Real Numbers: Important Questions Part 2 (Class X)

This is the second set of important questions on Real Numbers for Class 10. It covers LCM and HCF by prime factorization, why certain powers never end in zero, recognising composite numbers, proving surds are irrational, and two word problems.

LCM and HCF of 15 and 21

Factorize each number into primes: $15 = 3 \times 5, \qquad 21 = 3 \times 7$

The HCF is the product of the common prime factors, and the LCM is the product of the highest powers of all primes that appear: $\text{HCF} = 3, \qquad \text{LCM} = 3 \times 5 \times 7 = 105$

A true or false on the product of three numbers

The statement to judge is: for any three numbers, their product equals the product of their HCF and LCM. For two numbers $a$ and $b$ the identity $a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)$ does hold. However, this identity does not extend to three numbers: in general $a \times b \times c \neq \text{HCF}(a,b,c) \times \text{LCM}(a,b,c)$. For example, take $a = 2$, $b = 3$, $c = 4$. Then $a \times b \times c = 24$, while $\text{HCF}(2,3,4) = 1$ and $\text{LCM}(2,3,4) = 12$, so $\text{HCF} \times \text{LCM} = 12 \neq 24$. Since the proposed identity fails for three numbers, the statement is false.

Why $4^n$ never ends in zero

A number ends in $0$ only if it is divisible by $10 = 2 \times 5$, that is, only if its prime factorization contains a $5$. But $4^n = (2^2)^n = 2^{2n},$ which contains no factor of $5$. So $4^n$ never ends in the digit $0$ for any natural number $n$.

Why $6^n$ never ends in zero

The same idea applies. If $6^n$ ended in $0$ it would be divisible by $5$, so a $5$ would appear in its prime factorization. But $6^n = (2 \times 3)^n = 2^n \times 3^n,$ which has no factor of $5$. Therefore $6^n$ does not end in $0$ for any natural number $n$.

Showing expressions are composite

A number is composite if it can be written as a product of primes (other than itself and $1$). The trick is to take out a common factor.

First expression. $7 \times 11 \times 13 + 13 = 13\,(7 \times 11 + 1) = 13 \times 78 = 13 \times (2 \times 3 \times 13) = 2 \times 3 \times 13^2$ This is a product of primes, so the number is composite.

Second expression. Here $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 7! = 5040$, so $7! + 5 = 5\,(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times 1009$ Since $5$ is a factor, the number has a divisor other than $1$ and itself, so it is composite.

Proving $\sqrt{2}$ is irrational

We use proof by contradiction. Assume $\sqrt{2}$ is rational, so it can be written as $\sqrt{2} = \frac{a}{b}$ where $a$ and $b$ are co-prime integers with $b \neq 0$.

Squaring and rearranging: $2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$

So $2$ divides $a^2$, and by the theorem that if a prime $p$ divides $a^2$ then $p$ divides $a$, we get that $2$ divides $a$. Write $a = 2c$. Substituting back: $2b^2 = (2c)^2 = 4c^2 \implies b^2 = 2c^2$

So $2$ divides $b^2$, and hence $2$ divides $b$. Now $2$ is a common factor of both $a$ and $b$, which contradicts the assumption that they are co-prime. Therefore the assumption is wrong and $\sqrt{2}$ is irrational.

More irrationality proofs

$\sqrt{3}$ is irrational. By the same argument, assume $\sqrt{3} = \tfrac{a}{b}$ with $a, b$ co-prime. Then $3b^2 = a^2$, so $3$ divides $a$; writing $a = 3c$ gives $b^2 = 3c^2$, so $3$ divides $b$. The shared factor $3$ contradicts co-primeness, so $\sqrt{3}$ is irrational.

$\dfrac{1}{\sqrt{2}}$ is irrational. Rationalize: $\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ If this were rational, then $\sqrt{2} = 2 \times \tfrac{1}{\sqrt{2}}$ would be rational too. But $\sqrt{2}$ is irrational, a contradiction, so $\tfrac{1}{\sqrt{2}}$ is irrational.

$6 + \sqrt{2}$ is irrational. Assume it is rational and equals $\tfrac{a}{b}$. Then $\sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6b}{b},$ which would make $\sqrt{2}$ rational. That contradicts the irrationality of $\sqrt{2}$, so $6 + \sqrt{2}$ is irrational.

$\sqrt{2} + \sqrt{3}$ is irrational. Assume it is rational. Squaring: $(\sqrt{2} + \sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}$ If $5 + 2\sqrt{6} = \tfrac{a}{b}$, then $\sqrt{6} = \frac{a - 5b}{2b},$ which would be rational. Since $\sqrt{6}$ is irrational, this is a contradiction, so $\sqrt{2} + \sqrt{3}$ is irrational.

Word problem: stacking with HCF

A sweet stall has $420$ kaju barfis and $130$ badam barfis, to be stacked so each stack holds the same number and the number of stacks is the least possible. The number to put in each stack is the HCF of $420$ and $130$.

$420 = 2^2 \times 3 \times 5 \times 7, \qquad 130 = 2 \times 5 \times 13$

The HCF takes the common primes to their lowest powers: $\text{HCF} = 2 \times 5 = 10$

So each stack holds $10$ barfis.

Word problem: meeting again with LCM

Two people start together at the same point of a circular field and go the same way. One takes $18$ minutes per round and the other takes $12$ minutes. They next meet at the start after the LCM of the two times.

$12 = 2^2 \times 3, \qquad 18 = 2 \times 3^2$

The LCM takes the highest power of each prime: $\text{LCM} = 2^2 \times 3^2 = 4 \times 9 = 36$

So they meet again at the starting point after $36$ minutes.

Key takeaways

• The HCF uses the lowest powers of the common primes and the LCM uses the highest powers of all primes; for two numbers their product equals HCF times LCM.
• A number can end in $0$ only if a factor of $5$ appears in its prime factorization, so powers like $4^n$ and $6^n$ never do.
• To prove a surd is irrational, assume it is a fraction in lowest terms and derive a shared factor, contradicting co-primeness.