Real Numbers

Concept: Prime factorisation — divide the number successively by the smallest prime factor until the quotient is 1.

(i) 140
$$140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7$$
$$\boxed{140 = 2^2 \times 5 \times 7}$$

(ii) 156
$$156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$$
$$\boxed{156 = 2^2 \times 3 \times 13}$$

(iii) 3825
$$3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3 \times 3 \times 5 \times 85 = 3 \times 3 \times 5 \times 5 \times 17$$
$$\boxed{3825 = 3^2 \times 5^2 \times 17}$$

(iv) 5005
$$5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13$$
$$\boxed{5005 = 5 \times 7 \times 11 \times 13}$$

(v) 7429
$$7429 = 17 \times 437 = 17 \times 19 \times 23$$
$$\boxed{7429 = 17 \times 19 \times 23}$$

Concept: Express each number as a product of prime factors. HCF = product of the smallest powers of common prime factors. LCM = product of the greatest powers of all prime factors. Verification: LCM × HCF = product of the two numbers.

---
(i) 26 and 91

Prime factorisations:
$$26 = 2 \times 13, \quad 91 = 7 \times 13$$

Common prime factor: $13$
$$\text{HCF}(26, 91) = 13$$
$$\text{LCM}(26, 91) = 2 \times 7 \times 13 = 182$$

Verification:
$$\text{LCM} \times \text{HCF} = 182 \times 13 = 2366$$
$$26 \times 91 = 2366 \quad \checkmark$$

---
(ii) 510 and 92

Prime factorisations:
$$510 = 2 \times 255 = 2 \times 3 \times 85 = 2 \times 3 \times 5 \times 17$$
$$92 = 2 \times 46 = 2 \times 2 \times 23 = 2^2 \times 23$$

Common prime factor: $2$ (smallest power $= 2^1$)
$$\text{HCF}(510, 92) = 2$$
$$\text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$$

Verification:
$$\text{LCM} \times \text{HCF} = 23460 \times 2 = 46920$$
$$510 \times 92 = 46920 \quad \checkmark$$

---
(iii) 336 and 54

Prime factorisations:
$$336 = 2 \times 168 = 2^4 \times 3 \times 7$$
$$54 = 2 \times 27 = 2 \times 3^3$$

Common prime factors: $2$ and $3$ (smallest powers $2^1$ and $3^1$)
$$\text{HCF}(336, 54) = 2^1 \times 3^1 = 6$$
$$\text{LCM}(336, 54) = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$$

Verification:
$$\text{LCM} \times \text{HCF} = 3024 \times 6 = 18144$$
$$336 \times 54 = 18144 \quad \checkmark$$

Concept: HCF = product of smallest powers of common prime factors; LCM = product of greatest powers of all prime factors present.

---
(i) 12, 15 and 21

Prime factorisations:
$$12 = 2^2 \times 3, \quad 15 = 3 \times 5, \quad 21 = 3 \times 7$$

Common prime factor to all three: $3$ (smallest power $= 3^1$)
$$\text{HCF}(12, 15, 21) = 3$$

Greatest powers of all prime factors: $2^2,\ 3^1,\ 5^1,\ 7^1$
$$\text{LCM}(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 420$$

---
(ii) 17, 23 and 29

All three numbers are prime, so they have no common factor other than 1.
$$17 = 17, \quad 23 = 23, \quad 29 = 29$$
$$\text{HCF}(17, 23, 29) = 1$$
$$\text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = 11339$$

---
(iii) 8, 9 and 25

Prime factorisations:
$$8 = 2^3, \quad 9 = 3^2, \quad 25 = 5^2$$

There is no prime factor common to all three numbers.
$$\text{HCF}(8, 9, 25) = 1$$

Greatest powers of all prime factors: $2^3,\ 3^2,\ 5^2$
$$\text{LCM}(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$$

Given: $\text{HCF}(306, 657) = 9$

Formula used:
$$\text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)}$$

Calculation:
$$\text{LCM}(306, 657) = \frac{306 \times 657}{9}$$
$$= \frac{201042}{9} = 22338$$

$$\boxed{\text{LCM}(306, 657) = 22338}$$

Concept: A number ends with the digit 0 if and only if it has both 2 and 5 as prime factors (i.e., 10 is a factor of the number).

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

The only prime factors of $6^n$ are 2 and 3.

For $6^n$ to end in 0, it must be divisible by 10, which requires 5 as a prime factor. But 5 does not appear in the prime factorisation of $6^n$ for any natural number $n$.

By the Fundamental Theorem of Arithmetic, the prime factorisation of a number is unique. Since 5 is never a factor of $6^n$, it can never be divisible by 10.

Conclusion: $6^n$ cannot end with the digit 0 for any natural number $n$.

Concept: A composite number has at least one factor other than 1 and itself.

---
Expression 1: $7 \times 11 \times 13 + 13$

Take 13 as a common factor:
$$7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78$$

Further: $78 = 2 \times 3 \times 13$, so
$$13 \times 78 = 13 \times 2 \times 3 \times 13 = 2 \times 3 \times 13^2$$

Since the number has factors other than 1 and itself (e.g., 13 and 78), it is a composite number.

---
Expression 2: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$

Take 5 as a common factor:
$$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$$
$$= 5 \times (1008 + 1) = 5 \times 1009$$

Since 1009 is a prime number, $5 \times 1009$ has factors 1, 5, 1009, and 5045. It has factors other than 1 and itself, so it is a composite number.

Conclusion: Both expressions are composite numbers because each can be expressed as a product of two integers both greater than 1.

Given: Sonia takes 18 minutes per round; Ravi takes 12 minutes per round. They start together from the same point.

Concept: They will meet again at the starting point after a time that is the LCM of their individual times (the smallest time at which both complete a whole number of rounds).

Prime factorisations:
$$18 = 2 \times 3^2, \quad 12 = 2^2 \times 3$$

LCM:
$$\text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$$

Conclusion: Sonia and Ravi will meet again at the starting point after 36 minutes.

*(In 36 minutes, Sonia completes $36 \div 18 = 2$ rounds and Ravi completes $36 \div 12 = 3$ rounds.)*

Proof (by contradiction):

Assume that $\sqrt{5}$ is rational.

Then there exist integers $a$ and $b$ (with $b \neq 0$) such that
$$\sqrt{5} = \frac{a}{b}$$
where $a$ and $b$ are coprime (i.e., $\text{HCF}(a, b) = 1$).

Squaring both sides:
$$5 = \frac{a^2}{b^2} \implies a^2 = 5b^2 \quad \cdots (1)$$

This means $5$ divides $a^2$. Since 5 is prime, by the theorem *"if a prime $p$ divides $a^2$, then $p$ divides $a$"*, we get:
$$5 \text{ divides } a$$

So we can write $a = 5c$ for some integer $c$.

Substituting in (1):
$$(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2$$

This means $5$ divides $b^2$, and therefore $5$ divides $b$.

But now 5 divides both $a$ and $b$, which contradicts our assumption that $\text{HCF}(a, b) = 1$.

Conclusion: Our assumption was wrong. Therefore, $\sqrt{5}$ is irrational. $\blacksquare$

Proof (by contradiction):

We first note that $\sqrt{5}$ is irrational (proved in Q.1).

Assume that $3 + 2\sqrt{5}$ is rational.

Then there exist integers $a$ and $b$ ($b \neq 0$) such that:
$$3 + 2\sqrt{5} = \frac{a}{b}$$

Rearranging:
$$2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$$
$$\sqrt{5} = \frac{a - 3b}{2b}$$

Since $a$, $b$ are integers and $b \neq 0$, the right-hand side $\dfrac{a-3b}{2b}$ is a rational number.

This means $\sqrt{5}$ is rational — but this contradicts the fact that $\sqrt{5}$ is irrational.

Conclusion: Our assumption was wrong. Therefore, $3 + 2\sqrt{5}$ is irrational. $\blacksquare$

Preliminary fact: $\sqrt{2}$ and $\sqrt{5}$ are irrational (standard results).

---
(i) $\dfrac{1}{\sqrt{2}}$

Assume $\dfrac{1}{\sqrt{2}}$ is rational. Then there exist integers $a$, $b$ ($b \neq 0$, $\text{HCF}(a,b)=1$) such that:
$$\frac{1}{\sqrt{2}} = \frac{a}{b}$$
$$\sqrt{2} = \frac{b}{a}$$

Since $a$ and $b$ are integers ($a \neq 0$), $\dfrac{b}{a}$ is rational, which means $\sqrt{2}$ is rational. This contradicts the known fact that $\sqrt{2}$ is irrational.

Conclusion: $\dfrac{1}{\sqrt{2}}$ is irrational. $\blacksquare$

---
(ii) $7\sqrt{5}$

Assume $7\sqrt{5}$ is rational. Then there exist integers $a$, $b$ ($b \neq 0$) such that:
$$7\sqrt{5} = \frac{a}{b}$$
$$\sqrt{5} = \frac{a}{7b}$$

Since $a$ and $7b$ are integers ($7b \neq 0$), $\dfrac{a}{7b}$ is rational, which means $\sqrt{5}$ is rational. This contradicts the known fact that $\sqrt{5}$ is irrational.

Conclusion: $7\sqrt{5}$ is irrational. $\blacksquare$

---
(iii) $6 + \sqrt{2}$

Assume $6 + \sqrt{2}$ is rational. Then there exist integers $a$, $b$ ($b \neq 0$) such that:
$$6 + \sqrt{2} = \frac{a}{b}$$
$$\sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6b}{b}$$

Since $a$ and $b$ are integers ($b \neq 0$), $\dfrac{a-6b}{b}$ is rational, which means $\sqrt{2}$ is rational. This contradicts the known fact that $\sqrt{2}$ is irrational.

Conclusion: $6 + \sqrt{2}$ is irrational. $\blacksquare$