Gujarat Board GSEB Solutions Class 10 Maths Chapter 1 Number Systems Ex 1.5 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.5

Question 1.
Classify the following numbers as rational or irrational.
Solution:
(i) 2 – (sqrt{5})
(ii) (3 + (sqrt{23})) – (sqrt{23})
(iii) (frac{2 sqrt{7}}{7 sqrt{7}})
(iv) (frac{1}{sqrt{2}})
(v) 2π
Solution:
(i) 2 is a rational number and (sqrt{5}) is an irrational number, then 2 – (sqrt{5}) is an irrational number.
(The difference of rational number and irrational number is an irrational number)

(ii) (3 + (sqrt{23})) – (sqrt{23}) = 3 + (sqrt{23}) – (sqrt{23}) = 3 (3 is a rational number)

(iii) (frac{2 sqrt{7}}{7 sqrt{7}}) = (frac { 2 }{ 7 }) = which is a rational number.

(iv) (frac{1}{sqrt{2}})
Here, 1 is rational number and (sqrt{2}) is an irrational number.
∴ The quotient of a non-zero rational number with an irrational number we get an irrational number.
∴ (frac{1}{sqrt{2}}) is an irrational number.

(v) 2π
Here, 2 is a rational number (2 ≠ 0)
and π is an irrational number.
∴ π is an irrational number.
(Because the product of a non-zero rational number with an irrational number is an irrational number).

Question 2.
Simplify each of the following expressions:
(i) (3 + (sqrt{3})) (2 + (sqrt{2}))
(ii) (3 + (sqrt{3})) (3 – (sqrt{3}))
(iii) ((sqrt{5}) + (sqrt{2}))²
(iv) ( (sqrt{5}) – (sqrt{2})) ((sqrt{5}) + (sqrt{2}))
Solution:
(i) (3 + (sqrt{3})) (2 + (sqrt{2}))
= 3(2 + (sqrt{2})) + (sqrt{3})(2 + (sqrt{2}))
= 6 + 3(sqrt{2}) + 2(sqrt{3}) + (sqrt{3}) x (sqrt{2})
= 6 + 3(sqrt{2}) + 2(sqrt{3}) + (sqrt{6}) (-(sqrt{a}) x (sqrt{b}) = (sqrt{ab}))

(ii) (3 + (sqrt{3})) (3 – (sqrt{3}))
= 3² – ((sqrt{3}))²
[(a – (sqrt{a})) (a + (sqrt{b})) = a² – b]
= 9 – 3 = 6

(iii) ((sqrt{5}) + (sqrt{2}))²
= ( (sqrt{5}))² + 2 (sqrt{5}) x (sqrt{2}) + ( (sqrt{2}))²
= 5 + 2(sqrt{10}) + 2 ( (sqrt{a}) x (sqrt{b}) = (sqrt{ab}))
= 7 + 27(sqrt{10})

(iv) ((sqrt{5}) – (sqrt{2})) ( (sqrt{5}) + (sqrt{2}))
= ((sqrt{5}))² – ((sqrt{2}))² = 5 – 2 = 3

Question 3.
Recall, n is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = (frac { c }{ d }). This seems to contradict the fact that n is irrational. How will you resolve this contradiction?
Solution:
π = (frac { c }{ d })
There is no contradiction as either c or d is irrational and hence n is an irrational.

Question 4.
Represent (sqrt{9.3}) on the number line.
Solution:
We find (sqrt{9.3}) geometrically.
Let x = 79.3
1. Draw a number line and mark point A on it and take a distance AB = 9.3 units.
2. From point B mark a distance of 1 unit and mark the new point C.
3. Draw perpendicular bisector of AC and mark that point as O.
4. Draw a semicircle with centre O and radius OC.
5. Draw perpendicular at the point B which intersect the semicircle at point D then BD = (sqrt{9.3}).
6. Now from B as centre and radius equal to BD, draw an arc which intersects the number line at P. Thus, BP = (sqrt{9.3})units.

AOBD is a right angled triangle. Also the
radius of circle r = (frac { AC }{ 2 }) i.e., (frac { x+1 }{ 2 }) units.
∴ OC = OD = OA = (frac { x+1 }{ 2 }) units
Now OB = OD = OA = (frac { x+1 }{ 2 }) units
⇒ OB = x – [ (frac { x+1 }{ 2 }) ]
⇒ OB = x – [ (frac { x+1 }{ 2 }) ]
By Pythagoras theorem, we have
BD² = OD² – OB²
[ (frac { x+1 }{ 2 }) ]² – [ (frac { x-1 }{ 2 }) ]² = (frac { 4x }{ 4 }) = x
∴ BD² = x
⇒ BD = (sqrt{x})
Hence BD = (sqrt{9.3})
⇒ BD = 3.049

Question 5.
Rationalise the denominators of the following:
(i) (frac{1}{sqrt{7}})
(ii) (frac{1}{sqrt{7 – 6}})
(iii) (frac{1}{sqrt{7}-sqrt{6}})
(iii) (frac{1}{sqrt{5}+sqrt{2}})
(iv) (frac{1}{sqrt{7}-2})
Solution:
(i) We have (frac{1}{sqrt{7}})
Multiplying by (sqrt{7}) to numerator and denominator, we get
(frac{1}{sqrt{7}}) x (frac{sqrt{7}}{sqrt{7}}) x (frac{sqrt{7}}{7})

(ii) (frac{1}{sqrt{7 – 6}})
Multiplying the numerator and denominator by (sqrt{7}) + (sqrt{7}), we get

(iv) (frac{1}{sqrt{7}-2})
Multiplying the numerator and denominator by (sqrt{7}) + 2, we get