{LCM} = 22338 \end{aligned}\)

Solution:
A number ends with the digit 0 only when it is divisible by 10. Since \(10 = 2 \times 5\), any number ending in 0 must have both 2 and 5 as prime factors. Let us check whether \(6^n\) ever contains 5 as a factor.

Prime factorisation of \(6^n = (2 \times 3)^n\)

It can be observed that 5 is not in the prime factorisation of \(6^n\).
Hence, for any value of \(n\), \(6^n\) will not be divisible by 5.
Therefore, \(6^n\) cannot end with the digit 0 for any natural number \(n\).

Solution:
Numbers are of two types — prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself. The strategy here is to factor out a common term from each expression and check whether the result has more than two factors.

\(\begin{aligned} &7 \times 11 \times 13 + 13 \\ &= 13 \times (7 \times 11 + 1) \\ &= 13 \times (77 + 1) \\ &= 13 \times 78 \\ &= 13 \times 13 \times 6 \end{aligned}\)

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.

\(\begin{aligned} &7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \\ &= 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) \\ &= 5 \times (1008 + 1) \\ &= 5 \times 1009 \end{aligned}\)

1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Solution:
Since Ravi completes each lap faster than Sonia and they both travel in the same direction, Ravi will gradually lap Sonia. They will meet at the starting point exactly when Ravi has gained a full lap over Sonia. The first time this happens is after a duration equal to the LCM of their individual lap times — that is, the LCM of 18 minutes and 12 minutes.

\(18 = 2 \times 3 \times 3\)
\(12 = 2 \times 2 \times 3\)
LCM of 12 and \(18 = 2 \times 2 \times 3 \times 3 = 36\)

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Solution:
We use proof by contradiction — we assume the opposite of what we want to prove and show that it leads to an impossible situation.

Let \(\sqrt{5}\) be a rational number. Therefore, we can find two integers \(a, b\ (b \neq 0)\) such that \(\sqrt{5} = \dfrac{a}{b}\). Let \(a\) and \(b\) have a common factor other than 1. Then we can divide them by the common factor, and assume that \(a\) and \(b\) are co-prime.

\(a = \sqrt{5}\,b\)
\(\Rightarrow a^2 = 5b^2\)

Therefore, \(a^2\) is divisible by 5 and it can be said that \(a\) is divisible by 5. Let \(a = 5k\), where \(k\) is an integer:

\((5k)^2 = 5b^2\)
\(\Rightarrow 5k^2 = b^2\)

This means that \(b^2\) is divisible by 5 and hence, \(b\) is divisible by 5.
This implies that \(a\) and \(b\) have 5 as a common factor.
And this is a contradiction to the fact that \(a\) and \(b\) are co-prime.
Hence, \(\sqrt{5}\) cannot be expressed as \(\dfrac{a}{b}\), or it can be said that \(\sqrt{5}\) is irrational.

Solution:
We use proof by contradiction. We assume the expression is rational, then isolate the irrational part to reach a contradiction.

Let \(3 + 2\sqrt{5}\) be rational. Therefore, we can find two co-prime integers \(a, b\ (b \neq 0)\) such that:

\(3 + 2\sqrt{5} = \dfrac{a}{b}\)
\(\Rightarrow 2\sqrt{5} = \dfrac{a}{b} – 3\)
\(\Rightarrow \sqrt{5} = \dfrac{1}{2}\left(\dfrac{a}{b} – 3\right)\)

Since \(a\) and \(b\) are integers, \(\dfrac{1}{2}\left(\dfrac{a}{b} – 3\right)\) will also be rational and therefore, \(\sqrt{5}\) is rational.
This contradicts the fact that \(\sqrt{5}\) is irrational.
Hence, our assumption that \(3 + 2\sqrt{5}\) is rational is false. Therefore, \(3 + 2\sqrt{5}\) is irrational.

Solution:
In each part below, we assume the expression is rational and then show that this forces a known irrational number to be rational — a contradiction that proves our assumption wrong.

(i) Let \(\dfrac{1}{\sqrt{2}}\) be rational. Therefore, we can find two co-prime integers \(a, b\ (b \neq 0)\) such that:
\(\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}\)
\(\Rightarrow \sqrt{2} = \dfrac{b}{a}\)

\(\dfrac{b}{a}\) is rational as \(a\) and \(b\) are integers. Therefore, \(\sqrt{2}\) is rational, which contradicts the fact that \(\sqrt{2}\) is irrational.
Hence, our assumption is false and \(\dfrac{1}{\sqrt{2}}\) is irrational.

(ii) Let \(7\sqrt{5}\) be rational. Therefore, we can find two co-prime integers \(a, b\ (b \neq 0)\) such that:
\(7\sqrt{5} = \dfrac{a}{b}\)
\(\Rightarrow \sqrt{5} = \dfrac{a}{7b}\)

\(\dfrac{a}{7b}\) is rational as \(a\) and \(b\) are integers. Therefore, \(\sqrt{5}\) should be rational.
This contradicts the fact that \(\sqrt{5}\) is irrational. Therefore, our assumption that \(7\sqrt{5}\) is rational is false. Hence, \(7\sqrt{5}\) is irrational.

(iii) Let \(6 + \sqrt{2}\) be rational. Therefore, we can find two co-prime integers \(a, b\ (b \neq 0)\) such that:
\(6 + \sqrt{2} = \dfrac{a}{b}\)
\(\Rightarrow \sqrt{2} = \dfrac{a}{b} – 6\)

Since \(a\) and \(b\) are integers, \(\dfrac{a}{b} – 6\) is also rational and hence, \(\sqrt{2}\) should be rational.
This contradicts the fact that \(\sqrt{2}\) is irrational. Therefore, our assumption is false and hence, \(6 + \sqrt{2}\) is irrational.