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Chapter 1: Number Systems > Exercise 1.4

Exercise 1.4

1. Classify the following numbers as rational or irrational.

(i) 2−5 : .

(ii)12 : .

(iii)(3+23−23) : .

(iv)2π : .

(v)2777 : .

2. Simplify each of the following expressions:

(i) (3 + 3)(2 + 2)

Instruction

3+32+2

  • We use distributive property.
  • Applying: (a + b) (c + d) = ac + ad + bc + bd
  • (3 + 3)(2 + 2) = 3 × + 32 + 3 × + 3 × 2
  • = + 2 + 3 + 6
  • We have found the answer.

(ii) 3+33−3

3+33−3

  • We use identity property.
  • Applying: (a + b)(a - b) = a2−b2
  • 3+33−3 = -
  • = - =
  • We have found the answer.

(iii) 5+22

5+22

  • We use identity property.
  • Applying: a+b2 = a2 + 2ab + b2
  • We get: 5+22 = 52 + 2×5×2 + 22
  • = + 210 +
  • = + 210
  • We have found the answer.

(iv) 5−25+2

5−25+2

  • We use identity property.
  • Applying: a+b2 = a2 + 2ab + b2
  • We get: (5−2)(5+2) = -
  • = - =
  • We have found the answer.

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

π is defined as the ratio of the circumference of a circle to its diameter, that is, π = c/d. Hence, we see that π is a number as it is expressed in the form of p/q.

But, we know that π is an irrational number. In fact, the value of π is calculated as the and decimal number as π = 3.14159...and hence π is not exactly equal to 227.

In conclusion, π is an number.

4. Represent 9.3 on the number line.

Let AB be of a length '9.3' units on the numberline.

Now, extend AB upto a point C on the numberline such that BC is 1 unit.

Taking AC of length , find the mid point D. Thus, AD = CD = units.

Using D as the center, draw a semi-circle/circle and also extend AC further, to see where this circle intersects the numberline.

From point B, draw a perpendicular intersecting the circle. Join BE. Now, project it onto the numberline.

The length of the segment BE is equal to 9.3.

5. Rationalise the denominators of the following:

Instruction

(i)17

(i)17
Dividing and multiplying by 7, 17 = 17x77
=

(ii)17−6

(ii)17−6
To rationalise: dividing and multiplying by we get: 17−6 × 7+67+6
Using identity a+ba−b = a2−b2
= / 72−62
=

(iii) 15+2

(iii) 15+2
Dividing and multiplying by : 15+2 × 5−25−2
Using identity (a + b)(a - b) = a2−b2
= /
= 5−2/

(iv) 17−2

(iv) 17−2
Dividing and multiplying 17−2 × 7+27+2 = 7+27+27−2
Using identity (a - b)(a + b) = a2−b2
7−27+2 = 72 - 22 = - =
=