Real Numbers

Mathematics Class Tenth

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NCERT Exercise 1.3 - irrational numbers

Irrational Number

A number is called irrational if it cannot be written in the form of ^p/_q , where p and q are integers and q ≠ 0. For example - 2, 3, π, 0.101101110 . . . . , etc.

Theorem 1.3: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Proof:

Let the prime factorization of

a =p₁, p₂, . . . p_n where p₂, p₂, . . . . , p_n are primes

Therefore, a² = (p₁p₂ . . . p_n) (p₁p₂ . . . p_n)

=p₁²p₂² . . . p_n²

Now, as given p divides a². Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a². However using the uniqueness part of the Fundamental Theorem of Arithmetic, we realize that the only prime factors of a² are p₁, p₂, . . ., p_n.

So, pp is one of the p₁, p₁, . . ., p_n.

Now, since a = p₁ p₂ . . . p_n,

Thus, p divides a.

Theorem 1.4 : 2 is irrational.

Proof:

Let assume, to the contrary, that 2 is rational.

So, we can find integers r and s(s ≠ 0) such that 2 = ^r/_s

Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get 2 = ^a/_b, where a and b are coprime.

So, b 2 = a.

By squaring both sides, we get

2b² = a² -----(i).

Therefore, 2 divides a².

Now, by above theorem (1.3), it follows that 2 divides a.

So, it can be written that a = 2c for some integer c.

After substituting a = 2c in equation (i) we get

2b² = 4c²

⇒ b² = 2c²

This means that 2 divides b², and so 2 divides b.

Therefore, a and b have at least 2 common factors.

But, this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that 2 is rational.

Thus, 2 is irrational.

NCERT Exercise 1.3 solution class ten mathematics

Question: 1. Prove that 5 is irrational.

Solution:

Let assume, to the contrary, that 5 is rational.

So, we can find integers r and s(s!=0) such that 5 = ^r/_s

Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get 5 = ^a/_b, where a and b are coprime.

So, b5 = a.

By squaring both sides, we get

5b² = a² -----(i).

Therefore, 5 divides a².

So, it can be written that a = 5c for some integer c.

After substituting a = 5c in equation (i) we get

5b² = 25c²

⇒ b² = 5c²

This means that 5 divides b², and so 5 divides b.

Therefore, a and b have at least 5 as a common factor.

But, this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that 5 is rational.

Thus, 5 is irrational.

Question: 2. Prove that 3 + 25 is irrational.

Solution:

Let 3 + 25 is rational.

Therefore, we can find two integers a and b (b ≠ 0) such that

3 + 25 = ^a/_b

⇒ 25 = ^a/_b – 3

⇒ 5= ¹/₂ (^a/_b – 3)

Since, a and b are integers, hence ¹/₂ (^a/_b – 3) will also be rational.

And hence 5 is rational.

But this contradicts the fact that 5 is irrational. Thus, our assumption that 3 + 25 is rational is incorrect.

This shows that 3 + 25 is irrational.

Question (3) Prove that the following are irrationals:

(i) ¹/₂

Solution:

Let ¹/₂ is rational.

Therefore, we can find two integers a and b (b &n3; ) such that

¹/₂ = ^a/_b

⇒ 2 = ^b/_a

^b/_a is rational as a and b are integers.

Thus, 2 is rational which contradicts the fact that 2 is irrational.

This, is because we assume that, 2 is rational which is false.

Thus, ¹/₂ is irrational.

(ii) 75

Solution:

Let 75 is rational.

Thus, we can find two integers, a and b (b ≠ 0) such that 75 = ^a/_b for some integers a and b.

Thus, 5 = a(7b)

Now, since a and b are integers, thus ^a/_7b is rational.

Therefore, 5 should be rational.

But, this contradicts the fact that 5 is irrational. This is because our assumption 75 is as rational is false.

Thus, 75 is irrational.

(iii) 6 + 2

Solution:

Let 6 + 2 is rational.

Therefore, we can find two integers a and b (b &n3; ) such that

6+2 = ^a/_b

⇒ 2 = ^a/_{b –6}

Since, a and b are integers, ^a/_b – 6 is also rational and hence, 2 should be rational.

This contradicts the fact that 2 is irrational, because our assumption which says that 6 + 2 is rational is false.

Hence 6 + 2 is irrational.

NCERT Exercise 1.4 Class ten mathematics

Question (1) Without actually performing long division, state whether the following rational numbers will have a terminating decimal expansion.

(i) ¹³/₃₁₂₅

Solution:

¹³/₃₁₂₅ = ¹³/_5⁵

We know that if denominator q of a rational number ^p/_q is in the form of 5ⁿ then rational number have terminating decimal expression.

Now, since denominator of the given rational number is in the form of 5ⁿ, thus given rational number ¹³/₃₁₂₅ has terminating decimal expression.

Thus, Answer = yes Terminating.

(ii) ¹⁷/₈

Solution

¹⁷/₈ = ¹⁷/_2³

We know that if denominator q of a rational number ^p/_q is in the form of 2ⁿ then rational number have terminating decimal expression.

Now, since denominator of the given rational number is in the form of 2ⁿ, thus given rational number ¹⁷/₈ has terminating decimal expression.

Thus, Answer = Terminating.

(iii) ⁶⁴/₄₅₅

Solution

⁶⁴/₄₅₅ = ⁶⁴/_{5 × 7 × 13}

We know that if denominator q of a rational number ^p/_q is in the form of 2ⁿ × 5^m then rational number has terminating decimal expression otherwise rational number has non-terminating decimal expression.

Now, since denominator of the given rational number is not in the form of 2ⁿ × 5^m, thus given rational number ⁶⁴/₄₅₅ has non-terminating decimal expression.

Thus, Answer = Non-Terminating.

(iv) ¹⁵/₁₆₀₀

Solution

¹⁵/₁₆₀₀ = ¹⁵/_{2⁶ × 5²}

Now, since denominator of the given rational number is not in the form of 2ⁿ × 5^m, thus given rational number ¹⁵/₁₆₀₀ has non-terminating decimal expression.

Thus, Answer = Terminating.

(v) ²⁹/₃₄₃

Solution

²⁹/₃₄₃ = ²⁹/_7³

Now, since denominator of the given rational number is not in the form of 2ⁿ × 5^m, thus given rational number ²⁹/₃₄₃ has non-terminating decimal expression.

Thus, Answer = Non-terminating.

(vi) ²³/_{2³ 5²}

Solution

Given, ²³/_{2³ × 5²}

We know that if denominator q of a rational number ^p/_q is in the form of 2ⁿ × 5^m then rational number has terminating decimal expression.

Now, since denominator of the given rational number is in the form of 2ⁿ × 5^m, thus given rational number ²³/_{2³ 5²} has terminating decimal expression.

Thus, Answer = Terminating.

(vii) ¹²⁹/_{2² 5⁷ 7⁵}

Solution

Given, ¹²⁹/_{2² × 5⁷ × 7⁵}

Now, since denominator of the given rational number is not in the form of 2ⁿ × 5^m, thus given rational number ¹²⁹/_{2² 5⁷ 7⁵} has non-terminating decimal expression.

Thus, Answer = Non-Terminating.

(viii) ⁶/₁₅

Solution

Given, ⁶/₁₅

= ⁶/_{3 × 5}

Now, since denominator of the given rational number is not in the form of 2ⁿ × 5^m, thus given rational number ⁶/₁₅ has non-terminating decimal expression.

Thus, Answer = Non-Terminating.

(ix) ³⁵/₅₀

Solution

Given, ³⁵/₅₀

= ³⁵/_{2 × 5²}

Now, since denominator of the given rational number is in the form of 2ⁿ × 5^m, thus given rational number ³⁵/₅₀ has terminating decimal expression.

Thus, Answer = Terminating.

(x) ⁷⁷/₂₁₀

Solution

Given, ⁷⁷/₂₁₀

= ^{11 × 7}/_{30 × 7}

= ¹¹/₃₀ = ¹¹/_{2 × 5 × 3}

Now, since denominator of the given rational number is not in the form of 2ⁿ × 5^m, thus given rational number ⁷⁷/₂₁₀ has non-terminating decimal expression.

Thus, Answer = Non-Terminating.

Question (2) Write down the decimal expansions of those rational numbers in Question 1 which have terminating decimal expansions.

(i) ¹³/₃₁₂₅

Solution

¹³/₃₁₂₅ = ¹³/_5⁵

= ^{2⁵ × 13}/_{2⁵ × 5⁵}

= ^{32 × 13}/_10⁵

= ⁴¹⁶/_10⁵

= 0.00416 Answer

(ii) ¹⁷/₈

Solution

¹⁷/₈ = ¹⁷/_2³

= ^{17 × 5³}/_{2³ × 5³}

= ^{17 × 125}/_10³

= ²¹²⁵/_10³

= 2.125 Answer

(iv) ¹⁵/₁₆₀₀

Solution

¹⁵/₁₆₀₀ = ^{5 × 3}/_{2⁶ × 5²}

= ³/_{2⁶ × 5}

= ^{3 × 5⁵}/_{2⁶ × 5 × 5⁵}

= ^{3 × 3125}/_{2⁶ × 5⁶}

= ⁹³⁷⁵/_10⁶

= 0.009375 Answer

(vi) ²³/_{2³ 5²}

Solution

²³/_{2³ 5²} = ^{23 × 5}/_{2³ × 5² × 5}

= ¹¹⁵/_{2³ × 5³}

= ¹¹⁵/_10³

= 0.115 Answer

(viii) ⁶/₁₅

Solution

⁶/₁₅ = ^{2 × 3}/_{5 × 3}

= ²/₅ = ^{2 × 2}/_{2 × 5} = ⁴/₁₀

= 0.4 Answer

(ix) ³⁵/₅₀

Solution

³⁵/₅₀ = ^{5 × 7}/_{5 × 10}

= ⁷/₁₀ = 0.7 Answer

Question (3) The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form ^p/_q, what can you say about the prime factors of q?

(i) 43.123456789

Solution

Given,

43.123456789

Since the given number has terminating decimal expansion, thus given number is a rational number and can be written in the form of ^p/_q in which q is in the form of 2ⁿ5^m.

Thus, prime factor of denominator q is in the form of 2ⁿ5^m. Answer

(ii) 0.120120012000120000 . . . . .

Solution

Given, 0.120120012000120000 . . . . .

Since the given number has a non-terminating and non-recurring decimal expansion, thus given number is an irrational number.

(iii) 43.123456789

Solution

Given, 43.123456789

Since the given number has a non-terminating recurring decimal expansion.

Thus, this number is a rational number of the form ^p/_q but q is not in the form of 2ⁿ5^m.

This means that denominator q of the given number has a factor other than 2 or 5.

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