Chapter 4.2 - Solved Examples on Mathematical Induction

In the previous section, we saw problems involving inequalities. In this section we will see some advanced problems.

Solved example 4.1
For all n ≥ 1, prove that:
$\frac{1}{1 × 2}+\frac{1}{2 × 3}+\frac{1}{3 × 4}+\;.\;.\;.\;+\frac{1}{n(n+1)}=\frac{n}{n+1}$
Solution:
1. For all n ≥ 1, let P(n) denote the given statement.
• Then we can write:
$P(n):\,\frac{1}{1 × 2}+\frac{1}{2 × 3}+\frac{1}{3 × 4}+\;.\;.\;.\;+\frac{1}{n(n+1)}=\frac{n}{n+1}$
2. Basic step: (n=1)
$\begin{array}{ll}
\phantom{\Rightarrow~}P(1): & \frac{1}{1 × (1+1)} & {}={} & \frac{1}{1+1} \\
{\Rightarrow~} P(1): & \frac{1}{2} & {}={} & \frac{1}{2} \\
\end{array}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
$P(k): \frac{1}{1 × 2}+\frac{1}{2 × 3}+\frac{1}{3 × 4}+\;.\;.\;.\;+\frac{1}{k(k+1)}~=~\frac{k}{k+1}$
• Assume that, the above result is true.
(ii) n = k+1
$\begin{array}{ll}
\phantom{\Rightarrow~}P(k+1): & \frac{1}{1 × 2}+\frac{1}{2 × 3}+\frac{1}{3 × 4}+\;.\;.\;.\;+\frac{1}{k(k+1)}+\{\frac{1}{(k+1)[(k+1)+1]}\} & {}={} & \frac{k+1}{(k+1)+1} & {} \\
{\Rightarrow~} P(k+1): & \frac{k}{k+1}+\{\frac{1}{(k+1)(k+2)}\} & {}={} & \frac{k+1}{k+2} & {} \\
{}&\color {green}{\text{[Using result in (i)]}} & {}&{} & {} \\
{} & \frac{k(k+2)+1}{(k+1)(k+2)} & {}={} & \frac{k+1}{k+2} & {} \\
{} & \frac{k^2 + 2k +1}{(k+1)(k+2)} & {}={} & \frac{k+1}{k+2} & {} \\
{} & \frac{(k+1)^2}{(k+1)(k+2)} & {}={} & \frac{k+1}{k+2} & {} \\
{}&\color {green}{\text{[Using identity }(a+b)^2=a^2+2ab+b^2]} & {}&{} & {} \\
{} & \frac{k+1}{k+2} & {}={} & \frac{k+1}{k+2} & {} \\
\end{array}$
• So P(k+1) is true.
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for any positive integer n.

Solved example 4.2
For every positive integer n, prove that 7ⁿ - 3ⁿ is divisible by 4.
Solution:
1. For any positive integer n, let P(n) denote the given statement.
• Then we can write:
$P(n):7^n - 3^n \; \text{is divisible by 4}$
2. Basic step: (n=1)
$\begin{array}{ll}
P(1): & 7^1 - 3^1 & {} & \color {silver}{\text{is divisible by 4}} & {} \\
{} & 7-3 & {} & {} & {} \\
{} & 4 & {} & \color {silver}{\text{is divisible by 4}} & {} \\
\end{array}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
$\begin{alignat*}{3}
P(k)&: 7^k - 3^k &&\text{is divisible by 4} \\
\Rightarrow P(k)&: 7^k - 3^k = 4d &&\;\text{[d is a natural number]}
\end{alignat*}$
• Assume that, the above result is true.
(ii) n = k+1
$\begin{array}{ll}
P(k+1): & 7^{k+1} - 3^{k+1} & {} & {} & {} \\
{} & 7^k  × 7 ~-~3^k × 3 & {} & \color {silver}{\text{is divisible by 4}}~~\color {green}{\text{---- (a)}} & {} \\
{}& {} & {}& \color {green}{\text{[We want to prove (a)]}} & {} \\
P(k+1): & (4d+3^k)  × 7 ~-~3^k × 3 & {} & {} & {} \\
{}& {} & {}& \color {green}{\text{[Using the result in (i)]}} & {} \\
{} & 28d~+~7 × 3^k ~-~3^k × 3 & {} & {} & {} \\
{} & 28d~+~3^k(7 ~-~3) & {} & {} & {} \\
{} & 28d~+~3^k  × 4 & {} & {} & {} \\
{} & 4(7d~+~3^k) & {} &  \color {silver}{\text{is indeed divisible by 4}} & {} \\
{}& {} & {}& \color {green}{\text{[d and k are integers]}} & {} \\
{}& {} & {}& \color {green}{\text{[So }(7d~+~3^k)\text{ is an integer]}} & {} \\
\end{array}$
• We proved (a) by rearranging it. The rearrangement was done using P(k)
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for any positive integer n.

Solved example 4.3
Prove that (1+x)ⁿ ≥ (1+nx) for all natural number n, where x > -1.
Solution:
1. For all natural number n, let P(n) denote the given statement.
Then we can write: P(n): (1+x)ⁿ ≥ (1+nx), x > -1
2. Basic step: (n=1)
$\begin{array}{ll}
P(1): & (1+x)^1 & {}\ge{} & 1+ 1 × x & {} \\
{} & (1+x) & {}\ge{} & (1+x) & {} \\
\end{array}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
P(k): (1+x)^k ≥ (1+kx)
• Assume that, the above result is true.
(ii) n = k+1
$\begin{array}{ll}
P(k+1): & (1+x)^{k+1} & {}\ge{} & 1+(k+1)x & {} \\
{} & (1+x)^{k+1} & {}\ge{} & 1+kx+x~\color {green}{\text{---- (a)}} & {} \\
{}& {} & {}& \color {green}{\text{[We want to prove (a)]}} & {} \\
{}&{}& {}&{} & {} \\
P(k): & (1+x)^k & {}\ge{} & 1+kx~\color {green}{\text{---- (b)}} & {} \\
{} & (1+x)^k (1+x) & {}\ge{} & (1+kx)(1+x) & {} \\
{}&{}& {}& \color {green}{\text{[Multiplying by (1+x) on both sides]}} & {} \\
{} & (1+x)^{k+1} & {}\ge{} & (1+x+kx+kx^2)~\color {green}{\text{---- (c)}} & {} \\
{}&{}& {}&{} & {} \\
{} & (1+x+kx+kx^2) & {}\ge{} & 1+kx+x~\color {green}{\text{---- (d)}} & {} \\
{}&{} & {}& \color {green}{\text{[Comparing RHS of (b) and (c)]}} & {} \\
{}&{} & {}& \color {green}{\text{[}kx^2~\text{is the only diference]}} & {} \\
{}&{} & {}& \color {green}{\text{[k is a +ve number. Also x is greater than -1]}} & {} \\
{}&{} & {}& \color {green}{\text{[So}~kx^2 \ge 0 \text{]}} & {} \\
{}&{} & {}& \color {green}{\text{[So (d) is true]}} & {} \\
{}&{}& {}&{} & {} \\
{} & (1+x)^{k+1} & {}\ge{} & (1+x+kx+kx^2)~~\ge 1+x+kx~\color {green}{\text{---- (e)}}  & {} \\
{}&{} & {}& \color {green}{\text{[Using (c) and (d)]}} & {} \\
\Rightarrow & (1+x)^{k+1} & {}\ge{} & 1+x+kx & {} \\
{}&{} & {}& \color {green}{\text{[Picking first and last items from (e)]}} & {} \\
{} &{} & {}& \color {green}{\text{[Starting from (b), statement in (a) is proved]}} & {} \\
\end{array}$
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for all natural numbers.

Solved example 4.4
For every positive integer n, prove that (5ⁿ - 5) is divisible by 4.
Solution:
1. For any positive integer n, let P(n) denote the given statement.
• Then we can write:
$P(n): 5^n - 5 \; \text{is divisible by 4}$
2. Basic step: (n=1)
$\begin{array}{ll}
P(1): & 5^1~-~5 & {} & \color {silver}{\text{is divisible by 4}} & {} \\
{} & 5 - 5 & {} & {} & {} \\
{} & 0 & {} & \color {silver}{\text{is divisible by 4}} & {} \\
\end{array}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
$\begin{alignat*}{3}
P(k)&: 5^k ~-~5 &&\text{is divisible by 4} \\
\Rightarrow P(k)&: 5^k ~-~5 = 4d &&\;\text{[d is a natural number]}
\end{alignat*}$
• Assume that, the above result is true.
(ii) n = k+1
$\begin{array}{ll}
P(k+1): & 5^{k+1}~-~5 & {} & {} & {} \\
{} & 5^k  \times 5~-~5 & {} & \color {silver}{\text{is divisible by 4}}~~\color {green}{\text{---- (a)}} & {} \\
{}& {} & {}& \color {green}{\text{[We want to prove (a)]}} & {} \\
P(k+1): & [(4d~+~5) ×5]~-~5 & {} & {} & {} \\
{}& {} & {}& \color {green}{\text{[Using the result in (i)]}} & {} \\
{} & [20d~+~25]~-~5 & {} & {} & {} \\
{} & 20d~+~20 & {} & {} & {} \\
{} & 4(5d~+~5) & {} & \color {silver}{\text{is indeed divisible by 4}} & {}  & {} \\
\end{array}$
• We proved (a) by rearranging it. The rearrangement was done using P(k)
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for all natural numbers.

Solved example 4.5
For every positive integer n, prove that (2 × 7ⁿ) + (3 × 5ⁿ) - 5 is divisible by 24.
Solution:
1. For any positive integer n, let P(n) denote the given statement.
• Then we can write:
$P(n):(2 \times 7^n) + (3 \times 5^n)-5 \; \text{is divisible by 24}$
2. Basic step: (n=1)
$\begin{array}{ll}
P(1): & (2 \times 7^1) + (3 \times 5^1)-5 & {} & \color {silver}{\text{is divisible by 24}} & {} \\
{} & 14 + 15 - 5 & {} & {} & {} \\
{} & 24 & {} & \color {silver}{\text{is divisible by 24}} & {} \\
\end{array}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
$\begin{alignat*}{3}
P(k)&: (2 \times 7^k) + (3 \times 5^k)-5 &&\text{is divisible by 24} \\
\Rightarrow P(k)&: (2 \times 7^k) + (3 \times 5^k)-5 = 24d &&\;\text{[d is a natural number]}
\end{alignat*}$
• Assume that, the above result is true.
(ii) n = k+1
$\begin{array}{ll}
P(k+1): & (2 \times 7^{k+1}) + (3 \times 5^{k+1})-5 & {} & {} & {} \\
{} & (2 \times 7^k \times 7) + (3 \times 5^k \times 5)-5 & {} & \color {silver}{\text{is divisible by 24}}~~\color {green}{\text{---- (a)}} & {} \\
{}& {} & {}& \color {green}{\text{[We want to prove (a)]}} & {} \\
P(k+1): & [(24d~-3\times 5^k~+~5) \times 7]  & {} & {} & {} \\
{}& {}+ (3 \times 5^k \times 5)-5 & {}& \color {green}{\text{[Using the result in (i)]}} & {} \\
{} & [(24 × d × 7~-3 × 5^k × 7~+~5 × 7)] & {} & {} & {} \\
{} & {}+ (3 \times 5^k \times 5)-5 & {} & {} & {} \\
{} & {} & {} & {} & {} \\
{} & 24 × d × 7~-21 × 5^k & {} & {} & {} \\
{} & {}+~35 + 15 \times 5^k ~-~5 & {} & {} & {} \\
{} & {} & {} & {} & {} \\
{} & 24 × d × 7~-6 × 5^k~+~30 & {} & {} & {} \\
{} & 24 × d × 7~-6(5^k~-~5) & {} & {} & {} \\
{} & 24 × d × 7~-6(4m) & {} &\color {green}{\text{[From previous example,   }(5^k~-~5)\text{ is a multiple of 4]}} & {} \\
{}& {} & {}& \color {green}{\text{[m is an integer]}} & {} \\
{}& {} & {}& {} & {} \\
{} & 24 × d × 7~-24m & {} &  {} & {} \\
{} & 24(7d~-~m) & {} &  \color {silver}{\text{is indeed divisible by 24}} & {} \\
\end{array}$
• We proved (a) by rearranging it. The rearrangement was done using P(k)
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for all natural numbers.

Solved example 4.6
For all natural numbers n, prove that:
$$1^2+2^2+3^2+\;.\;.\;.\;+n^2>\frac{n^3}{3}$$
Solution:
1. For all natural numbers n, let P(n) denote the given statement.
• Then we can write:
$P(n):\,1^2+2^2+3^2+\;.\;.\;.\;+n^2>\frac{n^3}{3}$
2. Basic step: (n=1)
$\begin{array}{ll}
P(1): & 1^2 & {}>{} & \frac{1^3}{3} & {} \\
{} & 1 & {}>{} & \frac{1}{3} & {} \\
\end{array}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
$P(k): 1^2+2^2+3^2+\;.\;.\;.\;+k^2>\frac{k^3}{3}$
• Assume that, the above result is true.
(ii) n = k+1
$\begin{array}{ll}
P(k+1): & 1^2+2^2+3^2+\;.\;.\;.\;+k^2+\{(k+1)^2\} & {}>{} & \frac{(k+1)^3}{3} & {} \\
{} & 1^2+2^2+3^2+\;.\;.\;.\;+k^2+\{(k+1)^2\} & {}>{} & \frac{k^3+3k^2+3k+1}{3}~\color {green}{\text{---- (a)}} & {} \\
{}& {} & {}& \color {green}{\text{[We want to prove (a)]}} & {} \\
{}&{}& {}&{} & {} \\
P(k): & 1^2+2^2+3^2+\;.\;.\;.\;+k^2 & {}>{} & \frac{k^3}{3}~\color {green}{\text{---- (b)}} & {} \\
{} & 1^2+2^2+3^2+\;.\;.\;.\;+k^2+(k+1)^2 & {}>{} & \frac{k^3}{3}+(k+1)^2~\color {green}{\text{---- (c)}} & {} \\
{}&{}& {}& \color {green}{\text{[Adding}~ (k+1)^2~\text{on both sides]}} & {} \\
{} & {} & {}>{} & \frac{k^3+3(k+1)^2}{3} & {} \\
{} & {} & {}>{} & \frac{k^3+3(k^2+2k+1)}{3} & {} \\
\Rightarrow & 1^2+2^2+3^2+\;.\;.\;.\;+k^2+(k+1)^2 & {}>{} & \frac{k^3+3k^2+6k+3}{3}~\color {green}{\text{---- (d)}} & {} \\
{}&{}& {}&{} & {} \\
{} & \frac{k^3+3k^2+6k+3}{3} & {}>{} & \frac{k^3+3k^2+3k+1}{3}~\color {green}{\text{---- (e)}} & {} \\
{}&{} & {}& \color {green}{\text{[Comparing RHS of (d) and (a)]}} & {} \\
{}&{} & {}& \color {green}{\text{[The denominators are same. So compare numerators]}} & {} \\
{}&{} & {}& \color {green}{\text{['k' is an integer. So (6k+3) > (3k+1)]}} & {} \\
{}&{} & {}& \color {green}{\text{[So (e) is true]}} & {} \\
{}&{}& {}&{} & {} \\
\Rightarrow & 1^2+2^2+3^2+\;.\;.\;.\;+k^2+(k+1)^2 & {}>{} & \frac{k^3+3k^2+6k+3}{3}~~>~~\frac{k^3+3k^2+3k+1}{3}~\color {green}{\text{---- (f)}} & {} \\
{}&{} & {}& \color {green}{\text{[Using (d) and (e)]}} & {} \\
\Rightarrow & 1^2+2^2+3^2+\;.\;.\;.\;+k^2+(k+1)^2 & {}>{} & \frac{k^3+3k^2+3k+1}{3}~\color {green}{\text{---- (f)}} & {} \\
{}&{} & {}& \color {green}{\text{[Picking first and last items from (f)]}} & {} \\
{} &{} & {}& \color {green}{\text{[Starting from (b), statement in (a) is proved]}} & {} \\
\end{array}$
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for all natural numbers.

Solved example 4.7
Prove that for every natural number, (ab)ⁿ = aⁿbⁿ
Solution:
1. For all natural numbers n, let P(n) denote the given statement.
• Then we can write:
P(n): (ab)ⁿ = aⁿbⁿ
2. Basic step: (n=1)
$\begin{alignat*}{3}
P(1)&: (ab)^1 = a^1 \times b^1 \\
\Rightarrow P(1)&: ab = a \times b = ab
\end{alignat*}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
$\begin{alignat*}{3}
P(k)&: (ab)^k = a^k \times b^k \\
\Rightarrow P(k)&: (ab)^k = a^k b^k \\
\end{alignat*}$
• Assume that, the above result is true.
(ii) n = k+1
$\begin{alignat*}{3}
P(k+1)&: (ab)^{k+1} &&= a^{k+1} \times b^{k+1} \; \text{[We want to prove this to be true]} \\
\Rightarrow P(k+1)&: (ab)^{k+1} &&= a^k \times a \times b^k \times b \\
{}&{} &&= \{a^k \times b^k \} \times \{a \times  b \} \\
{}&{} &&= \{(ab)^k \} \times \{a \times  b \} \; \text{[Using the result in (i)]} \\
{}&{} &&= \{(ab)^k \} \times \{(ab) \} \\
\Rightarrow P(k+1)&: (ab)^{k+1} &&=  (ab)^{k+1} \; \text{[This is true]}
\end{alignat*}$
• So P(k+1) is true.
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for any positive integer n.

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• The link below gives the folder containing additional solved examples on this chapter.

Additional Solved Examples

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In the next chapter, we will see complex numbers.

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Chapter 4.1 - Problems Involving Inequalities

In the previous section, we saw the basics of mathematical induction. We saw some examples also. In this section we will see problems involving inequality.

Let us see an example. It can be written in 4 steps:
1. Consider any positive integer n. A mathematician tells us that, this n will be always less than twice it's value.
That is: n < 2n
• We want to prove this statement.
2. First we prove that, this statement is true when n = 1
• Put n = 1 in the LHS. We get LHS = 1
• Put n = 1 in the RHS. We get: RHS = 2 × 1 = 2
• LHS < RHS. So the statement is true when n = 1
◼ This step is called basic step.
3. Next we prove this:
If the statement is true for a positive integer ‘k’, it will be true for the next positive integer (k+1)
• This can be proved in 5 steps:
(i) Put n = k in the LHS. We get: LHS = k
• Put n = k in the RHS. We get: 2 × k = 2k
• We assume that, this LHS is less than RHS. That is., k < 2k
(ii) Put n = (k+1) in the LHS. We get: LHS = k+1
• Put n = (k+1) in the RHS. We get: RHS = 2(k+1) = 2k+2
(iii) We have to prove that LHS in (ii) is less than RHS in (ii)
• That is., we have to prove that: k+1 < 2k+2
   ♦ For proving this, we can use the statement in (i)
   ♦ Remember that, we assume the statement in (i) to be true.
(iv) Consider the statement in (i): k < 2k
• Adding '1' on both sides, we get: k+1 < 2k+1
• So k+1 will be obviously less than 2k+2
• Thus the statement in (iii) is proved.
• So we proved that:
If the statement is true for n = k, it will be true for n = (k+1)
◼ This step is called inductive step.
4. So we used basic step in (2) and inductive step in (3)
• In the basic step, we proved that:
The given statement is true when n = 1
• In the inductive step, we proved that:
If the given statement is true for n = k, it will be true for n = (k+1)
• Since both are proved, we can write:
The given statement is true.

---

Let us write the above steps in the condensed form:

Problem:
Prove that for any positive integer n, n  < 2n

Solution:
1. For any positive integer n, let P(n) denote the given statement.
Then we can write: P(n): n < 2n
2. Basic step: (n=1)
$\begin{array}{ll}
\phantom{\Rightarrow~}P(1): & 1 & {}<{} & (2 × 1) & {} \\
{\Rightarrow~} P(1): & 1 & {}<{} & 2 & {} \\
\end{array}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
P(k): k < 2k
• Assume that, the above result is true.
(ii) n = k+1
$\begin{array}{ll}
\phantom{\Rightarrow~}P(k+1): & k+1 & {}<{} & 2(k+1) & {} \\
{\Rightarrow~} P(k+1): & k+1 & {}<{} & 2k+2~\color {green}{\text{---- (a)}} & {} \\
{}& {} & {}& \color {green}{\text{[We want to prove (a)]}} & {} \\
{}&{}& {}&{} & {} \\
\phantom{\Rightarrow~}P(k): & k & {}<{} & 2k~\color {green}{\text{---- (b)}} & {} \\
{} & k+1 & {}<{} & 2k+1~\color {green}{\text{---- (c)}} & {} \\
{}&{}& {}& \color {green}{\text{[Adding '1' on both sides]}} & {} \\
{}&{}& {}&{} & {} \\
{} & 2k+1 & {}<{} & 2k+2~\color {green}{\text{---- (d)}} & {} \\
{}&{} & {}& \color {green}{\text{[Since 'k' is a +ve integer]}} & {} \\
{}&{}& {}&{} & {} \\
{} & k+1 & {}<{} & 2k+1 ~~ <~~2k+2 & {} \\
{}&{} & {}& \color {green}{\text{[Using (c) and (d)]}} & {} \\
{} & k+1 & {}<{} & 2k+2 & {} \\
{}&{} & {}& \color {green}{\text{[Picking first and third items]}} & {} \\
{}&{} & {}& \color {green}{\text{[Starting from (b), statement in (a) is proved]}} & {} \\
\end{array}$
• So P(k+1) is true.
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for any positive integer n.
5. We can write the conclusion in another way also:
• We want to prove that k+1 < 2k+2
• In (ii), we saw that, k+1 < 2k+1
• Since k is positive integer, 2k+1 will be less than 2k+2
• So we can write: k+1 < 2k+1 < 2k+2
• Picking the first and last expressions, we get: k+1 < 2k+2

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Let us see another example. It can be written in 4 steps:
1. Consider any positive integer n. A mathematician tells us that, this n will be always less than 2 raised to n.
That is: 2ⁿ > n
• We want to prove this statement.
2. First we prove that, this statement is true when n = 1
• Put n = 1 in the LHS. We get LHS = 2¹ × 1 = 2
• Put n = 1 in the RHS. We get: RHS = 1
• LHS > RHS. So the statement is true when n = 1
◼ This step is called basic step.
3. Next we prove this:
If the statement is true for a positive integer ‘k’, it will be true for the next positive integer (k+1)
• This can be proved in 5 steps:
(i) Put n = k in the LHS. We get: LHS = 2^k
• Put n = k in the RHS. We get: RHS = k
• We assume that, this LHS is greater than RHS. That is., 2^k > k
(ii) Put n = (k+1) in the LHS. We get: LHS = 2^k+1
• Put n = (k+1) in the RHS. We get: RHS = k+1
(iii) We have to prove that LHS in (ii) is greater than RHS in (ii)
• That is., we have to prove that: 2^k+1 > k+1
   ♦ For proving this, we can use the statement in (i)
   ♦ Remember that, we assume the statement in (i) to be true.
(iv) Statement in (i) is: 2^k > k
• Adding 1 on both sides, we get: 2^k +1 > k+1 
(v) Let us replace the '1' on the left side by 2^k. We get:
2^k + 2^k > k+1
• This is true because, k is a +ve integer. 2^k will be greater than 1.
(vi) The left side can be simplified. We get: 2 × 2^k > k+1
• This is same as: 2^k+1 > k+1
• So the statement in (iii) is true.
◼ This step is called inductive step.
4. So we used basic step in (2) and inductive step in (3)
• In the basic step, we proved that:
The given statement is true when n = 1
• In the inductive step, we proved that:
If the given statement is true for n = k, it will be true for n = (k+1)
• Since both are proved, we can write:
The given statement is true.

---

Let us write the above steps in the condensed form:

Problem:
Prove that for any positive integer n, 2ⁿ > n

Solution:
1. For any positive integer n, let P(n) denote the given statement.
Then we can write: P(n): 2ⁿ > n
2. Basic step: (n=1)
$\begin{array}{ll}
\phantom{\Rightarrow~}P(1): & 2^1 & {}>{} & 1 & {} \\
{\Rightarrow~} P(1): & 2 & {}>{} & 1 & {} \\
\end{array}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
P(k): 2^k > k
• Assume that, the above statement is true.
(ii) n = k+1
$\begin{array}{ll}
P(k+1): & 2^{k+1} & {}>{} & k+1~\color {green}{\text{---- (a)}} & {} \\
{}& {} & {}& \color {green}{\text{[We want to prove (a)]}} & {} \\
{}&{}& {}&{} & {} \\
P(k): & 2^k & {}>{} & k~\color {green}{\text{---- (b)}} & {} \\
{} & 2^k+1 & {}>{} & k+1~\color {green}{\text{---- (c)}} & {} \\
{}&{}& {}& \color {green}{\text{[Adding '1' on both sides]}} & {} \\
{}&{}& {}&{} & {} \\
{} & 2^k & {}>{} & 1~\color {green}{\text{---- (d)}} & {} \\
{}&{} & {}& \color {green}{\text{[Since 'k' is a +ve integer]}} & {} \\
{} & 2^k+2^k & {}>{} & k+1~\color {green}{\text{---- (e)}} & {} \\
{}&{} & {}& \color {green}{\text{[Using (c) and (d)]}} & {} \\
\Rightarrow & 2 × 2^k & {}>{} & k+1 & {} \\
\Rightarrow & 2^{k+1} & {}>{} & k+1 & {} \\
{} &{} & {}& \color {green}{\text{[Starting from (b), statement in (a) is proved]}} & {} \\
\end{array}$
• So P(k+1) is true.
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for any positive integer n.

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Let us see another problem involving inequality. It can be written in 4 steps:
1. Consider any positive integer n. A mathematician tells us that, '4' raised to this n will be always greater than (n+1).
That is: 4ⁿ > n+1
• We want to prove this statement.
2. First we prove that, this statement is true when n = 1
• Put n = 1 in the LHS. We get LHS = 4¹ = 4
• Put n = 1 in the RHS. We get: RHS = 1+1 = 2
• LHS > RHS. So the statement is true when n = 1
◼ This step is called basic step.
3. Next we prove this:
If the statement is true for a positive integer ‘k’, it will be true for the next positive integer (k+1)
• This can be proved in 5 steps:
(i) Put n = k in the LHS. We get: LHS = 4^k
• Put n = k in the RHS. We get: RHS = k+1
• We assume that, this LHS is greater than RHS. That is., 4^k > k+1
(ii) Put n = (k+1) in the LHS. We get: LHS = 4^(k+1)
• Put n = (k+1) in the RHS. We get: RHS = (k+1)+1
(iii) We have to prove that LHS in (ii) is greater than RHS in (ii)
• That is., we have to prove that: 4^(k+1) > (k+1)+1
   ♦ For proving this, we can use the statement in (i)
   ♦ Remember that, we assume the statement in (i) to be true.
(iv) In (i), it is said that, 4^k is greater than (k+1)
• So (4^k +1) will be greater than [(k+1)+1]
• So (4^k + 4^k) will be greater than [(k+1)+1] (since k is a positive number, 4^k will be greater than 1)
⇒ (2 × 4^k) will be greater than [(k+1)+1]
• So (4^k × 4^k) will be greater than [(k+1)+1] (since k is a positive number, 4^k will be greater than 2)
⇒ 4^(k+1) will be greater than [(k+1)+1]
Thus we proved that: 4^(k+1) > (k+1)+1
• That is:
If the statement is true for n = k, it will be true for n = (k+1)
◼ This step is called inductive step.
4. So we used basic step in (2) and inductive step in (3)
• In the basic step, we proved that:
The given statement is true when n = 1
• In the inductive step, we proved that:
If the given statement is true for n = k, it will be true for n = (k+1)
• Since both are proved, we can write:
The given statement is true.

---

Let us write the above steps in the condensed form:

Problem:
Prove that for any positive integer n, 4ⁿ > n+1

Solution:
1. For any positive integer n, let P(n) denote the given statement.
Then we can write: P(n): 4ⁿ > n+1
2. Basic step: (n=1)
$\begin{array}{ll}
\phantom{\Rightarrow~}P(1): & 4^1 & {}>{} & 1+1 & {} \\
{\Rightarrow~} P(1): & 4 & {}>{} & 2 & {} \\
\end{array}$
• So P(n) is true for n=1
3. Inductive step: (n=k) and (n= k+1)
(i) n= k
P(k): 4^k > k+1
• Assume that, the above result is true.
(ii) n = k+1
$\begin{array}{ll}
P(k+1): & 4^{k+1} & {}>{} & k+1+1~\color {green}{\text{---- (a)}} & {} \\
{}& {} & {}& \color {green}{\text{[We want to prove (a)]}} & {} \\
{}&{}& {}&{} & {} \\
P(k): & 4^k & {}>{} & k+1~\color {green}{\text{---- (b)}} & {} \\
{} & 4^k+1 & {}>{} & k+1+1~\color {green}{\text{---- (c)}} & {} \\
{}&{}& {}& \color {green}{\text{[Adding '1' on both sides]}} & {} \\
{}&{}& {}&{} & {} \\
{} & 4^k & {}>{} & 1~\color {green}{\text{---- (d)}} & {} \\
{}&{} & {}& \color {green}{\text{[Since 'k' is a +ve integer]}} & {} \\
{} & 4^k+4^k & {}>{} & k+1+1~\color {green}{\text{---- (e)}} & {} \\
{}&{} & {}& \color {green}{\text{[Using (c) and (d)]}} & {} \\
\Rightarrow & 2 × 4^k & {}>{} & k+1+1~\color {green}{\text{---- (f)}} & {} \\
{}&{}& {}&{} & {} \\
{} & 4 & {}>{} & 2~\color {green}{\text{---- (g)}} & {} \\
{} & 4 × 4^k & {}>{} & k+1+1 & {} \\
{}&{} & {}& \color {green}{\text{[Using (f) and (g)]}} & {} \\
\Rightarrow & 4^{k+1} & {}>{} & k+1+1 & {} \\
{} &{} & {}& \color {green}{\text{[Starting from (b), statement in (a) is proved]}} & {} \\
\end{array}$
• So P(k+1) is true.
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for any positive integer n.

---

Let us see another problem involving inequality. We will directly write the solution in the condensed form.

Problem:
Prove that for positive integer n, 2n+1 < 2ⁿ, n ≥ 3

Solution:
1. For positive integer n ≥ 3, let P(n) denote the given statement.
Then we can write: P(n): 2n+1 < 2ⁿ, n ≥ 3
2. Basic step: (n=3)
$\begin{array}{ll}
P(3): & [(2 × 3)+1] & {}<{} & 2^3 & {} \\
{} & [6+1] & {}<{} & 8 & {} \\
{} & 7 & {}<{} & 8 & {} \\
\end{array}$
• So P(n) is true for n=3
3. Inductive step: (n=k) and (n= k+1)
(i) n= k [positive integer k ≥ 3]
P(k): 2k+1 < 2^k
• Assume that, the above result is true.
(ii) n = k+1
$\begin{array}{ll}
P(k+1): & 2(k+1)+1 & {}<{} & 2^{k+1} & {} \\
{} & 2k+2+1 & {}<{} & 2^{k+1}~\color {green}{\text{---- (a)}} & {} \\
{}& {} & {}& \color {green}{\text{[We want to prove (a)]}} & {} \\
{}&{}& {}&{} & {} \\
P(k): & 2k+1 & {}<{} & 2^k~\color {green}{\text{---- (b)}} & {} \\
{} & 2k+2+1 & {}<{} & 2^k+2~\color {green}{\text{---- (c)}} & {} \\
{}&{}& {}& \color {green}{\text{[Adding '2' on both sides]}} & {} \\
{}&{}& {}&{} & {} \\
{} & 2^k & {}>{} & 2~\color {green}{\text{---- (d)}} & {} \\
{}&{} & {}& \color {green}{\text{[Since k ≥ 3]}} & {} \\
{} & 2k+2+1 & {}<{} & 2^k+2^k & {} \\
{}&{} & {}& \color {green}{\text{[Using (c) and (d)]}} & {} \\
\Rightarrow & 2k+2+1 & {}<{} & 2 × 2^k & {} \\
\Rightarrow & 2k+2+1 & {}<{} & 2^{k+1} & {} \\
{} &{} & {}& \color {green}{\text{[Starting from (b), statement in (a) is proved]}} & {} \\
\end{array}$
• That is., P(k+1) is true.
4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for any positive integer n

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In the next section, we will see more solved examples.

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Chapter 1.1 - Roster Form And Set-Builder Form

In the previous section, we saw an introduction to sets. In this section, we will see the two methods for representing a set.

◼ There are two methods for representing a set.
    ♦ Roster form
        ✰ This is also known as Tabular form.
    ♦ Set-builder form

We will first see roster form. Some basic features can be written in 5 steps:
1. In roster form, all elements of the set are clearly written.
2. The elements are separated by commas.
3. The elements are enclosed within braces.
4. The order in which the elements are written is not important.
5. Repeating elements are written only once.

Let us see some examples:
(a) Consider the following set:
• The set of all even positive integers which are less than 7.
    ♦ Clearly, the elements are 2, 4, and 6.
    ♦ We must clearly write all those three elements.
    ♦ We must separate them with commas.
    ♦ We must enclose them within braces.
• So the set will appear as: {2, 4, 6}
• If we decide to name the set as A, we can write: A = {2, 4, 6}
• Since the order is not important, we can write A = {4, 6, 2} also. Different arrangements like this are allowed.

(b) Consider the following set:
• The set of all natural numbers which completely divide 42
    ♦ Clearly, the elements are 1, 2, 3, 6, 7, 14, 21, and 42
    ♦ We must clearly write all those 8 elements.
    ♦ We must separate them with commas.
    ♦ We must enclose them within braces.
• So the set will appear as: {1, 2, 3, 6, 7, 14, 21, 42}
• If we decide to name the set as B, we can write: B = {1, 2, 3, 6, 7, 14, 21, 42}
• Since the order is not important, we can write B = {21, 42, 7, 14, 3, 6, 1, 2} also. Different arrangements like this are allowed.

(c) Consider the following set:
• The set of all vowels in English alphabet.
    ♦ Clearly, the elements are a, e, i, o, and u.
    ♦ We must clearly write all those 5 elements.
    ♦ We must separate them with commas.
    ♦ We must enclose them within braces.
• So the set will appear as: {a, e, i, o, u}
• If we decide to name the set as V, we can write: V = {a, e, i, o, u}
• Since the order is not important, we can write V = {i, o, a, e, u} also. Different arrangements like this are allowed.

(d) Consider the following set:
• The set of all natural odd numbers.
    ♦ Clearly, the elements are 1, 3, 5, 7, ...
        ✰ These numbers continue indefinitely. We cannot write them all.
    ♦ In such cases, we write the first 4 elements, followed by a comma and 3 dots.
    ♦ We must enclose them within braces.
• So the set will appear as: {1, 3, 5, 7, ...}
• If we decide to name the set as D, we can write: D = {1, 3, 5, 7, ...}
• Here the order is important.

(e) Consider the following set:
• The set of all letters in the word 'SCHOOL'
    ♦ Clearly, the letters are S, C, H, O, O and L.
        ✰ We must write 'O' only once
• So the set will appear as: {S, C, H, O, L}
• If we decide to name the set as E, we can write: E = {S, C, H, O, L}
• Since the order is not important, we can write E = {S, H, C, L, O} also. Different arrangements like this are allowed.

---

Next we will see the set-builder form. It can be explained in 7 steps:
1. We know that, all the elements of a set will possess a common property.
   ♦ All the elements which possess that property should be included into that set.
         ✰ None of such elements should be excluded.
   ♦ Also, objects which do not possess that property should not be included into that set.
2. This gives us an idea to specify a set:
• A set can be specified by writing down the property.
   ♦ A person who reads the property can easily form the set.
3. Let us see an example:
• The property possessed by every element of a set is this:
It is a vowel of English alphabet.
• A person who reads this property will understand that, elements in the set are: a, e, i, o and u.
4. Another example:
• The property possessed by every element of a set is this:
It is positive even integer less than 10.
• A person who reads this property will understand that, elements in the set are: 2, 4, 6, and 8.
5. Mathematicians have specified the form in which the property is to be written.
• By making such a specification, we can ensure that, people all over the world will follow the same form.
◼ It is called the set-builder form.
6. We can familiarize ourselves with the form through some examples:
Example 1:
• The set V of vowels in English alphabet is written in set builder form as:
V = {x : x is a vowel in English alphabet}
• It is read as:
The set of all x such that x is a vowel of the English alphabet.
   ♦ The braces stand for: The set of all.
   ♦ The colon ':' stands for: such that.
• So we can read in the order as shown in fig.1.1 below:

Fig.1.1

 

Example 2:

Fig.1.2

• A person who reads this set-builder form can easily write the corresponding roster form as: A = {4, 5, 6, 7, 8, 9}
7. So now we know how to write the set-builder form.
• Note that, ‘x’ is a variable. We know that, a variable can have different values.
• However, the values taken up by ‘x’ should satisfy the property given after the colon ‘:’.
• Remember that, instead of 'x', we can use, 'y', 'z' etc., also.

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Let us see the previous examples that we wrote for roster form. We should be able to write each of them in set-builder form also: 

(a) The set of all even positive integers which are less than 7.
• In roster form, we wrote: A = {2, 4, 6}
• In set-builder form, this will be:
A = {x : x is an even positive integer less than 7}

(b) The set of all natural numbers which completely divide 42
• In roster form, we wrote: B = {1, 2, 3, 6, 7, 14, 21, 42}
• In set-builder form, this will be:
A = {x : x is a natural number which completely divide 42}

(c) The set of all vowels in English alphabet.
• In roster form, we wrote: V = {a, e, i, o, u}
• In set-builder form, this will be:
V = {x : x is a vowel in English alphabet}

(d) The set of all natural odd numbers.
• In roster form, we wrote: D = {1, 3, 5, 7, ...}
• In set-builder form, this will be:
V = {x : x is a natural odd number}  

(e) The set of all letters in the word 'SCHOOL'
• In roster form, we wrote: E = {S, C, H, O, L}
• In set-builder form, this will be:
V = {x : x is a letter in the word 'SCHOOL'}

Solved example 1.2
Write the solution set of the equation x² + x - 2 = 0 in roster form.
Solution:
• The given equation can be written as: $\mathbf\small{\rm{x^2+x=-2}}$
⇒ $\mathbf\small{\rm{x^2+x+\left(\frac{1}{2} \right)^2=-2+\left(\frac{1}{2} \right)^2}}$
• We take the half of the coefficient of x. We then add it's square to both sides. Thus we get:
$\mathbf\small{\rm{\left(x+\frac{1}{2} \right)^2=\frac{9}{4}}}$
⇒ $\mathbf\small{\rm{\left(x+\frac{1}{2} \right)=\pm\frac{3}{2}}}$
⇒ x = 1, -2
• So the solution set of the given equation in the roster form is: {1, -2}.

Solved example 1.3
Write the set {x : x is a positive integer and x² < 40}in the roster form.
Solution:
1. The positive integers are: 1, 2, 3, 4,...
• All integers from 1 to 6 will satisfy x² < 40
2. So the values which x can take are: 1, 2, 3, 4, 5 and 6
• Thus the set in roster form is: {1, 2, 3, 4, 5, 6}

Solved example 1.4
Write the set A = {1, 4, 9, 16, 25, . . .} in set builder form.
Solution:
1. It is clear that, the given set contains infinite number of elements. They are the squares of natural numbers 1, 2, 3, 4, . . .
2. So in set-builder form, it will be:
A = {x : x is square of a natural number}
3. Alternatively, we can write: A = {x : x = n², where n ∈ N}

Solved example 1.5
Write the set $\mathbf\small{\rm{\{\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},\frac{6}{7}\}}}$ in the set-builder form.
Solution:
1. We see that,
   ♦ each x is a fraction
   ♦ All numerators and denominators are natural numbers
• So they belong to the set N
2. If we denote the numerator as n, the denominator will be (n+1)
• So each x is $\mathbf\small{\rm{\frac{n}{n+1}}}$
3. Thus the set-builder form is: {x : $\mathbf\small{\rm{n=\frac{n}{n+1}}}$ and 1 ≤ n ≤ 6}

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The link below gives a 4 more solved examples:

Solved examples 1.6 to 1.9 

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In the next section, we will see the Empty set

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Chapter 1 - Sets

All students of mathematics should have a good knowledge about set concept. It is essential for learning other topics like geometry, sequence, probability etc.,

Sets and their representations

• Let us first see an example. It can be written in 5 steps:
1. Consider the following group:
The group which contains all odd natural numbers which are less than 10.
• Obviously, that group will contain 1, 3, 5, 7 and 9
2. It is a well-defined group.
• It is well-defined because, doubts of any kind will not arise while we form that group.
3. No member can be excluded.
• None of the numbers 1, 3, 5, 7 and 9 can be excluded from the group
• If any of those numbers is/are excluded, the definition ‘all odd natural numbers which are less than 10’ will become invalid.
4. No other member can be included.
• No number other than 1, 3, 5, 7 and 9 must be included in the group.
• If any other number is included, the definition ‘all odd natural numbers which are less than 10’ will become invalid.
5. So indeed it is a well-defined group.

Let us see another example. It can be written in 5 steps:
1. Consider the following group:
The group which contains all vowels in the English alphabet.
• Obviously, that group will contain a, e, i, o and u
2. It is a well-defined group.
• It is well-defined because, doubts of any kind will not arise while we form that group.
3. No member can be excluded.
• None of the letters a, e, i, o and u can be excluded from the group.
• If any of those letters is/are excluded, the definition ‘all vowels in the English alphabet’ will become invalid.
4. No other member can be included.
• No letter other than a, e, i, o and u must be included in the group.
• If any other letter is included, the definition ‘all vowels in the English alphabet’ will become invalid.
5. So indeed it is a well-defined group.

Let us see one more example. It can be written in 5 steps:
1. Consider the following group:
The group which contains the solution of the equation x² + x - 6 = 0
• Obviously, that group will contain 2 and -3.
2. It is a well-defined group.
• It is well-defined because, doubts of any kind will not arise while we form that group.
3. No member can be excluded.
• None of values 2 and -3 can be excluded from the group.
• If any of those values is/are excluded, the definition ‘solution of the equation x² + x - 6 = 0’ will become invalid.
4. No other value can be included.
• No value other than 2 and -3 must be included in the group.
• If any other value is included, the definition ‘solution of the equation x² + x - 6 = 0’ will become invalid.
5. So indeed it is a well-defined group.

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Let us see an example of a group which is not well-defined. It can be written in 3 steps:
1. Consider the following group:
The group which contains the best ten novels in the world.
2. This group is not well-defined because, a novel which one person considers as very good may fail to impress another person.
• So different individuals will be forming different groups.
3. Indeed, such a group is not well-defined.

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◼ Now we can write the definition of set: A set is a well-defined collection of objects.

• So the well-defined groups that we saw above can be referred to as:
    ♦ The set of all odd natural numbers which are less than 10.
    ♦ The set of all vowels in the English alphabet.
    ♦ The set of all solution of the equation x² + x - 6 = 0

---

• We see that, inside a set, there will be some well-defined objects.
• Those objects can be referred to, in any one of the three different ways given below:
    ♦ Elements of the set.
    ♦ Objects of the set.
    ♦ Members of the set.
◼ Sets are denoted by upper case letters A, B, C, X, Y, Z etc.,
◼ Elements are denoted by lower case letters a, b, c, x, y, z etc.,

---

• If ‘a’ is an element of set A, we say that: a belongs to A.
    ♦ The Greek letter epsilon is used to denote ‘belongs to’
        ✰ So we write: a ∈ A.
        ✰ And read it as ‘a belongs to A’.
• If ‘b’ does not belong to A,
        ✰ We write: b ∉ A.
        ✰ And read it as ‘b does not belong to A’.
◼ An example:
• Let V be the set of vowels in English alphabet.
        ✰ We can write: i ∈ V.
        ✰ And read it as ‘i belongs to V’.
• We know that 'd' is not a vowel.
        ✰ We can write: d ∉ V.
        ✰ And read it as ‘d does not belong to V’.
◼ Another example:
• The factors of 30 are: 1, 2, 3, 5, 6, 10, 15 and 30.
• Let us make a set P which contains all the prime factors of 30.
    ♦ Then the elements in that set will be: 2, 3 and 5.
        ✰ We can write: 5 ∈ P.
        ✰ And read it as ‘5 belongs to P’.
• We know that '15' is not an element of P.
        ✰ We can write: 15 ∉ P.
        ✰ And read it as ‘15 does not belong to P’.

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Let us see some predefined sets which we will be using very often in our math classes.
1. We know that the natural numbers are: 1, 2, 3, 4, . . .
    ♦ The set containing all the natural numbers is denoted as N.
        ✰ We can write: 5 ∈ N, 7 ∈ N, 235 ∈ N etc.,
        ✰ Also we can write: ¹⁄₂ ∉ N, -5 ∉ N, √2 ∉ N etc., 
2. We know that the integers are: . . . -4, -3, -2, -1, 0, 1, 2, 3, . . .
• Integers do not contain decimal or fractional part.
    ♦ The set containing all the integers is denoted as Z
        ✰ We can write: -7 ∈ Z, -5 ∈ Z, 235 ∈ Z etc.,
        ✰ Also we can write: ¹⁄₂ ∉ Z, -0.35 ∉ Z, √2 ∉ Z etc., 
3. We know that rational numbers are those numbers which can be written in the form ^p⁄_q, where p and q are integers and q is not equal to zero.
    ♦ The set containing all the rational numbers is denoted as Q
• All integers can be written in the form ^p⁄_q. So they are rational numbers.
        ✰ We can write: -7 ∈ Q, -5 ∈ Q, 235 ∈ Q etc.,
• All fractions are already in the form ^p⁄_q. So they are rational numbers.
        ✰ We can write: ¹⁄₂ ∈ Q, ^-2⁄₃ ∈ Q, ³⁄₁₀₀ ∈ Q etc.,
• All terminating decimals can be written in the form ^p⁄_q. So they are rational numbers.
   ♦ Example: 0.35, -0.215 etc., are terminating decimals.
        ✰ We can write: -7 ∈ Q, -5 ∈ Q, 235 ∈ Q etc.,
• All non-terminating but repeating decimals can be written in the form ^p⁄_q. So they are rational numbers.
   ♦ Example: 0.3333..., -0.1111...  etc., are non-terminating but repeating decimals.
   ♦ Recall that: 0.3333... = ¹⁄₃, -0.1111... = ^-1⁄₉
        ✰ We can write: 0.3333... ∈ Q, -0.1111... ∈ Q etc.,
4. Non-terminating and non-repeating decimals cannot be written in the form ^p⁄_q. So they are not rational numbers.
   ♦ Example: 3.14159..., 1.41425... etc., are non terminating and non repeating.
   ♦ Recall that: 3.14159... = π, 1.41425... = √2
        ✰ We can write: π ∉ Q, √2 ∉ Q, √5 ∉ Q etc.,
• Such numbers are called irrational numbers.
    ♦ The set containing all irrational numbers is denoted as T
        ✰ We can write: π ∈ T, √2 ∈ T, √5 ∈ T etc.,
5. We know that real numbers are those numbers which can be marked on a number line.
• It is easy to mark natural numbers and integers on a number line.
◼  What about rational numbers ?
• Consider ¹⁄₃ as an example. It can be marked as follows:
   ♦ On the number line, divide the portion between 0 and 1 into three equal parts.
   ♦ The end of the first part will be the position of ¹⁄₃
◼  What about irrational numbers ?
• In our earlier math classes, we have already seen how to mark √2, √5 √7 etc., (See fig.22.4).
◼ So we can write:
All natural numbers, integers, rational numbers and irrational numbers can be marked on a number line.
◼ In other words:
All natural numbers, integers, rational numbers and irrational numbers are real numbers.
    ♦ The set containing all the real numbers is denoted as R.
        ✰ We can write: 2 ∈ R, -0.34 ∈ R, 0.3333... ∈ R, √2 ∈ R etc.,
6. We have seen that Z contains all integers only.
• But if we decide to make a set with positive integers only, then that set will be denoted as Z⁺
7. We have seen that Q contains all rational numbers only.
• But if we decide to make a set with positive rational numbers only, then that set will be denoted as Q⁺
8. We have seen that R contains all real numbers only.
• But if we decide to make a set with positive real numbers only, then that set will be denoted as R⁺

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Solved example 1.1
Which of the following are sets ? Justify your answer.
(i) The collection of all the months of a year beginning with the letter J.
(ii) The collection of ten most talented writers of India.
(iii) A team of eleven best-cricket batsmen of the world.
(iv) The collection of all boys in your class.
(v) The collection of all natural numbers less than 100.
(vi) A collection of novels written by the writer Munshi Prem Chand.
(vii) The collection of all even integers.
(viii) The collection of questions in this Chapter.
(ix) A collection of most dangerous animals of the world.
Solution:
Part (i):
• The collection is well-defined.
   ♦ The objects in the collection are: January, June and July.
         ✰ None of these three months can be excluded from the collection.
         ✰ No other month can be included in the collection.
◼ So it is a set.

Part (ii):
• The collection is not well-defined.
   ♦ A writer whom one person considers as very good may fail to impress another person.
   ♦ So different individuals will be forming different collections.
◼ So it is not a set.

Part (iii):
• The team is not well-defined.
   ♦ A player whom one person considers as very good may fail to impress another person.
   ♦ So different individuals will be forming different teams.
◼ So it is not a set.

Part (iv):
• The collection is well-defined.
   ♦ The objects in the collection are: All the boys in the class.
         ✰ None of those boys can be excluded from the collection.
         ✰ No other boy can be included in the collection.
◼ So it is a set.

Part (v):
• The collection is well-defined.
   ♦ The objects in the collection are: 1, 2, 3, 4, ... , 99
         ✰ None of those numbers can be excluded from the collection.
         ✰ No other number can be included in the collection.
◼ So it is a set.

Part (vi):
• The collection is well-defined.
   ♦ The objects in the collection are: Novels written by the writer Munshi Prem Chand.
         ✰ None of his novels can be excluded from the collection.
         ✰ No other novel can be included in the collection.
◼ So it is a set.

Part (vii):
• The collection is well-defined.
   ♦ The objects in the collection are:  . . . -8, -6, -4, -2, 2, 4, 6, . . .
         ✰ None of those numbers can be excluded from the collection.
         ✰ No other number can be included in the collection.
◼ So it is a set.

Part (viii):
• The collection is well-defined.
   ♦ The objects in the collection are: All the questions in this chapter.
         ✰ None of those questions can be excluded from the collection.
         ✰ No other question can be included in the collection.
◼ So it is a set.

Part (ix):
• The collection is not well-defined.
   ♦ An animal which one person considers as 'very dangerous' may be considered as 'not so dangerous' by another person.
   ♦ So different individuals will be forming different collections.
◼ So it is not a set.

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In the next section, we will see the two methods for representing a set.

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