Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

## Practice Set 2.2 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.

Show that 4√2 is an irrational number.

Solution:

Let us assume that 4√2 is a rational number .

So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

4√2 = \(\frac { a }{ b }\)

∴ √2 = \(\frac { a }{ 4b }\)

Since, a and b are integers, \(\frac { a }{ 4b }\) is a rational number and so √2 is a rational number.

Alternate Proof:

Let us assume that 4√2 is a rational number.

So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

Since, 32 divides a^{2}, so 32 divides ‘a’ as well.

So, we write a = 32c, where c is an integer.

∴ a^{2} = (32c)^{2} … [Squaring both the sides]

∴ 32b^{2} = 32 x 32c^{2} …[From(i)]

∴ b^{2} = 32c^{2}

∴ c^{2} = \(\frac { { b }^{ 2 } }{ 32 }\)

Since, 32 divides b^{2}, so 32 divides ‘b’.

∴ 32 divides both a and b.

a and b have at least 32 as a common factor.

But this contradicts the fact that a and b have no common factor other than 1.

∴ Our assumption that 4√2 is a rational number is wrong.

∴ 4√2 is an irrational number.

Question 2.

Prove that 3 + √5 is an irrational number.

Solution:

Let us assume that 3 + √5 is a rational number.

So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

Since, a and b are integers, \(\frac { a }{ b }\) – 3 is a rational

number and so √5 is a rational number.

But this contradicts the fact that √5 is an irrational number.

∴ Our assumption that 3 – √5 is a rational number is wrong.

3 + √5 is an irrational number.

Question 3.

Represent the numbers √5 and √10 on a number line.

Solution:

i. Draw a number line and take point A at 2.

Draw AB perpendicular to the number line such that AB = 1 unit.

In ∆OAB, m∠OAB = 90°

∴ (OB)^{2} = (OA)^{2} + (AB)^{2} … [Pythagoras theorem]

= (2)^{2} + (1)^{2}

∴ (OB)^{2} = 5

∴ OB = √5 units. … [Taking square root of both sides]

With O as centre and radius equal to OB, draw an arc to intersect the number line at C.

The coordinate of the point C is √5 .

ii. Draw a number line and take point Pat 3.

Draw PR perpendicular to the number line such that PR = 1 unit.

In ∆OPR, m∠OPR = 90°

∴ (OR)^{2} = (OP)^{2} + (PR)^{2} … [Pythagoras theorem]

= (3)^{2} + (1)^{2}

∴ (OR)^{2} = 10

∴ OR= √10units. … [Taking square root of both sides]

With O as centre and radius equal to OR, draw an arc to intersect the number line at Q.

The coordinate of the point Q is √10 .

Question 4.

Write any three rational numbers between the two numbers given below.

i. 0.3 and – 0.5

ii. – 2.3 and – 2.33

iii. 5.2 and 5.3

iv. – 4.5 and – 4.6

Solution:

i. 0.3 = 0.30 and -0.5 = -0.50

We know that,

0. 30 >0.29 >….. >0.10>.. > – 0.10>…. > -0.30>…> -0.50

∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

Alternate Method:

A rational number between two rational numbers a and b

∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

ii. -2.3 = -2.300 and -2.33 = -2.330

We know that,

-2.300 > -2.301>… > -2.310>…> -2.320>…> -2.330

∴ the three rational numbers between -2.3 and -2.33 are -2.310, -2.320 and -2.325.

iii. 5.2 = 5.20 and 5.3 = 5.30

We know that,

5.20 < 5.21 < 5.22 < 5.23 < … < 5.30

∴ the three rational numbers between 5.2 and 5.3 are 5.21, 5.22 and 5.23.

iv. -4.5 = -4.50 and -4.6 = -4.60 We know that,

-4.50 > -4.51 > -4.52 >… > – 4.55 >…>- 4.60

∴ the three rational numbers between -4.5 and -4.6 are -4.51, -4.52 and -4.55.