Step 1: Mark a point O

Choose a point O on your paper. This will be the centre of your square root spiral.

Step 2: Draw line OA of 1 cm horizontally

From point O, draw a straight line OA horizontally to the right. The length should be 1 cm.

Step 3: Draw perpendicular line AB of 1 cm

From point A (the end of OA), draw a line AB vertically upwards. The length of AB should also be 1 cm.

Step 4: Join OB (length √2)

Join point O to point B (the end of AB). The length of OB should be √2 cm.

Step 5: Draw perpendicular line from B and mark C

From point B, draw another line perpendicular to OB (going upwards) of 1 cm. Mark the endpoint of this line as C.

Step 6: Join OC (length √3)

Join point O to point C. The length of OC should be √3 cm.

Step 7: Repeat the process

Repeat steps 5 and 6 to continue the spiral:

From point C, draw a perpendicular line of 1 cm upwards and mark the endpoint as D.

Join O to D. The length OD should be √4 cm.

Continue this process, each time increasing the length of the perpendicular line by 1 cm and joining the new point to O to form the next segment of the spiral, where the length of each segment from O increases by 1 each time (resulting in √2, √3, √4, etc.).

Exercise-1.3

1.Write the following in decimal form and say what kind of decimal expansion each has

Ans: (i) Divide 36 by 100.

So, 36/100=0.36 and it is a terminating decimal number.

(ii) Dividing 1 by 11, we have

Thus, the decimal expansion of 1/11 is non-terminating repeating.

(iii) We have 4 1/8 = 33/8
Dividing 33 by 8, we get

∴ 4 1/8 = 4.125. Thus, the decimal expansion is terminating.

(iv) Dividing 3 by 13, we get

Here, the repeating block of digits is 230769
∴ 3/13 = 0.23076923… = 0.230769¯
Thus, the decimal expansion of 3/13 is non-terminating repeating.

(v) Dividing 2 by 11, we get

Here, the repeating block of digits is 18.
∴ 2/11 = 0.1818… = 0.18¯
Thus, the decimal expansion of 2/11 is non-terminating repeating.

(vi) Dividing 329 by 400, we get

329/400  = 0.8225. Thus, the decimal expansion of 329/400 is terminating.

2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, and 6/7 are without actually doing the long division? If so, how?

[Hint: Study the remainders carefully while finding the value of 1/7 .]

Ans:

3. Express the following in the form p/q, where p and q are integers and q ≠ 0.

(i) 

Ans:

Assume that  x = 0.666…

Then,10x = 6.666…

10x = 6 + x

9x = 6

x = 2/3

Ans:

= (4/10)+(0.777/10)

Assume that x = 0.777…

Then, 10x = 7.777…

10x = 7 + x

x = 7/9

(4/10)+(0.777../10) = (4/10)+(7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9×10) = 7/90 )

= (36/90)+(7/90) = 43/90

Ans:

Assume that  x = 0.001001…

Then, 1000x = 1.001001…

1000x = 1 + x

999x = 1

x = 1/999

4. Express 0.99999…. in the form p/q. Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense.

Ans:

Assume that x = 0.9999…..Eq (a)

Multiplying both sides by 10,

10x = 9.9999…. Eq. (b)

Eq.(b) – Eq.(a), we get

10x = 9.9999…

-x = -0.9999…

___________

9x = 9

x = 1

The difference between 1 and 0.999999 is 0.000001, which is negligible.

Hence, we can conclude that 0.999 is too much near 1; therefore, 1 as the answer can be justified.

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.

Ans:

1/17

Dividing 1 by 17:

There are 16 digits in the repeating block of the decimal expansion of 1/17.

6. Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and have terminating decimal representations (expansions). Can you guess what property q must satisfy?

Ans:

We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion terminates. For example:

1/2 = 0. 5, denominator q = 2¹

7/8 = 0. 875, denominator q =2³

4/5 = 0. 8, denominator q = 5¹

We can observe that the terminating decimal may be obtained in the situation where the prime factorisation of the denominator of the given fractions has the power of only 2 or only 5 or both.

7. Write three numbers whose decimal expansions are non-terminating and non-recurring.

Ans:

We know that all irrational numbers are non-terminating and non-recurring. The three numbers with decimal expansions that are non-terminating and non-recurring are:

1. √3 = 1.732050807568
2. √26 =5.099019513592
3. √101 = 10.04987562112

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

Ans:

Three different irrational numbers are:

1. 0.73073007300073000073…
2. 0.75075007300075000075…
3. 0.76076007600076000076…

9. Classify the following numbers as rational or irrational according to their type:

(i)√23

Ans:

√23 = 4.79583152331…

Since the number is non-terminating and non-recurring, it is an irrational number.

(ii)√225

Ans:

√225 = 15 = 15/1

Since the number can be represented in p/q form, it is a rational number.

(iii) 0.3796

Ans:

Since the number,0.3796, is terminating, it is a rational number.

(iv) 7.478478

Ans:

The number 7.478478at  is non-terminating but recurring; hence, it is a rational number.

(v) 1.101001000100001…

Ans:

Since the number 1.101001000100001… is non-terminating and non-repeating (non-recurring), it is an irrational number.

Exercise-1.4

1. Classify the following numbers as rational or irrational:

(i) 2 –√5

Ans:

We know that √5 = 2.2360679…

Here, 2.2360679…is non-terminating and non-recurring.

Now, substituting the value of √5 in 2 –√5, we get,

2-√5 = 2-2.2360679… = -0.2360679

Since the number – 0.2360679… is non-terminating non-recurring, 2 –√5 is an irrational number.

(ii) (3 +√23)- √23

Ans:

(3 +√23) –√23 = 3+√23–√23

= 3

= 3/1

Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational.

(iii) 2√7/7√7

Ans:

2√7/7√7 = ( 2/7)× (√7/√7)

We know that (√7/√7) = 1

Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7

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

(iv) 1/√2

Ans:

Multiplying and dividing the numerator and denominator by √2, we get,

(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)

We know that √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0.7071..

Since the number 0.7071… is non-terminating non-recurring, 1/√2 is an irrational number.

(v) 2π

Ans:

We know that the value of = 3.1415

Hence, 2 = 2×3.1415.. = 6.2830…

Since the number 6.2830… is non-terminating non-recurring, 2 is an irrational number.

2. Simplify each of the following expressions:

(i) (3+√3)(2+√2)

Ans:

(3+√3)(2+√2 )

Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)

= 6+3√2+2√3+√6

(ii) (3+√3)(3-√3 )

Ans:

(3+√3)(3-√3 ) = 3²-(√3)² = 9-3

= 6

(iii) (√5+√2)²

Ans:

(√5+√2)² = √5²+(2×√5×√2)+ √2²

= 5+2×√10+2 = 7+2√10

(iv) (√5-√2)(√5+√2)

Ans:

(√5-√2)(√5+√2) = (√5²-√2²) = 5-2 = 3

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?

Ans:

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

4. Represent (√9.3) on the number line.

Ans:

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

Step 3: Draw a semi-circle of radius OC with centre O.

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, is a right-angled triangle.

Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1

OB = OC – BC

⟹ (10.3/2)-1 = 8.3/2

Using Pythagoras’ theorem,

We get,

OD²=BD²+OB²

⟹ (10.3/2)² = BD²+(8.3/2)²

⟹ BD^{2 =} (10.3/2)²-(8.3/2)²

⟹ (BD)² = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)

⟹ BD² = 9.3

⟹ BD =  √9.3

Thus, the length of BD is √9.3.

Step 6: Taking BD as the radius and B as the centre, draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O, as shown in the figure.

5. Rationalize the denominators of the following:

(i) 1/√7

Ans:

Multiply and divide 1/√7 by √7

(1×√7)/(√7×√7) = √7/7

(ii) 1/(√7-√6)

Ans:

Multiply and divide 1/(√7-√6) by (√7+√6)

[1/(√7-√6)]×(√7+√6)/(√7+√6) = (√7+√6)/(√7-√6)(√7+√6)

= (√7+√6)/√7²-√6² [denominator is obtained by the property, (a+b)(a-b) = a²-b²]

= (√7+√6)/(7-6)

= (√7+√6)/1

= √7+√6

(iii) 1/(√5+√2)

Ans:

Multiply and divide 1/(√5+√2) by (√5-√2)

[1/(√5+√2)]×(√5-√2)/(√5-√2) = (√5-√2)/(√5+√2)(√5-√2)

= (√5-√2)/(√5²-√2²) [denominator is obtained by the property, (a+b)(a-b) = a²-b²]

= (√5-√2)/(5-2)

= (√5-√2)/3

(iv) 1/(√7-2)

Ans:

Multiply and divide 1/(√7-2) by (√7+2)

1/(√7-2)×(√7+2)/(√7+2) = (√7+2)/(√7-2)(√7+2)

= (√7+2)/(√7²-2²) [denominator is obtained by the property, (a+b)(a-b) = a²-b²]

= (√7+2)/(7-4)

= (√7+2)/3

Exercise-1.5