# Number system – Exercise 1.5 – Class IX

1. 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 ;

1. 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

1. 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.

1. 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.

1. 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