Chapter 1: Real Numbers

To say that a number “ends with the digit 0” in the decimal system means that it is a multiple of 10.

Therefore, if for some natural number $$n$$ the number $$4^n$$ were to end with the digit 0, then

$$10 \mid 4^n$$    (10 divides $$4^n$$).

Because $$10 = 2 \times 5$$, divisibility by 10 requires simultaneous divisibility by both 2 and 5. In other words, we would need

$$2 \mid 4^n \quad\text{and}\quad 5 \mid 4^n.$$

The first condition $$2 \mid 4^n$$ is always true, since $$4^n = (2^2)^n = 2^{2n}$$ contains the prime factor 2.

The crucial point is the second condition. Let us examine the prime-factorisation of $$4^n$$ explicitly:

$$4^n = (2^2)^n = 2^{2n}.$$

This expression is a power of 2 only; its prime-factorisation contains no other primes. In particular, it contains no factor 5. Hence

$$5 \nmid 4^n \quad\text{for every natural } n.$$

Since divisibility by 5 fails, divisibility by 10 fails as well. Consequently $$4^n$$ can never end with the digit 0 for any natural number $$n$$.

A quick check of the first few powers confirms this reasoning:

• $$4^1 = 4$$ (last digit 4)
• $$4^2 = 16$$ (last digit 6)
• $$4^3 = 64$$ (last digit 4)
• $$4^4 = 256$$ (last digit 6)

The last digit keeps alternating between 4 and 6, never becoming 0.

Therefore, there is no natural number $$n$$ for which $$4^n$$ ends with the digit 0.

Step 1 · Prime factorise each number

6 = $$2\times3$$   (both 2 and 3 are primes)
20 = $$2\times2\times5 =\;2^{2}\times5$$

Step 2 · Identify common and uncommon prime factors

• Common prime factor = $$2$$ (appears in both).
• Uncommon prime factors = $$3$$ (only in 6) and $$5$$ (only in 20).

Step 3 · Highest Common Factor (HCF)

The HCF is the product of only the common prime factors, each taken with the smallest power in which it occurs.
Common factor $$2$$ occurs as $$2^{1}$$ in 6 and $$2^{2}$$ in 20 ⇒ take $$2^{1}$$.
Hence $$\text{HCF}=2$$.

Step 4 · Lowest Common Multiple (LCM)

The LCM is the product of all different prime factors, each with its highest power occurring in either number:

Highest power of $$2$$ is $$2^{2}$$,
Highest power of $$3$$ is $$3^{1}$$,
Highest power of $$5$$ is $$5^{1}$$.

Therefore $$\text{LCM}=2^{2}\times3\times5=4\times3\times5=60$$.

Result

The HCF of 6 and 20 is $$2$$ and their LCM is $$60$$.

A prime-factorisation table for each number is the neatest way to begin.

Collecting the prime factors we have

$$96 = 2^5 \times 3\!, \qquad 404 = 2^2 \times 101$$

Highest Common Factor (HCF)

• The common prime is only $$2$$.
• Take the smallest power of this common prime: $$2^2$$.

Therefore $$\text{HCF}(96,\,404) = 2^2 = 4$$.

Lowest Common Multiple (LCM)

• For each distinct prime that appears in either factorisation, take its highest power:
• $$2^5,\; 3^1,\; 101^1$$

Hence

$$\text{LCM}(96,\,404) = 2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 96 \times 101 = 9696$$

One may verify with the product–HCF relation:

$$96 \times 404 = 38784, \quad \frac{38784}{\text{HCF}} = \frac{38784}{4} = 9696,$$ agreeing with the result.

We factorise each number into primes.

• For $$6$$: $$6 = 2 \times 3$$.
• For $$72$$: divide successively by the smallest primes:
  $$72 = 2 \times 36 = 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 9 = 2 \times 2 \times 2 \times 3 \times 3$$,
  so $$72 = 2^{3} \times 3^{2}$$.
• For $$120$$: $$120 = 2 \times 60 = 2 \times 2 \times 30 = 2 \times 2 \times 2 \times 15 = 2 \times 2 \times 2 \times 3 \times 5$$,
  so $$120 = 2^{3} \times 3 \times 5$$.

Step 2 — Highest Common Factor (HCF)

Take each prime that appears in every factorisation and choose its smallest power.

• Prime $$2$$: minimum exponent $$1$$.
• Prime $$3$$: minimum exponent $$1$$.

Therefore $$\operatorname{HCF} = 2^{1} \times 3^{1} = 6$$.

Step 3 — Lowest Common Multiple (LCM)

Take each prime that appears in any factorisation and choose its highest power.

• Prime $$2$$: highest exponent $$3$$.
• Prime $$3$$: highest exponent $$2$$.
• Prime $$5$$: highest exponent $$1$$.

Hence $$\operatorname{LCM} = 2^{3} \times 3^{2} \times 5 = 8 \times 9 \times 5 = 360$$.

Thus, the required HCF is $$6$$ and the required LCM is $$360$$.

Statement to prove: $$\sqrt{3}$$ is irrational.

In other words, we must show that there are no two integers $$p$$ and $$q\;(q\neq0)$$ such that $$\sqrt{3}=\dfrac{p}{q}$$ when the fraction $$\dfrac{p}{q}$$ is written in its lowest (simplest) form.

Proof by contradiction

1. Assume the contrary. Suppose $$\sqrt{3}$$ is rational. Then we can write

   $$\sqrt{3}=\dfrac{p}{q},$$

   where $$p$$ and $$q$$ are integers, $$q\neq0$$, and $$\gcd(p,q)=1$$ (so the fraction is already in its lowest terms).

2. Square both sides.

   $$3 = \dfrac{p^{2}}{q^{2}}$$

   Multiply by $$q^{2}$$:

   $$3q^{2}=p^{2}$$

3. Conclude a divisibility fact. From $$3q^{2}=p^{2}$$ we see that $$p^{2}$$ is a multiple of 3. In particular,

   $$3 \mid p^{2}.$$

   A prime that divides a square divides the base integer itself, so

   $$3 \mid p.$$

4. Write $$p$$ with factor 3 and substitute. Because 3 divides $$p$$, let

   $$p=3k$$ for some integer $$k$$.

   Substitute into $$3q^{2}=p^{2}$$:

   $$3q^{2}=(3k)^{2}=9k^{2}$$

   Simplify by dividing both sides by 3:

   $$q^{2}=3k^{2}$$

5. Obtain another divisibility fact. The equation $$q^{2}=3k^{2}$$ shows that $$q^{2}$$ is a multiple of 3, so

   $$3 \mid q^{2}\;\Rightarrow\;3 \mid q.$$

6. Contradiction. We have shown that 3 divides both $$p$$ and $$q$$. That contradicts our starting condition $$\gcd(p,q)=1$$ (the numbers were supposed to have no common factor other than 1).

7. Therefore our original assumption, namely that $$\sqrt{3}$$ is rational, is false. Hence $$\sqrt{3}$$ is irrational.

Conclusion: The square root of 3 cannot be expressed as a ratio of two integers in simplest form.

Step 1 – Assume the contrary
Suppose $$5 - \sqrt{3}$$ is a rational number.

Step 2 – Write it as a fraction
Then there exist coprime integers $$p$$ and $$q$$, with $$q \neq 0$$, such that

$$5 - \sqrt{3} = \frac{p}{q}$$

Step 3 – Isolate $$\sqrt{3}$$
Rearrange to get

$$\sqrt{3} = 5 - \frac{p}{q}$$

Combine the terms on the right-hand side over the common denominator $$q$$:

$$\sqrt{3} = \frac{5q - p}{q}$$

Step 4 – Analyse the right-hand side
The numerator $$5q - p$$ and the denominator $$q$$ are both integers, so the fraction $$\frac{5q - p}{q}$$ is a rational number.

Step 5 – Reach a contradiction
Thus we have shown that $$\sqrt{3}$$ equals a rational number. However, it is a standard result (proved earlier in the chapter) that $$\sqrt{3}$$ is irrational. This contradicts our conclusion in Step 4.

Step 6 – Conclude
Our initial assumption must therefore be false. Hence $$5 - \sqrt{3}$$ cannot be rational; that is,

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

Goal : Prove that the number $$3\sqrt{2}$$ is irrational.

Method : Use proof by contradiction (the contrapositive of Euclid’s Division Lemma argument used in the textbook).

Suppose, for the sake of contradiction, that $$3\sqrt{2}$$ is a rational number.

Then we can write it as a quotient of two coprime positive integers:

$$3\sqrt{2}=\frac{p}{q},$$ where $$p,q\in\mathbb{Z}^{+}$$ and $$\gcd(p,q)=1.$$

Isolate $$\sqrt{2}$$ by dividing both sides by $$3$$:

$$\sqrt{2}=\frac{p}{3q}.$$

Square both sides to remove the square root:

$$2=\left(\frac{p}{3q}\right)^{2}=\frac{p^{2}}{9q^{2}}.$$

Cross–multiply to obtain a relation between integers:

$$p^{2}=18q^{2}=2\cdot9q^{2}.$$

Step 1 : From $$p^{2}=2\cdot(9q^{2})$$ we see that $$p^{2}$$ is divisible by $$2$$.
Therefore $$p$$ itself is divisible by $$2$$ (because if the square of an integer is even, the integer is even).
Hence write $$p=2k$$ for some integer $$k$$.

Step 2 : Substitute $$p=2k$$ back into the equation $$p^{2}=18q^{2}$$:

$$(2k)^{2}=18q^{2}\;\Longrightarrow\;4k^{2}=18q^{2}.$$

Divide both sides by $$2$$:

$$2k^{2}=9q^{2}.$$

The left side $$2k^{2}$$ is even; hence the right side $$9q^{2}$$ is also even.
But $$9$$ is odd, so the parity of $$9q^{2}$$ is the same as that of $$q^{2}$$. Therefore $$q^{2}$$ (and hence $$q$$) is even.

Step 3 : We have shown both $$p$$ and $$q$$ are even, so they share a common factor $$2$$. This contradicts our initial assumption that $$p$$ and $$q$$ are coprime (i.e. $$\gcd(p,q)=1$$).

Since assuming $$3\sqrt{2}$$ was rational leads to a contradiction, the assumption is false.

Therefore, $$3\sqrt{2}$$ is irrational.

The Highest Common Factor (HCF) and the Lowest Common Multiple (LCM) of two positive integers are connected by the relation

$$a \times b = \mathrm{HCF}(a,b) \times \mathrm{LCM}(a,b).$$

For the given numbers we have $$a = 306$$, $$b = 657$$ and $$\mathrm{HCF}(306,657) = 9$$. Substituting these values:

$$306 \times 657 = 9 \times \mathrm{LCM}(306,657).$$

Rearranging to isolate the LCM,

$$\mathrm{LCM}(306,657) = \dfrac{306 \times 657}{9}.$$

First multiply:

$$306 \times 657 = 201042.$$

Now divide by the HCF:

$$\mathrm{LCM}(306,657) = \dfrac{201042}{9} = 22338.$$

Therefore, $$\mathrm{LCM}(306,657) = 22338.$$

We want to know whether the number $$6^n$$, where $$n$$ is any natural (positive integer) exponent, can have the digit 0 in the units (ones) place.

Step 1 · Recall when a number ends in 0
A whole number ends with the digit 0 if and only if it is a multiple of 10. Equivalently, its prime–factor form must contain both the primes $$2$$ and $$5$$, because $$10 = 2 \times 5$$. In particular, the power of $$5$$ in the factorisation must be at least 1.

Step 2 · Prime–factorise $$6^n$$
Write $$6$$ first: $$6 = 2 \times 3$$ (its prime factors). Raising to the power $$n$$ gives

$$6^n = (2 \times 3)^n = 2^n \times 3^n.$$

Step 3 · Check for a factor 5
The factorisation $$2^n \times 3^n$$ clearly contains the prime $$2$$ (indeed, $$2^n$$), but it contains no factor $$5$$ at all, no matter what natural number $$n$$ we choose.

Step 4 · Conclude about divisibility by 10
Because $$6^n$$ lacks the prime $$5$$, it is not divisible by $$5$$, hence not divisible by $$10$$. Therefore $$6^n$$ can never end with the digit 0.

Final statement
For every natural number $$n$$, the number $$6^n$$ does not end with the digit 0.

First number: $$7 \times 11 \times 13 + 13$$

Take the common factor $$13$$:

$$7 \times 11 \times 13 + 13 = 13 \bigl(7 \times 11 + 1\bigr)$$

Work out the bracket:

$$7 \times 11 = 77 \;\;\Rightarrow\;\; 7 \times 11 + 1 = 78$$

Hence the number equals

$$13 \times 78$$

Because it has a factor $$13\,(>1)$$ and another factor $$78\,(>1)$$, the number is not prime; therefore it is a composite number.

Second number: $$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$$

The product $$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ is $$7!$$, so the expression is $$7! + 5$$.

Now $$7!$$ already contains the factor $$5$$ (because one of the terms in the product is $$5$$). Hence write $$7! = 5k$$ for some whole number $$k$$.

Then

$$7! + 5 = 5k + 5 = 5(k + 1)$$

Thus the given number is a multiple of $$5$$. Since it clearly exceeds $$5$$, both factors $$5$$ and $$(k+1)$$ are greater than $$1$$. Consequently the number is composite.

Let the time taken by Sonia to complete one round be $$T_S = 18\text{ min}$$ and the time taken by Ravi be $$T_R = 12\text{ min}$$.

For both of them to be simultaneously at the starting point again, we need a time $$t$$ which is a common multiple of both $$18$$ and $$12$$. The smallest such time is their lowest common multiple, LCM$$(18,12)$$.

Step 1 – Prime factorisation

• $$18 = 2 \times 3^2$$
• $$12 = 2^2 \times 3$$

Step 2 – Build the LCM

Take each prime involved with its highest power appearing in either number:

• Prime $$2$$: highest power is $$2^2$$
• Prime $$3$$: highest power is $$3^2$$

Hence

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

Step 3 – Interpret the result

The smallest time after the start at which both will be back together at the starting point is therefore $$t = 36\text{ minutes}$$.

Proof by contradiction

1. Assume, for the sake of contradiction, that $$\sqrt{5}$$ is a rational number. Then it can be written in the form $$\sqrt{5}=\dfrac{p}{q}$$ where $$p,q\in\mathbb{Z}$$, $$q\ne0$$ and $$p$$ and $$q$$ have no common factor other than $$1$$ (that is, the fraction is in its lowest terms).
2. Square both sides:
   $$5=\dfrac{p^{2}}{q^{2}}$$
   Multiplying by $$q^{2}$$ gives
   $$p^{2}=5q^{2}$$
3. The equation $$p^{2}=5q^{2}$$ shows that $$p^{2}$$ is a multiple of $$5$$. If a prime divides the square of an integer, it must also divide the integer itself. Therefore $$5$$ divides $$p$$.
4. Since $$5$$ divides $$p$$, we can write $$p=5k$$ for some integer $$k$$. Substitute this into $$p^{2}=5q^{2}$$:
   $$(5k)^{2}=5q^{2}\quad\Longrightarrow\quad25k^{2}=5q^{2}$$
   Divide both sides by $$5$$:
   $$q^{2}=5k^{2}$$
5. The equation $$q^{2}=5k^{2}$$ shows that $$q^{2}$$ is also a multiple of $$5$$, so $$5$$ divides $$q$$.
6. Thus both $$p$$ and $$q$$ are divisible by $$5$$. This contradicts our initial statement that $$p$$ and $$q$$ have no common factor other than $$1$$.
7. The assumption that $$\sqrt{5}$$ is rational leads to a contradiction. Hence our assumption is false, and we conclude that $$\sqrt{5}$$ is irrational.

Step 1 ― Recall a known fact
It is already proved in the textbook that $$\sqrt{5}$$ is irrational. (A short proof is repeated in Step 4.)

Step 2 ― Assume the contrary
Suppose, for the sake of contradiction, that $$3+2\sqrt{5}$$ is a rational number.
Let us denote it by $$r$$, so we write

$$3+2\sqrt{5}=r$$ where $$r$$ is rational.

Step 3 ― Isolate $$\sqrt{5}$$
Subtract $$3$$ from both sides:

$$2\sqrt{5}=r-3$$

Next divide by $$2$$:

$$\sqrt{5}=\dfrac{r-3}{2}$$

Step 4 ― Use the nature of rational numbers
The right-hand side $$\dfrac{r-3}{2}$$ is a quotient of rational numbers (because the set of rational numbers is closed under subtraction and division by a non-zero rational). Hence the right-hand side is rational.
Therefore our equation says that $$\sqrt{5}$$ is rational.

Step 5 ― Contradiction
But Step 1 states that $$\sqrt{5}$$ is irrational. This contradiction arises from our assumption that $$3+2\sqrt{5}$$ is rational.

Step 6 ― Conclude
Because the assumption leads to a contradiction, the assumption is false. Hence $$3+2\sqrt{5}$$ cannot be rational; it is irrational.

(Optional check: why is $$\sqrt{5}$$ irrational?)
Let $$\sqrt{5}=\dfrac{p}{q}$$ with $$p,q$$ coprime integers and $$q\neq0$$. Squaring gives $$5q^2=p^2$$, which implies $$5$$ divides $$p^2$$ and therefore divides $$p$$. Write $$p=5k$$, then $$5q^2=25k^2\;\Rightarrow\;q^2=5k^2$$, so $$5$$ divides $$q$$. Thus both $$p$$ and $$q$$ are multiples of $$5$$, contradicting their being coprime. Hence $$\sqrt{5}$$ is irrational.

Therefore, by contradiction, $$3+2\sqrt{5}$$ is irrational.