Number Systems (Exercise 1.5)

Number Systems

Chapter 1

Exercise 1.5

EX 1.5 QUESTION 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:

(i) 2 –√5

Irrational number.

(ii) (3 +√23)- √23

⇒ 3 +√23- √23 = 3

Rational number.

(iii) 2√7/7√7

⇒  = =

Rational number.

(iv) 1/√2

Irrational number.

(v) 2π

Irrational number.

EX 1.5 QUESTION 2.

Simplify each of the following expressions:

(i) (3+√3)(2+√2)

(ii) (3+√3)(2+√2 )

(iii) (√5+√2)²

(iv) (√5-√2)(√5+√2)

Solution:

(i) (3+√3)(2+√2 )

(3×2)+(3×√2)+(√3×2)+(√3×√2)

= 6+3√2+2√3+√6

(ii) (3+√3)(2+√2 )

(3+√3)(2+√2 )

= 3²-(√3)²

= 9-3

= 6

(iii) (√5+√2)²

(√5+√2)² =

√5²+(2×√5×√2)+ √2²

= 5+2×√10+2

= 7+2√10

(iv) (√5-√2)(√5+√2)

(√5-√2)(√5+√2)

= (√5²-√2²)

= 5-2

= 3

EX 1.5 QUESTION 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:

When we measure the length of a line with a scale or a tape, we get only an approximate rational number. Therefore, the value of c and d both are irrational.

c/d is an irrational number then π is irrational.

EX 1.5 QUESTION 4.

Represent (√9.3) on the number line.

Solution:

Draw a line segment AB = 9.3 units

Now produce AB to C such that BC = 1 unit.

Draw the perpendicular bisector of AC which intersect  AC at O.

Draw a semicircle taking O as centre and AO as the radius. Draw BD ⊥ AC.

Draw an arc taking B as centre and BD as radius meeting AC produced at E such that BE = BD = √9.3 units.

EX 1.5 QUESTION 6.

Rationalize the denominators of the following:

(i) 1/√7

(ii) 1/(√7-√6)

(iii) 1/(√5+√2)

(iv) 1/(√7-2)

Solution: