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 ;

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2. 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

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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.

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4. 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.

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5. 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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