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
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 \(\sqrt{7}\).