By using Baudhãyana-Pythagoras Theorem at every step,we get

\(\text { T1: } \sqrt{(1+1}=\sqrt{2} \text { (irrational) }\)

\(\text { T2: } \sqrt{(2+1}=\sqrt{3} \text { (irrational) }\)

\(\text { T3: } \sqrt{(3+1}=\sqrt{4}=2 \text { (rational) }\)

\(\text { T4: } \sqrt{(4+1}=\sqrt{5} \text { (irrational) }\)

\(\text { T5: } \sqrt{(5+1}=\sqrt{6} \text { (irrational) }\)

\(\text { T6: } \sqrt{(6+1}=\sqrt{7} \text { (irrational) }\)

\(\text { T7: } \sqrt{(7+1}=\sqrt{8}=2 \sqrt{2} \text { (irrational) }\)

\(\text { T8: } \sqrt{(8+1}=\sqrt{9}=3 \text { (rational) }\)

\(\text { T9: } \sqrt{(9+1}=\sqrt{10} \text { (irrational) }\)

\(\text { T10: } \sqrt{(10+1}=\sqrt{11} \text { (irrational) }\)

Construction of \(\sqrt{5}\) on the Number Line

1. On the number line, mark O(0) and A(2) (since OA = 2).

2.At A,draw a perpendicular AB of length 1 unit.

3.Join O to B. Then OB \(=\sqrt{2^2+1^2}=\sqrt{5}\) by Pythagoras theorem

4. With centre O and radius OB,cut the number line at P.

Then OP = \(\sqrt{5}\)

Hence, the point P on the number line represents \(\sqrt{5}\)

(A) Natural Numbers → Integers → Rational Numbers → Real Numbers

Examples:

Integer but not a natural number: -3

Rational number but not an integer\(\frac{1}{2}\)

Real number but not a rational number: \(\sqrt{2}\)

(B) r₁ and r₂ be two rational numbers such that r₁ < r₂

Consider:

\(m=r_1+\frac{\left(r_2-r_1\right) \sqrt{2}}{2}\)

Since \(\sqrt{2}\) is irrational, m is also irrational.

Also,

\(0<\frac{\sqrt{2}}{2}\)

Therefore, r₁ < m < r₂

Hence, an irrational number lies between r₁ and r₂.

(C) Irrational numbers are not closed under addition. C counter-example\(\sqrt{2}+(-\sqrt{2})=0 \in Q .\)

Both addends are irrational, but their sum is rational.

(A) Prove that \(\sqrt{10}\) is irrational.

Assume that \(\sqrt{10}=\frac{p}{q}\)

where p and q are co-prime integers and q ≠ 0.

Squaring both sides, p² =10q²

Since 10 = 2 x 5, both 2 and 5 divide p².
Therefore, 2 divides p and 5 also divides p. Hence, 10 divides p.

Let p = 10m

Substituting in the equation:

(10m)² = 10q²

100m² = 10q²

q² = 10m²

Thus, 10 also divides q.

So both p and q are divisible by 10, which contradicts the assumption that p and q are co-prime.

Hence, \(\sqrt{10}\) is irrational.

(B) For \(\sqrt{4}: p^2=4 q^2\)

Taking p = 2k,

(2k)² = 4q²

4K² = 4q²

k = q

Therefore,

\(\sqrt{4}\) = 2 which is rational,

Similarly, for \(\sqrt{9}:\)

p² = 9q²

Taking p = 3k,

9k² = 9q²

So k = q, giving

\(\sqrt{9}=3\)

which is also rational.

The proof by contradiction works only when the number is not a perfect square. For perfect squares such as and 9, the steps simplify to a rational number instead of producing a contradiction.

On dividing \(\frac{1}{13}=0.07 \overline{6923}\)

This is a 6-digit repeating block

Other multiples \(\frac{2}{13}=0 . \overline{153846}, \frac{3}{13}=0 . \overline{230769}\)

\(\frac{4}{13}=0 . \overline{307692}, \text { etc. }\)

All fractions of the form \(\frac{k}{13}, k=1 \text { to } 12\) share the same digits, but in rotated (cyclic) order.

This happens because \(\frac{k}{13}\) generates a cyclic repeating

decimal of period 6, and its multiples rearrange these digits

Two cyclic groups

Group 1: Cyclic form of 0769 23

\(\frac{1}{13}, \frac{3}{13}, \frac{4}{13}, \frac{9}{13}, \frac{10}{13}, \frac{12}{13}\)

Group 2: Cyclic form of 153846

\(\frac{2}{13}, \frac{5}{13}, \frac{6}{13}, \frac{7}{13}, \frac{8}{13}, \frac{11}{13}\)

Given:

X+y+z=0, xy + yz + zx = 0.........(i)

Use identity

Substitute values in eq. (i)

\(\begin{aligned} 0^2 & =x^2+y^2+z^2+2(0) \\ 0 & =x^2+y^2+z^2 \end{aligned}\)

Since \(x^2, y^2, z^2 \geqq 0\) for all real (rational) numbers, their sum can be zero only when each term is zero:

\(\begin{gathered} x^2=0, y^2=0, z^2=0 \\ x=0, y=0, z=0 \end{gathered}\)

Hence, proved.

(A) Convert 12.6 into a rational number

\(12.6=\frac{126}{10}\)

Simplify:

\(\frac{126}{10}=\frac{63}{5}\)

(B) Convert 0.01120 into a rational number

\(0.0120=\frac{120}{10000}\)

Simplify:

\(\frac{120}{10000}=\frac{3}{250}\)

(C) Let, X= 2.16251625...

10000 x= 21625.1625...

On subtracting, we have

10000 x -X=21625.1625. -2.1625.

9999 x = 21623

\(\begin{aligned} &\text { So, }\\ &x=\frac{21623}{9999} \end{aligned}\)

(D) Let,X= 1.23535...

Multiply by 10 and 1000, we have
10x = 12.3535.. 1000x = 1235.3535..

Subtract:

1000x - 10x = 1235.3535.. - 12.3535..

\(\begin{aligned} &990 x=1223\\ &\text { So, }\\ &x=\frac{1223}{990} \end{aligned}\)

Given, rational numbers are \(\frac{3}{5} \text { and } \frac{4}{5} \text {. }\)

Write the equivalent fractions of the given fractions, by multiplying numerator and denominator with (5 + 1), we get,

\(\begin{aligned} & \frac{3}{5} \times \frac{6}{6}=\frac{18}{30} \\ & \frac{4}{5} \times \frac{6}{6}=\frac{24}{30} \end{aligned}\)

Thus, the five required rational numbers between

\(\frac{3}{5} \text { and }\) \(\frac{4}{5} \text { are } \frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}, \frac{23}{30} .\)

To represent \(\frac{5}{4}\) on the number line, we will follow the following steps:

Step 1: Draw a number line - First, draw a horizontal line with zero (0) at the centre.

Step 2: Locate 1 and 2 on the number line - Since \(\frac{5}{4}\) is greater than 1, mark the numbers 1 and 2 on the number line.

Step 3: Divide the unit between 1 and 2 into 4 parts Since \(\frac{5}{4}\) a fraction with a denominator of 4, divide \(\frac{1}{4} .\)

Step 4: Mark the points - The points between 1 and 2 will be

\(1 \frac{1}{4}\) (after the first division)

\(1 \frac{2}{4} \text { or } 1 \frac{1}{2}\) (after second division)

\(1 \frac{3}{4}\) (after the third division)

2 (at the end of the unit)

Step 5: Find \(\frac{5}{4}-\text { Since } \frac{5}{4}\) is same as \(1 \frac{1}{4}\) mark this point as \(\frac{5}{4} .\)