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
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:
