GSEB Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.1

   

Gujarat Board GSEB Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.1

Question 1.
Is zero a rational number? Can you write it in the form of \(\frac { p }{ q }\), where p and q are integers and q≠0?
Solution:
Yes, zero is a rational number, because 0 can be written in the form of \(\frac { p }{ q }\), where p and q are integers and q≠0.
We can write
\(\frac { 0 }{ 1 }\) = \(\frac { 0 }{ 2 }\) = \(\frac { 0 }{ 3 }\), 3 etc

GSEB Solutions Class 9 Maths Chapter Chapter 1 Number Systems Ex 1.1

Question 2.
Find six rational numbers between 3 and 4.
Solution:
Infinitely many rational numbers can exist between 3 and 4.
Rational number between 3 and 4
= \(\frac { 1 }{ 2 }\)(a + b) (Where a = 3 and b = 4)
= \(\frac { 1 }{ 2 }\)(3 + 4) = \(\frac { 7 }{ 2 }\)
Rational number between 3 and \(\frac { 7 }{ 2 }\)
= \(\frac { 1 }{ 2 }\)(3 + \(\frac { 7 }{ 2 }\)) (Where a = 3 and b = \(\frac { 7 }{ 2 }\))
= \(\frac { 1 }{ 2 }\)(\(\frac { 6+7 }{ 2 }\)) = \(\frac { 13 }{ 4 }\)
Rational number between 3 and \(\frac { 13 }{ 4 }\)
= \(\frac { 1 }{ 2 }\)(3 + \(\frac { 13 }{ 4 }\)) (Where a = 3 and b = \(\frac { 13 }{ 4 }\))
= \(\frac { 1 }{ 2 }\)(\(\frac { 12+13 }{ 4 }\)) = \(\frac { 25 }{ 8 }\)
Rational number between 3 and \(\frac { 25 }{ 8 }\)
= \(\frac { 1 }{ 2 }\)(3 + \(\frac { 25 }{ 8 }\)) (Where a = 3 and b = \(\frac { 25 }{ 8 }\))
= \(\frac { 1 }{ 2 }\)(\(\frac { 24+25 }{ 8 }\)) = \(\frac { 1 }{ 2 }\) × \(\frac { 49 }{ 8 }\) = \(\frac { 49 }{ 16 }\)
Rational number between 3 and \(\frac { 49 }{ 16 }\)
= \(\frac { 1 }{ 2 }\)(3 + \(\frac { 49 }{ 16 }\)) (Where a = 3 and b = \(\frac { 49 }{ 16 }\))
= \(\frac { 1 }{ 2 }\)(\(\frac { 48+49 }{ 16 }\)) = \(\frac { 97 }{ 32 }\)
Rational number between 3 and \(\frac { 97 }{ 32 }\)
= \(\frac { 1 }{ 2 }\)(3 + \(\frac { 97+32 }{ 8 }\)) (Where a = 3 and b = \(\frac { 97 }{ 32 }\))
= \(\frac { 1 }{ 2 }\)(\(\frac { 193 }{ 32 }\)) = \(\frac { 193 }{ 64 }\)
∴ Six rational numbers between 3 and 4 are
\(\frac { 193 }{ 64 }\), \(\frac { 97 }{ 32 }\) , \(\frac { 49 }{ 16 }\), \(\frac { 25 }{ 8 }\), \(\frac { 13 }{ 4 }\), \(\frac { 7 }{ 2 }\)
Alternative method:
n = 6 (to be find)
∴ n + 1 = 6 + 1 = 7
Hence, 3 = \(\frac { 3 }{ 1 }\) = \(\frac { 3×7}{ 1×7 }\) = \(\frac { 21 }{ 7 }\)
and 4 = \(\frac { 4 }{ 1 }\) = \(\frac { 4×7 }{ 1×7 }\) = \(\frac { 28 }{ 7 }\)
Six rational numbers between 3 and 4 are
\(\frac { 22 }{ 7 }\), \(\frac { 23 }{ 7 }\), \(\frac { 24 }{ 7 }\), \(\frac { 25 }{ 7 }\), \(\frac { 26 }{ 7 }\), \(\frac { 27 }{ 7 }\)

GSEB Solutions Class 9 Maths Chapter Chapter 1 Number Systems Ex 1.1

Question 3.
Find five rational numbers between \(\frac { 3 }{ 5 }\) and \(\frac { 4 }{ 5 }\)
Solution:
We have to find 5 rational numbers between \(\frac { 3 }{ 5 }\) and \(\frac { 4 }{ 5 }\)
Here, n = 5 ∴ n + 1 = 5 + 1 = 6
∴ Multiplying by 6 to the numerator and denominator.
\(\frac { 3 }{ 5 }\) = \(\frac { 3 }{ 5 }\) x \(\frac { 6 }{ 6 }\) = \(\frac { 18 }{ 30 }\)
and \(\frac { 4 }{ 5 }\) = \(\frac { 4 }{ 5 }\) x \(\frac { 6 }{ 6 }\) = \(\frac { 24 }{ 30 }\)
Hence rational numbers between \(\frac { 18 }{ 30 }\) and \(\frac { 24 }{ 30 }\)
are
\(\frac { 19 }{ 30 }\) < \(\frac { 20 }{ 30 }\) < \(\frac { 21 }{ 30 }\), \(\frac { 22 }{ 30 }\), \(\frac { 23 }{ 30 }\)

GSEB Solutions Class 9 Maths Chapter Chapter 1 Number Systems Ex 1.1

Question 4.
State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
Solution:
(i) True, because the collection (set) of whole numbers contains all the natural numbers

(ii) False, – 1 is an integer but it is not a whole number.

(iii) False, \(\frac { 2 }{ 3 }\) is a rational number but it is not a whole number.

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