Real Number System

(i) Yes, because natural numbers are 1, 2, 3, …, and so on

And whole numbers are 0, 1, 2, 3, …,

Therefore, Every natural number is a whole number.

(ii) No,

0 is a whole number, but not a natural number.

(iii) Yes, Any integer 'p' can be represented in the form and 1≠0,

Therefore, every integar is a rational number.

(iv) No, is a rational number, but not a whole number.

(v) No, is a rational number, but not an integer.

(vi) No, negative integers are not whole numbers,

For example, -1 is an integer but not a whole number.

Yes, 0 can be re-written as , where q is any non-negative integer and therefore 0 is a rational number.

As,

-5 < -4 < -3 < -2

i. , the decimal expansion is terminating.

ii. , the decimal expansion is non-terminating and recurring.

iii. , the decimal expansion is non-terminating and recurring.

iv. , the decimal expansion is terminating.

v. , the decimal expansion is non-terminating and recurring.

the decimal expansion is non-terminating and recurring.

vi. the decimal expansion is non-terminating and recurring.

vii. , the decimal expansion is non-terminating and recurring.

viii. , the decimal expansion is terminating.

i. , As the rational number can be represented in form , therefore its terminating.

ii. , As the rational number can't be represented in form , therefore its non-terminating and recurring.

iii. , As the rational number can be represented in form , therefore its terminating.

iv. , As the rational number can't be represented in form , therefore its non-terminating and recurring.

i. let

⇒ 100x = 18.181818 … = 18 + 0.181818…

⇒ 100x = 18 + x

⇒ 99x = 18

ii. let

⇒ 1000x = 427.427427 … = 18 + 0.427427427…

⇒ 1000x = 427 + x

⇒ 999x = 427

iii. let

⇒ 10000x = 1.00010001 … = 1 + 0.00010001…

⇒ 10000x = 1 + x

⇒ 9999x = 1

iv. As,

Let x = 0.454545…

⇒ 1 + x = 1 + 0.454545..

⇒ 1+ x = 1.45454545

⇒ 100 + 100x = 145.454545… = 145 + 0.454545…

⇒ 100 + 100x = 145 + x

⇒ 99x = 45

And

v. As,

Let x = 0.3333…

⇒ 7 + x = 7 + 0.3333…

⇒ 7 + x = 7.3333…

⇒ 70 + 10x = 73.3333… = 73 + 0.3333…

⇒ 70 + 10x = 73 + x

⇒ 9x = 3

And

vi. let

⇒ 1000x = 416.416416 … = 416 + 0.416416…

⇒ 1000x = 416 + x

⇒ 999x = 416

No of divisions in repeating block = 6

Also,

Step 1:

Draw a number line. Mark points O and A such that O represents the number zero and

A represents the number 2. i.e., OA = 2 unit

Step 2:

Draw AB ⊥ OA such that AB = 1unit.

Step 3:

Join OB

In right triangle OAB, by Pythagorean theorem,

OB² = OA² + AB²

OB² = 2² + 1²

OB² = 4 + 1

OB = √5

Step 4:

With O as centre and radius OB, draw an arc to intersect the number line at C on the right side of O. Clearly OC = OB = √5 . Thus, C corresponds to √5 on the number line.

As, √3 = 1.73205…

And √5 = 2.23606…

We have to find three non-terminating and non-recurring numbers between 1.73205… and 2.23606…

Three possible answers are

1.740400400040000…

1.750500500050000…

1.760600600060000…

We have to find two non-terminating and non-recurring numbers between 3 and 3.5

Two possible answers are

3.10100100010000…

3.20200200020000…

We have to find two non-terminating and non-recurring numbers between 0.15 and 0.16

Two possible answers are

0.15100100010000…

0.15200200020000…

Clearly,

and

We have to find two non-terminating and non-recurring numbers between 0.571428571428… and 0.714285714285…

Two possible answers are

0.580800800080000…

0.590900900090000…

Clearly,

As, √3 = 1.73205…

We have to find two non-terminating and non-recurring numbers between 1.73205… and 2

Two possible answers are

1.740400400040000…

1.750500500050000…

Clearly,

1.1011001110001… < 2 < 2.1011001110001…

As, 2 is a rational number.

Therefore, 2 is a required rational number.

Now, We have to find a non-terminating and non-recurring number between 1.1011001110001… and 2.1011001110001…

And one of the possible answer is

1.20200200020000…

Clearly,

0.12122122212222…<0.13<0.14<0.2122122122212222…

And 0.13, 0.14 are terminating, therefore rational numbers.

By definition, a rational number is one which follows either of the following 2 conditions:

1. It can be expressed inform where p and q are integers (q≠0).

2. Its decimal expansion is either terminating or non-terminating recurring.

Ex. – 0.35 (terminating decimal expansion) → Rational number

0.7777…. = (non-terminating but recurring i.e. the number does not terminate but its digits follow a pattern.) → Rational number

π = 3.14…… (Non-terminating non recurring) → Irrational number

Hence, the answer.

As explained in the above question, for a number to be rational its decimal expansion has to be either terminating or non-terminating and recurring.

For example –

π = 3.14…… (Non-terminating non recurring) → Irrational number

Hence, the answer.

To convert into its equivalent decimal expansion, we can either proceed by long division method or multiply the denominator and (hence) numerator by a number such that the denominator turns out to be 100.

Now, let 4×a = 100

Solving, we get a = 25.

Hence multiplying numerator and denominator by 25.

∴ (-3/4) × (25/25) = (-75/100) = -0.75 (Placing decimal point before 2 digits).

Hence, the answer.

Let x = 0. = 0.3333... → (eq)1

Since there is only one repeating digit after decimal point, multiply equation 1 by 10.

10x = 3.3333… = 3 + 0.3333… = 3 + x → (eq)2

(eq)2 –(eq)1

10x-x = 3.3333… - 0.3333… = 3

∴ 9x = 3

∴ x = 1/3

Hence, the answer.

Real numbers is the set which includes both rational and irrational numbers.

∴ Rational numbers and Irrational numbers are 2 independent subsets of Real numbers. Hence every real number is not a rational number (since real numbers also consists of irrational numbers.)

Option (A) is correct because every natural number can be expressed in form (q≠0 i.e a rational number).

Option (C) is correct because every whole number can be expressed in form (q≠0 i.e a rational number).

Option (D) is correct because every integer can be expressed in form (q≠0 i.e a rational number).

Natural numbers ⊂ Whole numbers ⊂ Integers ⊂ Rational Numbers ⊂ Real numbers

Real numbers ⊃ (Rational numbers + Irrational numbers).

Hence, only option B is incorrect.

If a rational number (q≠0) can be expressed in the form , where p ∈ Z, and m,n ∈ W, then the rational number will have a terminating decimal expansion.

Otherwise, the rational number will have a non-terminating and recurring decimal

expansion.

Now, 32 = 2⁵ × 5⁰ Hence, it has terminating decimal expansion.Hence, option A is correct.

Option B is incorrect ∵ 9 = 3² and it cannot be expressed in the form 2^m × 5ⁿ.

Option C is incorrect ∵ 15 = 3 × 5 and it cannot be expressed in the form 2^m × 5ⁿ.

Option D is incorrect ∵ 12 = 3 × 2² and it cannot be expressed in the form 2^m × 5ⁿ.

Hence, only option A is correct.

Option A is correct since π = 3.141592… i.e. the decimal expansion is non-terminating and non-recurring.

Option B is incorrect because = ±3 and it is an integer (and hence a rational number, Integers ⊂ Rational Numbers).

Option C is incorrect as it is expressed in the form (p = 1 and q = 4) and is a rational number (by definition, q≠0).

Option D is incorrect as it is expressed in the form (p = 1 and q = 5) and is a rational number (by definition, q≠0).

Hence, option A is correct.

i.√(2 + √3 ) is irrational ∵ √3 is irrational and 2 is rational and sum of a rational and irrational number is always irrational. Hence 2 + √3 is irrational. Hence, its square root is also irrational.

ii.√(4 + √25) is rational ∵ √25( = 5) is rational and 4 is also rational.

Hence, √(4 + 5) = √9 = 3 is rational.

iii. ∛(5 + √7) is irrational ∵ √7 is irrational and 5 is rational and sum of a rational and irrational number is always irrational. Hence 5 + √7 is irrational. Hence, its cube root is also irrational.

iv.√(8-∛8 ) is irrational. Here both ( = 2) and 8 are rational. Hence

8 – 8^1/3 = 8 – 2 = 6 is rational. But √6 is irrational.

Hence, the answer.