Given universal set = - 6 , - 5 3 4 , 4 , - 3 5 , - 3 8 , 0 , 4 5 , 1 , 1 2 3 , 8 , 3 . 01 , π , 8 . 47 From the given set , find set of rational numbers

We need to find the set of rational numbers. Rational numbers are numbers of the form p q , where q ≠ 0 U = - 6 , - 5 3 4 , 4 , - 3 5 , - 3 8 , 0 , 4 5 , 1 , 1 2 3 , 8 , 3 . 01 , π , 8 . 47 Clearly, - 5 3 4 , - 3 5 , - 3 8 , 4 5 and 1 2 3 are of the form p q Hence, they are rational numbers. Since the set of integers is a subset of rational numbers, - 6 , 0 and 1 are also rational numbers. Thus, decimal numbers 3 . 01 and 8 . 47 are also rational numbers because they are terminating decimals. Hence, from the above set, the set of rational numbers is Q, and Q = - 5 3 4 , - 3 5 , - 3 8 , 4 5 , 1 2 3 , - 6 , 0 , 1 , 3 . 01 , 8 . 47

Given universal set = - 6 , - 5 3 4 , 4 , - 3 5 , - 3 8 , 0 , 4 5 , 1 , 1 2 3 , 8 , 3 . 01 , π , 8 . 47 From the given set , find set of irrational numbers

We need to find the set of irrational numbers. Irrational numbers are numbers that are not rational or An irrational number is a type of real number which cannot be represented as a simple fraction. U = - 6 , - 5 3 4 , 4 , - 3 5 , - 3 8 , 0 , 4 5 , 1 , 1 2 3 , 8 , 3 . 01 , π , 8 . 47 The set of irrational numbers is the set of complement of the rational numbers over real numbers. Here the set of irrational numbers is U − Q = 8 , π

Given universal set = - 6 , - 5 3 4 , 4 , - 3 5 , - 3 8 , 0 , 4 5 , 1 , 1 2 3 , 8 , 3 . 01 , π , 8 . 47 From the given set , find set of integers.

We need to find the set of integers. Set of integers consists of zero, the natural numbers and their additive inverses. The set of integers is Z . Z = … − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , … Given Universal set is - 6 , - 5 3 4 , 4 , - 3 5 , - 3 8 , 0 , 4 5 , 1 , 1 2 3 , 8 , 3 . 01 , π , 8 . 47 Here the set of integers is U ∩ Z = − 6 , 4 , 0 , 1

Given universal set = - 6 , - 5 3 4 , 4 , - 3 5 , - 3 8 , 0 , 4 5 , 1 , 1 2 3 , 8 , 3 . 01 , π , 8 . 47 From the given set , find set of non-negative integers.

An integer is a number that has no fractional part, and no digits after the decimal point. An integer can be positive, negative or zero. The set of integers is Z . Z = … − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , … Given Universal set is - 6 , - 5 3 4 , 4 , - 3 5 , - 3 8 , 0 , 4 5 , 1 , 1 2 3 , 8 , 3 . 01 , π , 8 . 47 0 and 1 are the only non-negative integers, from the given set. Set of non-negative integers is 0 , 1 .

Use method of contradiction to show that 3 and 5 are irrational number.

Let us suppose that 3 and 5 are rational numbers Therefore, 3 = a b and 5 = x y (Where a , b ∈ 7 and b , y ≠ 0 x , y ) Squaring both sides 3 = a 2 b 2 , 5 = x 2 y 2 3 b 2 = a 2 , 5 y 2 = x 2 ⇒ a 2 and x 2 are odd as 3 b 2 and 5 y 2 are odd . ⇒ a and x are odd. . . . ( 1 ) Let a = 3 c , x = 5 z a 2 = 9 c 2 , x 2 = 25 z 2 3 b 2 = 9 c 2 , 5 y 2 = 25 z 2 (From equation . . . . ( 1 ) ⇒ b 2 = 3 c 2 , y 2 = 5 z 2 ⇒ b 2 and y 2 are odd as 3 c 2 and 5 z 2 are odd . ⇒ b and y are odd . . . ( 2 ) From equation ( 1 ) and ( 2 ) we get a , b , x , y are odd integers. i.e., a , b , and x , y have common factors 3 and 5 this contradicts our assumption that a b a n d x y are rational i.e, a , b and x , y do not have any common factors other than. ⇒ a b a n d x y is not rational ⇒ 3 and 5 are irrational.

Prove that the given number is irrational: 3 + 2

To prove 3 + 5 is an irrational number. Let 3 + 5 be a rational number. We know that a rational number can be written in the form of p q where p and q are coprime integers So, 3 + 5 = p q ⇒ 3 = p q - 5 On squaring both the sides, ⇒ 3 2 = p q - 5 2 ⇒ 3 = p 2 q 2 - 2 5 p q + 5 ⇒ 2 5 p q = p 2 q 2 + 5 - 3 ⇒ 2 5 p q = p 2 q 2 + 2 ⇒ 5 p q = p 2 2 q 2 + 2 2 ⇒ 5 p q = p 2 2 q 2 + 1 ⇒ 5 p q = p 2 + 2 q 2 2 q 2 ⇒ 5 = p 2 + 2 q 2 2 q 2 × q p ⇒ 5 = p 2 + 2 q 2 2 p q Here when p & q are integer then p 2 + 2 q 2 2 p q is a rational number. It means 5 is also a rational number. This is a contradiction as 5 is an irrational number. This is a contradiction and has arisen due to our wrong assumption. Hence our assumption is false. Therefore, 3 + 5 is an irrational number.

E M B I B E

Concise Mathematics I.C.S.E - Class IX>Chapter 1 - Rational and Irrational Numbers>EXERCISE>Q 3

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All rational numbers are real number.

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Important Questions on Rational and Irrational Numbers

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Concise Mathematics I.C.S.E - Class IX>Chapter 1 - Rational and Irrational Numbers>EXERCISE>Q 3

All real numbers are rational number.

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Concise Mathematics I.C.S.E - Class IX>Chapter 1 - Rational and Irrational Numbers>EXERCISE>Q 3

Some real numbers are rational number.

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Concise Mathematics I.C.S.E - Class IX>Chapter 1 - Rational and Irrational Numbers>EXERCISE>Q 4

Given universal set $= \{- 6, - 5 \frac{3}{4}, \sqrt{4}, - \frac{3}{5}, - \frac{3}{8}, 0, \frac{4}{5}, 1, 1 \frac{2}{3}, \sqrt{8}, 3.01, π, 8.47\}$ From the given set , find set of rational numbers

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9th ICSE

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Concise Mathematics I.C.S.E - Class IX>Chapter 1 - Rational and Irrational Numbers>EXERCISE>Q 4

Given universal set $= \{- 6, - 5 \frac{3}{4}, \sqrt{4}, - \frac{3}{5}, - \frac{3}{8}, 0, \frac{4}{5}, 1, 1 \frac{2}{3}, \sqrt{8}, 3.01, π, 8.47\}$ From the given set , find set of irrational numbers

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9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 1 - Rational and Irrational Numbers>EXERCISE>Q 4

Given universal set $= \{- 6, - 5 \frac{3}{4}, \sqrt{4}, - \frac{3}{5}, - \frac{3}{8}, 0, \frac{4}{5}, 1, 1 \frac{2}{3}, \sqrt{8}, 3.01, π, 8.47\}$ From the given set , find set of integers.

EASY

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 1 - Rational and Irrational Numbers>EXERCISE>Q 4

Given universal set $= \{- 6, - 5 \frac{3}{4}, \sqrt{4}, - \frac{3}{5}, - \frac{3}{8}, 0, \frac{4}{5}, 1, 1 \frac{2}{3}, \sqrt{8}, 3.01, π, 8.47\}$ From the given set , find set of non-negative integers.