## MP Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 1.5

Question 1.

Classify the following numbers as rational or irrational –

- 2 – √5
- (3 + \(\sqrt{23}\) ) – \(\sqrt{23}\)
- \(\frac { 2\sqrt { 7 } }{ 7\sqrt { 7 } } \)
- \(\frac{1}{sqrt { 2 }}\)
- 2π

Solution:

- 2 – \(\sqrt{5}\) is an irrational number
- (3 + \(\sqrt{23}\) ) – \(\sqrt{23}\) = 3 is a rational number
- \(\frac { 2\sqrt { 7 } }{ 7\sqrt { 7 } } \) = \(\frac{2}{7}\) is a rational number
- \(\frac{1}{\sqrt { 2 }}\) = is an irrational number
- 2π is an irrational number.

Question 2.

Simplify each of the following expressions –

- (3 + \(\sqrt{3}\) ) (2 +\(\sqrt{2}\))
- (3 + \(\sqrt{3}\) ) (3 – \(\sqrt{3}\))
- (\(\sqrt{5}\) + \(\sqrt{2}\))
^{2} - (\(\sqrt{5}\) – \(\sqrt{2}\)) (\(\sqrt{5}\) + \(\sqrt{2}\))

Solution:

1. (3 + \(\sqrt{3}\) ) (2 +\(\sqrt{2}\))

= 6 + 3\(\sqrt{2}\) + 2\(\sqrt{3}\) + \(\sqrt{6}\)

2. (3 + \(\sqrt{3}\)) (3 – \(\sqrt{3}\))

= (3)^{2} – (\(\sqrt{3}\) )^{2} = 9 – 3 = 6

3. (\(\sqrt{5}\) + \(\sqrt{2}\) )^{2}

= (\(\sqrt{5}\))^{2} + 2\(\sqrt{5}\) x \(\sqrt{2}\) + (\(\sqrt{2}\))^{2}

= 5 + 2\(\sqrt{10}\) + 2 = 7 + 2\(\sqrt{10}\)

4. (\(\sqrt{5}\) – \(\sqrt{2}\)) (\(\sqrt{5}\) + \(\sqrt{2}\))

= (\(\sqrt{5}\))^{2} – (\(\sqrt{2}\))^{2} = 5 – 2 = 3

Question 3.

Recall, 7t 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 contradiet the fact n is irrational. How will you resolve this contradiction?

Solution:

No contradiction will be there. Whenever we measure any length with any device, we only get an approx, rational value and so cannot realise that either c or d is irrational.

Question 4.

Represent \(\sqrt{9.3}\) on the number line.

Solution:

Steps:

- Draw a line segment AC = 9.3 cm and extend it to B such that CB – 1 cm.
- Draw the perpendicular bisector of AS and mark the mid point O of AB.
- With O as centre and OB as radius, draw a semicircle.
- At point C, draw perpendicular CD which intersect the semicircle at D such that CD = \(\sqrt{9.3}\).
- With C as centre and CD as radius, draw an arc which intersect AB produced at E.

Proof:

OA = OB = \(\frac{10.3}{2}\) = 5.15cm

OD = 5.15 cm

OC = OB – BC = 5.15 – 1 = 4.15 cm

CD = \(\sqrt { OD^{ 2 }-OC^{ 2 } } \)

= \(\sqrt{9.3}\) cm

Question 5.

Rationalise the denominators of the following:

- \(\frac{1}{\sqrt { 7 }}\)
- \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
- \(\frac{1}{\sqrt{5}+\sqrt{2}}\)
- \(\frac{1}{\sqrt{7}-2}\)

Solution:

1. \(\frac{1}{\sqrt { 7 }}\)

2. \(\frac{1}{\sqrt{7}-\sqrt{6}}\)

3. \(\frac{1}{\sqrt{5}+\sqrt{2}}\)

4. \(\frac{1}{\sqrt{7}-2}\)

Laws of Exponents for real numbers:

If a, n and m are natural numbers, then

- a
^{m}x a^{n}= a^{m+n} - (a
^{m})^{n}= a^{mn} - \(\frac { a^{ m } }{ b^{ n } } \) where m > n
- a
^{m}x b^{m}– (ab)^{m}

Example 1.

- 2
^{3}x 2^{4}= 2^{3+4}= 2^{7} - (3
^{2})^{3}= 3^{2×3}= 3^{6} - \(\frac { 5^{ 5 } }{ 5^{ 2 } } \) = 5
^{5-2}= 5^{3} - 2
^{3}x 3^{3}= (2 x 3)^{3}= 6^{3}

Example 2.

1. 6^{\(\frac{2}{5}\)} x 6^{\(\frac{3}{5}\)}

Solution:

6^{\(\frac{2}{5}\)} x 6^{\(\frac{3}{5}\)} = 6^{\(\frac{5}{5}\)} = 6

2. 6^{\(\frac{1}{2}\)} x 7^{\(\frac{1}{2}\)}

Solution:

6^{\(\frac{1}{2}\)} x 7^{\(\frac{1}{2}\)} = (6 x 7)^{\(\frac{1}{2}\)} = 42^{\(\frac{1}{2}\)}

3. \(\frac{5^{6 / 7}}{5^{2 / 3}}\)

Solution:

Example 3.

Simplify

Solution: