CBSE - Number Systems

Yes. Zero is a rational number as it can be represented as 0/1 or 0/2.

There are infinite rational numbers in between 3 and 4.
3 and 4 can be represented as 24/8 and 32/8 respectively.
Therefore, six rational numbers between 3 and 4 are
25/8, 26/8, 27/8, 28/8, 29/8, 30/8.

There are infinite rational numbers in between 3/5 and 4/5
3/5 = 3×6 / 5×6 = 18 / 30
4/5 = 4×6 / 5×6 = 24/30
Therefore, five rational numbers between 3/5 and 4/5 are
19/30, 20/30, 21/30, 22/30, 23/30.

(i) Every natural number is a whole number.

ANS: True, since the collection of whole numbers contains all natural numbers.

(ii) Every integer is a whole number.
ANS: False, as integers may be negative but whole numbers are always positive.

(iii) Every rational number is a whole number.

ANS: False, as rational numbers may be fractional but whole numbers may not be.

(i) Every irrational number is a real number.

ANS: True, since the collection of real numbers is made up of rational and irrational numbers.

(ii) Every point on the number line is of the form√m, where m is a natural number.
ANS: False, since positive number cannot be expressed as square roots.

(iii) Every real number is an irrational number.
ANS: False, as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.

No, the square roots of all positive integers are not irrational. For example √4 = 2

Step 1: Let AB be a line of length 2 unit on number line.
Step 2: At B, draw a perpendicular line BC of length 1 unit. Join CA.
Step 3: Now, ABC is a right angled triangle. Applying Pythagoras theorem,
AB² + BC² = CA²
⇒ 2² + 1² = CA²
⇒ CA² = 5
⇒ CA = √5
Thus, CA is a line of length √5 unit.
Step 4: Taking CA as a radius and A as a centre draw an arc touching
the number line. The point at which number line get intersected by
arc is at √5 distance from 0 because it is a radius of the circle
whose centre was A.
Thus, √5 is represented on the number line as shown in the figure.

Let x = 0.9999…
10x = 9.9999…
10x = 9 + x
9x = 9
x = 1
The difference between 1 and 0.999999 is 0.000001 which is negligible. Thus, 0.999 is too much near 1, Therefore, the 1 as answer can be justified. 

We observe that when q is 2, 4, 5, 8, 10... then the decimal expansion is terminating. 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 observed that terminating decimal may be obtained in the situation where prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both.

Three numbers whose decimal expansions are non-terminating non-recurring are:

0.303003000300003...

0.505005000500005...

0.7207200720007200007200000…

Three different irrational numbers are:

0.73073007300073000073…

0.75075007300075000075…

0.76076007600076000076…

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

Answer

(i) 2 - √5 = 2 - 2.2360679… = - 0.2360679…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(ii) (3 + √23) - √23 = 3 + √23 - √23 = 3 = 3/1
Since the number is rational number as it can represented in p/q form.

(iii) 2√7/7√7 = 2/7
Since the number is rational number as it can represented in p/q form.

(iv) 1/√2 = √2/2 = 0.7071067811...
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(v) 2π = 2 × 3.1415… = 6.2830…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(i) (3 + √3) (2 + √2)

(ii) (3 + √3) (3 - √3)

(iii) (√5 + √2)²
(iv) (√5 - √2) (√5 + √2)

Answer

(i) (3 + √3) (2 + √2)

⇒ 3 × 2 + 2 + √3 + 3√2+ √3 ×√2

⇒ 6 + 2√3 +3√2 + √6

(ii) (3 + √3) (3 - √3) [∵ (a + b) (a - b) = a² - b²]

⇒ 3² - (√3)²
⇒ 9 - 3

⇒ 6

(iii) (√5 + √2)² [∵ (a + b)² = a² + b² + 2ab]
⇒ (√5)² + (√2)² + 2 ×√5 × √2
⇒ 5 + 2 + 2 × √5× 2 

⇒ 7 +2√10

(iv) (√5 - √2) (√5 + √2) [∵ (a + b) (a - b) = a² - b²]

⇒ (√5)² - (√2)² 

⇒ 5 - 2

⇒ 3

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 realise that either c or d is irrational. The value of π is almost equal to 22/7 or 3.142857... Therefore, the fraction  c/d  is irrational. Hence, π is irrational.

Step 1: Draw a line segment of unit 9.3. Extend it to C so that BC is of 1 unit.
Step 2: Now, AC = 10.3 units. Find the centre of AC and name it as O.
Step 3: Draw a semi circle with radius OC and centre O.
Step 4: Draw a perpendicular line BD to AC at point B which intersect the semicircle at D. Also, Join OD.
Step 5: Now, OBD 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,
OD² = BD² + OB²
⇒ (10.3/2)² = BD2 + (8.3/2)²
⇒ BD² = (10.3/2)² - (8.3/2)²
⇒ BD² = (10.3/2 – 8.3/2) (10.3/2 + 8.3/2)
⇒ BD² = 9.3
⇒ BD² =  √9.3
Thus, the length of BD is √9.3.
Step 6: Taking BD as radius and B as 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.