**Classify the following numbers as rational or irrational:**

**(i) 2 − ****√5 **

**(ii) (3 + ****√23) − ****√23 **

**(iii) ^{2}**

^{√7}/_{7}

_{√7}**(iv) ^{1}/**

_{√2}**(v) 2π**

Solution:

Irrational: (i) 2 − √5 ; (iv)^{1}/_{√2} ; (v) 2π

Rational: (ii) (3 + √23) − √23 = 3 ; (iii)^{2}^{√7}/_{7}_{√7} = ^{2}/_{7 };

**Simplify each of the following expressions:**

**(i) (3 + ****√3) ( 2 + ****√2) **

**(ii) (3 + ****√3) ( 3 − ****√3)**

**(iii) (****√5+ √2) ^{2}**

**(iv) (****√5 − ****√2 ) (****√5 + ****√2 )**

Solution:

(i) (3 + √3) ( 2 + √2)

= 3×2 + 2√2 + 2√3 + √2√3

= 6 + 2√2 + 2√3 + √6

(ii) (3 + √3) ( 3 − √3)

=3×3 – 3√3 + 3√3 – √3. √3

= 9 – 3

= 6

(iii) (√5+ √2)^{2}

= (√5)^{2} + 2. √5. √2 + (√2)^{2}

= 5 + 2√10 + 2

= 7 + 2√10

(iv) (√5 − √2 ) (√5 + √2 )

= √5. √5 +√5. √2 – √2. √5 – √2.√2

= 5 – 2

= 3

**Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π =**^{c}/_{d}**. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?**

Solution:

We can never obtain an exact value of c and d so that C OR D IS IRRATIONAL AND PRODUCT or division of rational with irrational is always irrational. Therefore, the fraction ^{c}/_{d} is irrational hence π is irrational.

**Represent √(9.3) on the number line.**

Solution:

Draw a line segment AB of length 9.3 units. Extends the line 1 unit more such that BC = 1 unit. Find the mid point of AC. Draw a semicircle to AB. intersecting the semicircle at point D. Here BD = √(9.3) to represent it in the number line draw an arc DE such that BE = BD. Here BE is the required line.

**Rationalize the denominators of the following:**

**(i) ^{1}/ _{√}_{7}**

**(ii) ^{1}/_{√7 − }_{√6}**

**(iii) ^{1}/_{√5 + }_{√2}**

**(iv) ^{1}/_{√7 – 2}**

Solution:

(i)^{1}/ _{√}_{7} = ^{1}/_{√7} x ^{√7}/_{√7} = ^{√7}/_{7}

(ii) ^{1}/_{√7 − }_{√6} = ^{1}/_{√7 − }_{√6} x ^{√7 + }^{√6}/_{√7 + }_{√6} = ^{√7 + }^{√6}/_{7 – 6} =√7 + √6

(iii)^{1}/_{√5 + }_{√2} = ^{1}/_{√5 + }_{√2} x ^{√5 – }^{√2}/_{√5 – }_{√2} = ^{√5 – }^{√2}/_{5 – 2 }= ^{√5 – }^{√2}/_{3}

(iv)^{1}/_{√7 – 2} =^{1}/_{√7 – 2} x ^{√7 + 2}/_{√7 + 2} = ^{√7 + 2}/_{7-4} = ^{√7 + 2}/_{3}

## 2 responses to “Number system – Exercise 1.5 – Class IX”

[…] Number system – Exercise 1.5 […]

LikeLike

[…] Number system – Exercise 1.5 […]

LikeLike