## Chapter 1 – Rational and Irrational Numbers

**Practice set 1.1**

**1. Show the following numbers on a number line. Draw a separate number line for each example. **

**(1)** \(\large \frac {3}{2}\), \(\large \frac {5}{2}\), \(\large \frac {– 3}{2}\)

**Solution:**

Here, the denominator is 2, so each unit is divided into 2 parts.

∴ The number line is,

**(2)** \(\large \frac {7}{5}\), \(\large \frac {–2}{5}\), \(\large \frac {–4}{5}\)**Solution:**

Here, the denominator is 5, so each unit is divided into 5 parts.

∴ The number line is,

**(3)** \(\large \frac {–5}{8}\), \(\large \frac {11}{8}\)

**Solution:**

Here, the denominator is 8, so each unit is divided into 8 parts.

∴ The number line is,

**(4)** \(\large \frac {13}{10}\), \(\large \frac {–17}{10}\)

**Solution:**

Here, the denominator is 10, so each unit is divided into 10 parts.

∴ The number line is,

**2. Observe the number line and answer the questions.**

**(1) Which number is indicated by point B? **

**Ans:** \(\large \frac {–10}{4}\)

**(2) Which point indicates the number 1 \(\large \frac {3}{4}\)?**

**Ans:** Point C

**(3) State whether the statement, ‘the point D denotes the number ‘\(\large \frac {5}{2}\)’ is true or false.**

**Ans:** True

Reason:

Point D = \(\large \frac {10}{4}\)

After simplification, we get,

Point D = \(\large \frac {5}{2}\)

**Practice set 1.2**

**1. Compare the following numbers.**

**(1) –7, –2 ****Solution:**

–7 < –2

**Ans:** –7 < –2

**(2) 0, \(\large \frac {–9}{5}\)**

**Solution:**

0 > \(\large \frac {–9}{5}\) …*[Zero is greater than all negative numbers]*

**Ans:** 0 > \(\large \frac {–9}{5}\)

**(3) \(\large \frac {8}{7}\), 0**

**Solution:**

\(\large \frac {8}{7}\) > 0 …*[Positive numbers are greater than zero]*

**Ans:**\(\large \frac {8}{7}\) > 0

**(4) \(\large \frac {–5}{4}\), \(\large \frac {1}{4}\)**

**Solution:**

\(\large \frac {–5}{4}\) < \(\large \frac {1}{4}\) …*[Denominators are same and –5 < 1]*

**Ans: \(\large \frac {–5}{4}\) < \(\large \frac {1}{4}\)**

**(5) \(\large \frac {40}{29}\), \(\large \frac {141}{29}\)**

**Solution:**

\(\large \frac {40}{29}\) < \(\large \frac {141}{29}\) …*[Denominators are same and 40 < 141]*

**Ans:** \(\large \frac {40}{29}\) < \(\large \frac {141}{29}\)

**(6) \(\large \frac {-17}{20}\), \(\large \frac {-13}{20}\)**

**Solution:**

\(\large \frac {-17}{20}\), \(\large \frac {-13}{20}\) …*[Denominators are same and –17 < –13]*

**Ans:** \(\large \frac {-17}{20}\), \(\large \frac {-13}{20}\)

**(7) \(\large \frac {15}{12}\), \(\large \frac {7}{6}\)**

**Solution:**

\(\large \frac {15}{12}\), \(\large \frac {7}{6}\)

15 × 6, 7 × 12

90 < 84

∴ \(\large \frac {15}{12}\) < \(\large \frac {7}{6}\)

**Ans:** \(\large \frac {15}{12}\) < \(\large \frac {7}{6}\)

**(8) \(\large \frac {-25}{8}\), \(\large \frac {-9}{4}\)**

**Solution:**

\(\large \frac {-25}{8}\), \(\large \frac {-9}{4}\)

–25 × 4, –9 × 8

–100 < –72

∴ \(\large \frac {-25}{8}\) < \(\large \frac {-9}{4}\)

**Ans:** \(\large \frac {-25}{8}\) < \(\large \frac {-9}{4}\)

**(9) \(\large \frac {12}{15}\), \(\large \frac {3}{5}\)**

**Solution:**

\(\large \frac {12}{15}\), \(\large \frac {3}{5}\)

12 × 5, 3 × 15

60 > 45

∴ \(\large \frac {12}{15}\) > \(\large \frac {3}{5}\)

**Ans:** \(\large \frac {12}{15}\) > \(\large \frac {3}{5}\)

**(10) \(\large \frac {-7}{11}\), \(\large \frac {-3}{4}\)**

**Solution:**

\(\large \frac {7}{11}\), \(\large \frac {3}{4}\)

–7 × 4, –3 × 11

–28 < –33

∴ \(\large \frac {-25}{8}\) < \(\large \frac {-9}{4}\)

**Ans:** \(\large \frac {-25}{8}\) < \(\large \frac {-9}{4}\)

**Practice set 1.3**

**1. Write the following rational numbers in decimal form.**

**(1) \(\large \frac {9}{37}\)**

**Solution:**

**∴ \(\large \frac {9}{37}\) = \(0.\overline{243}\)**

**(2) \(\large \frac {18}{42}\)**

**Solution:**

**\(\large \frac {18}{42}\) = \(\large \frac {3\,×\,6}{7\,×\,6}\)**

**∴ \(\large \frac {18}{42}\) = \(\large \frac {3}{7}\)**

**∴ \(\large \frac {18}{42}\) = \(0.\overline{428571}\)**

**(3) \(\large \frac {9}{14}\)**

**Solution:**

**∴ \(\large \frac {9}{14}\) = \(0.6\overline{428571}\)**

**(4) \(\large \frac {–\,103}{5}\)**

**Solution:**

**\(\large \frac {103}{5}\)** = 20.6

**∴ \(\large \frac {–\,103}{5}\) = – 20.6**

**(5) – \(\large \frac {11}{13}\)**

**Solution:**

**\(\large \frac {11}{13}\)** = \(0.\overline{846153}\)

**∴ – \(\large \frac {11}{13}\) = – \(0.\overline{846153}\)**

**Practice set 1.4**

**Practice set 1.4**

**1. The number \(\sqrt{2}\) is shown on a number line. Steps are given to show 3 on the number line using \(\sqrt{2}\). Fill in the boxes properly and complete the activity.**

**Activity :**

**· The point Q on the number line shows the number _____**

** **

**· A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.**

** **

**· Right angled D ORQ is obtained by drawing seg OR.**

**· l (OQ) = \(\sqrt{2}\), l(QR) = 1**

** **

**∴ By Pythagoras theorem,[l(OR)]² = [l(OQ)]² + [l(QR)]²**

**∴ [l(OR)]² = ____² + ____²**

**∴ [l(OR)]² = ____ + ____**

**∴ [l(OR)]² = ____**

**∴ l(OR) = ____**

** **

**Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number \(\sqrt{3}\).**

**Solution:**

**· The point Q on the number line shows the number \(\sqrt{2}\)**

** **

**· A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.**

** **

**· Right angled D ORQ is obtained by drawing seg OR.**

** **

**· l (OQ) = \(\sqrt{2}\), l(QR) = 1**

** **

**∴ By Pythagoras theorem,[l(OR)]² = [l(OQ)]² + [l(QR)]²**

**∴ [l(OR)]² = \((\sqrt{2})\)² + (1)²**

**∴ [l(OR)]² = 2 + 1**

**∴ [l(OR)]² = 3**

**∴ l(OR) = \(\sqrt{3}\)**

** **

**Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number \(\sqrt{3}\).**

**2. Show the number \(\sqrt{5}\) on the number line.**

**Solution:**

**(1) Draw a number line and take a point Q at 2 such that l(OQ) = 2 units.**

** **

**(2) Draw a line QR perpendicular to the number line through the point Q such that l(QR) = 1 unit.**

** **

**(3) Draw seg OR.**

** **

**(4) ∆OQR formed is a right angled triangle.By Pythagoras theorem,[l(OR)]² = [l(OQ)]² + [l(QR)]²∴ [l(OR)]² = 2² + 1²∴ [l(OR)]² = 4 + 1∴ [l(OR)]² = 5∴ l(OR) = \(\sqrt{5}\) units …**

*[Taking square root of both sides]*

** **

**(5) Draw an arc with centre O and radius OR. Mark the point of intersection of the number line and arc as C.**

**The point F shows the number \(\sqrt{5}\).**

**3. Show the number \(\sqrt{7}\) on the number line.**

**Solution:**

**(1) Draw a number line and take a point Q at 2 such that l(OQ) = 2 units.**

** **

**(2) Draw a line QR perpendicular to the number line through the point Q such that l(QR) = 1 unit.**

** **

**(3) Draw seg OR.**

** **

**(4) ∆OQR formed is a right angled triangle.By Pythagoras theorem,[l(OR)]² = [l(OQ)]² + [l(QR)]²∴ [l(OR)]² = 2² + 1²∴ [l(OR)]² = 4 + 1∴ [l(OR)]² = 5∴ l(OR) = \(\sqrt{5}\) units …**

*[Taking square root of both sides]*

** **

**(5) Draw an arc with centre O and radius OR. Mark the point of intersection of the number line and arc as C. The point C shows the number \(\sqrt{5}\).**

**(6) Similarly, draw a line CD perpendicular to the number line through the point C such that l(CD) = 1 unit.**

** **

**(7) Draw seg OD.**

** **

**(8) ∆OCD formed is a right angled triangle.By Pythagoras theorem,[l(OD)]² = [l(OC)]² + [l(CD)]²∴ [l(OD)]² = \((\sqrt{5})\)² + 1²∴ [l(OD)]² = 5 + 1∴ [l(OD)]² = 6∴ l(OD) = \(\sqrt{6}\) units …**

*[Taking square root of both sides]*

** **

**(9) Draw an arc with centre O and radius OD. Mark the point of intersection of the number line and arc as E. The point E shows the number \(\sqrt{6}\).**

** **

**(10) Similarly, draw a line EP perpendicular to the number line through the point E such that l(EP) = 1 unit.**

** **

**(11) Draw seg OP.**

** **

**(12) ∆OEP formed is a right angled triangle.By Pythagoras theorem,[l(OP)]² = [l(OE)]² + [l(EP)]²∴ [l(OP)]² = \((\sqrt{6})\)² + 1²∴ [l(OP)]² = 6 + 1∴ [l(OP)]² = 7∴ l(OP) = \(\sqrt{7}\) units …**

*[Taking square root of both sides]*

** **

**(13) Draw an arc with centre O and radius OP. Mark the point of intersection of the number line and arc as F. The point F shows the number \(\sqrt{7}\.**

**The point F shows the number \(\sqrt{7}\).**