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# Class 9 NCERT Solutions- Chapter 1 Number System – Exercise 1.5

• Last Updated : 21 Dec, 2020

### Question 1: Classify the following numbers as rational or irrational:

(i) 2 â€“âˆš5

(ii) (3 +âˆš23)- âˆš23

(iii) 2âˆš7 / 7âˆš7

(iv) 1/âˆš2

(v) 2Ï€

Solution:

(i) 2 â€“âˆš5

As âˆš5 = 2.2360678â€¦ which is non-terminating and non-recurring. It is an irrational number.

When we substitute the value of âˆš5 in equation 2 â€“âˆš5, we get,

2-âˆš5 = 2-2.2360678â€¦

2-âˆš5 = -0.2360678

Since the number, â€“ 0.2360678â€¦, is a non-terminating and non-recurring,

Therefore, 2 â€“âˆš5 is an irrational number.

(ii) (3 +âˆš23)- âˆš23

(3 +âˆš23) â€“âˆš23 = 3+âˆš23â€“âˆš23

= 3

Since, the number 3 is rational number

Therefore, (3 +âˆš23)- âˆš23 is rational.

(iii) 2âˆš7 / 7âˆš7

2âˆš7 / 7âˆš7 = (2/7)Ã— (âˆš7/âˆš7)

2âˆš7 / 7âˆš7 = (2/7)Ã— (âˆš7/âˆš7)

= (2/7)Ã—1 [As (âˆš7/âˆš7) = 1]

= 2/7

Since the number, 2/7 is in p/q form

Therefore, 2âˆš7/7âˆš7 is rational.

(iv) 1/âˆš2

As, âˆš2 = 1.41421â€¦ which is non-terminating and non-recurring. It is a rational number.

When we divide 1/âˆš2 we get,

1/âˆš2 = 1/1.41421…

=0.70710…

Since the number, 0.7071..is a non-terminating and non-recurring,

Therefore, 1/âˆš2 is an irrational number.

(v) 2Ï€

The value of Ï€ is 3.1415…

When we substitute the value of Ï€ in equation 2Ï€, we get,

2Ï€ = 2 Ã— 3.1415…  = 6.2831…

Since the number, 6.2831â€¦, is non-terminating non-recurring,

Therefore, 2Ï€ is an irrational number.

### Question 2: Simplify each of the following expressions:

(i) (3+âˆš3)(2+âˆš2)

(ii) (3+âˆš3)(3-âˆš3)

(iii) (âˆš5+âˆš2)2

(iv) (âˆš5-âˆš2)(âˆš5+âˆš2)

Solution:

(i) (3+âˆš3)(2+âˆš2)

After opening the brackets, we get,

(3+âˆš3)(2+âˆš2)= (3Ã—2)+(3Ã—âˆš2)+(âˆš3Ã—2)+(âˆš3Ã—âˆš2)

(3+âˆš3)(2+âˆš2) = 6+3âˆš2+2âˆš3+âˆš6

(ii) (3+âˆš3)(3-âˆš3)

After opening the brackets, we get,

(3+âˆš3)(3-âˆš3) = 32-(âˆš3)2

= 9-3

(3+âˆš3)(3-âˆš3) = 6

(iii) (âˆš5+âˆš2)2

After opening the brackets, we get,

(âˆš5+âˆš2)2 = âˆš52+(2Ã—âˆš5Ã—âˆš2)+ âˆš22       [By using the formula (a + b)2 = a2 + 2ab + b2]

= 5+2Ã—âˆš10+2

(âˆš5+âˆš2)2 = 7+2âˆš10

(iv) (âˆš5-âˆš2)(âˆš5+âˆš2)

After opening the brackets, we get,

(âˆš5-âˆš2)(âˆš5+âˆš2) = (âˆš52-âˆš22

= 5-2

= 3

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

Solution:

Given Ï€ = c/d = 22/7 which is equal to 3.142… which is non-terminating non-recurring decimal.

Therefore, Ï€ is irrational.

### Question 4: Represent (âˆš9.3) on the number line.

Solution:

To represent âˆš9.3 on the number line, follow the following steps,

Step 1: Draw a 9.3 units long line segment, name the line as AB.

Step 2: Extend AB to C such that BC=1 unit.

Step 3: Now, AC = 10.3 units. Let the centre of AC be O.

Step 4: Draw a semi-circle with radius OC and centre O.

Step 5: Draw a BD perpendicular to AC at point B which is intersecting the semicircle at D.

Step 6: Join BD.

Step 7: Taking BD as radius and B as the centre point and draw an arc which touches the line segment.

The point where it intersects the line segment is at a distance of âˆš9.3 from B as shown in the figure.

### Question 5: Rationalize the denominators of the following:

(i) 1/âˆš7

(ii) 1/(âˆš7-âˆš6)

(iii) 1/(âˆš5+âˆš2)

(iv) 1/(âˆš7-2)

Solution:

(i) 1/âˆš7

Multiply and divide 1/âˆš7 by âˆš7 we get,

(1Ã—âˆš7)/(âˆš7Ã—âˆš7) = âˆš7/7

= âˆš7/7

(ii) 1/(âˆš7-âˆš6)

Multiply and divide 1/(âˆš7-âˆš6) by (âˆš7+âˆš6) we get,

[1/(âˆš7-âˆš6)]Ã—(âˆš7+âˆš6)/(âˆš7+âˆš6) = (âˆš7+âˆš6)/(âˆš7-âˆš6)(âˆš7+âˆš6)

= (âˆš7+âˆš6)/âˆš72-âˆš62 [As, (a+b)(a-b) = a2-b2]

= (âˆš7+âˆš6)/(7-6)

= (âˆš7+âˆš6)/1

= âˆš7+âˆš6

(iii) 1/(âˆš5+âˆš2)

Multiply and divide 1/(âˆš5+âˆš2) by (âˆš5-âˆš2) we get,

[1/(âˆš5+âˆš2)]Ã—(âˆš5-âˆš2)/(âˆš5-âˆš2) = (âˆš5-âˆš2)/(âˆš5+âˆš2)(âˆš5-âˆš2)

= (âˆš5-âˆš2)/(âˆš52-âˆš22) [As, (a+b)(a-b) = a2-b2]

= (âˆš5-âˆš2)/(5-2)

= (âˆš5-âˆš2)/3

(iv) 1/(âˆš7-2)

Multiply and divide 1/(âˆš7-2) by (âˆš7+2) we get,

1/(âˆš7-2)Ã—(âˆš7+2)/(âˆš7+2) = (âˆš7+2)/(âˆš7-2)(âˆš7+2)

= (âˆš7+2)/(âˆš72-22) [As, (a+b)(a-b) = a2-b2]

= (âˆš7+2)/(7-4)

= (âˆš7+2)/3

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