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 – \(\sqrt{3}\))

= 3^{2} – (\(\sqrt{3}\))^{2}

[(a – \(\sqrt{a}\)) (a + \(\sqrt{b}\)) = a^{2} – 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

BD^{2} = OD^{2} – OB^{2}

[ \(\frac { x+1 }{ 2 }\) ]^{2} – [ \(\frac { x-1 }{ 2 }\) ]^{2} = \(\frac { 4x }{ 4 }\) = x

∴ BD^{2} = 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