GSEB Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.5

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