- 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
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