KC Sinha Mathematics Solution Class 10 Chapter 1 Real numbers Exercise 1.3

Q1 Q2 Q3 Q4 A Q4 B Q4 C Q5 A Q5 B Q5 C Q5 D Q5 E Q5 F

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4

Exercise 1.3

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Question 1 

Prove that √2 is irrational.

Sol :
Let us assume that √2 is rational. So, we can find integers p and q (≠ 0) such that √2 $=\frac{p}{q}$  .
Suppose p and q have a common factor other than 1.
Then, we divide by the common factor to get √2 = $\frac{a}{b}$ , where a and b are coprime.
So, b√2 = a.
Squaring on both sides, we get
2b² = a²
Therefore, 2 divides a².
Now, by Theorem which states that Let p be a prime number. If p divides a² , then p divides a, where a is a positive integer,
⇒2 divides a².
So, we can write a = 2c for some integer c
Substituting for a, we get 2b² = 4c² ,i.e. b² = 2c² .
This means that 2 divides b², and so 2 divides b (again using the above Theorem with p = 2). Therefore, a and b have at least 2 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 2 is rational.
So, we conclude that √2 is irrational.

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Question 2 

Prove that √3 is irrational.

Sol :
Let us assume that √3 is rational.
Hence, √3 can be written in the form $\frac{a}{b}$
where a and b (≠ 0) are co-prime (no common factor other than 1).
Hence, √3 = $\frac{a}{b}$
So, b√3 = a.
Squaring on both sides, we get
3b² = a²
$\frac{a^{2}}{3}$ = b²
Hence, 3 divides a².
By theorem: Let p is a prime number and p divides a² , then p divides a, where a is a positive integer,
⇒ 3 divides a also …(1)
Hence, we can say a = 3c for some integer c
Now, we know that 3b² = a²
Putting a = 3c
3b² = (3c)²
3b² = 9c²
b² = 3c²
Hence, 3 divides b²
By theorem: Let p is a prime number and p divides a², then p divides a, where a is a positive integer,
So, 3 divides b also …(2)
By (1) and (2)
3 divides both a and b
Hence, 3 is a factor of a and b
So, a and b have a factor 3
Therefore, a and b are not co-prime.
Hence, our assumption is wrong
Therefore, by contradiction √3 is irrational.

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Question 3 

Prove that $\frac{1}{\sqrt{5}}$ is irrational.

Sol :
Let us assume that $\frac{1}{\sqrt{5}}$ be a rational number.
Then, it will be of the form $\frac{a}{b}$ where a and b are co-prime and b≠0.
Now, $\frac{a}{b}=\frac{1}{\sqrt{5}}$

⇒$\frac{a}{b}=\frac{1 \times \sqrt{5}}{\sqrt{5 \times \sqrt{5}}}$

⇒$\frac{5 a}{b}=\sqrt{5}$

Since, 5a is an integer and b is also an integer
So, $\frac{5 a}{b}$ is a rational number

⇒√5 is a rational number
But this contradicts to the fact that √5 is an irrational number.

Therefore, our assumption is wrong.
Hence, $\frac{1}{\sqrt{5}}$ is an irrational number.

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Question 4 A 

Prove that following numbers are not rational :

(6)^1/3
Sol :
Suppose 6^1/3 is rational.
Then, 6^1/3 = $\frac{n}{m}$ for some integers n and m which are co-prime.
So, 6 = $\frac{n^{3}}{m^{3}}$
⇒ 6m³ = n³
So, n³ must be divisible by 6
⇒ n must be divisible by 6.
Let n = 6p for some integer p
This gives
$6=\frac{(6 p)^{3}}{m^{3}}$
$1=\frac{6^{2} p^{3}}{m^{3}}$
⇒m³ is divisible by 6
Hence, m must be divisible by 6.
But n and m where co-prime.
So, we have a contradiction.
Hence, (6)^1/3 is irrational

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Question 4 B 

Prove that following numbers are not rational :

3√3
Sol :
Let us assume that 3√3 be a rational number.
Then, it will be of the form $\frac{a}{b}$ where a and b are co-prime and b≠0.
Now, $\frac{a}{b}=3 \sqrt{3}$
⇒$\frac{a}{3 b}=\sqrt{3}$
Since, a is an integer and 3b is also an integer (3b ≠ 0)
So, $\frac{a}{3 b}$ is a rational number
⇒√3 is a rational number
But this contradicts to the fact that √3 is an irrational number.
Therefore, our assumption is wrong.
Hence, 3√3 is an irrational number.

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Question 4 C 

Prove that following numbers are not rational :

5√3
Sol :
Let us assume that 5√3 be a rational number.
Then, it will be of the form $\frac{a}{b}$ where a and b are co-prime and b≠0.
Now, $\frac{a}{b}=5 \sqrt{3}$
⇒$\frac{a}{5 b}=\sqrt{3}$
Since, a is an integer and 3b is also an integer (5b ≠ 0)
So, $\frac{a}{5 b}$ is a rational number
⇒√3 is a rational number
But this contradicts to the fact that √3 is an irrational number.
Therefore, our assumption is wrong.
Hence, 5√3 is an irrational number.

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Question 5 A 

Prove that following numbers are irrational :

6 + √2
Sol :
Let us assume 6+√2 is rational
⇒6+√2 can be written in the form $\frac{a}{b}$ where a and b are co-prime.
Hence, $6+\sqrt{2}=\frac{a}{b}$
$\sqrt{2}=\frac{a}{b}-6$
$\sqrt{2}=\frac{a-6 b}{b}$
$\sqrt{2}=\left(\frac{a-6 b}{b}\right)$

$\begin{array}{ll}\sqrt{2}&=\left(\frac{a-6 b}{b}\right)\\\text{Irrational}&\text{Rational}\end{array}$

Since, rational ≠ irrational
This is a contradiction.
∴, Our assumption is incorrect.
Hence, 6+√2 is irrational.

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Question 5 B 

Prove that following numbers are irrational :

5 – √3
Sol :
Let us assume 5-√3 is rational
⇒5-√3 can be written in the form $\frac{a}{b}$ where a and b are co-prime.
Hence, $5-\sqrt{3}=\frac{a}{b}$
$-\sqrt{3}$ = $\frac{a}{b}$ – 5
-$-\sqrt{3}$ = $\frac{a-5 b}{b}$
$\sqrt{3}$ =- $\left(\frac{a-5 b}{b}\right)$

$\begin{array}{ll}\sqrt{3}&=\left(\frac{5 b-a}{b}\right)\\\text{Irrational}&\text{Rational}\end{array}$

Since, rational ≠ irrational
This is a contradiction.
$\therefore$, Our assumption is incorrect.
Hence, 5-√3 is irrational.

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Question 5 C 

Prove that following numbers are irrational :

2 + √2
Sol :
Let us assume 2 +√2 is rational
⇒2+√2 can be written in the form $\frac{a}{b}$ where a and b are co-prime.
Hence, 2 + $\sqrt{2}=\frac{a}{b}$
$\sqrt{2}=\frac{a}{b}-2$
$\sqrt{2}=\frac{a-2 b}{b}$
$\sqrt{2}=\left(\frac{a-2 b}{b}\right)$

$\begin{array}{ll}\sqrt{2}&=\left(\frac{a-2 b}{b}\right)\\\text{Irrational}&\text{Rational}\end{array}$

Since, rational ≠ irrational
This is a contradiction.
∴, Our assumption is incorrect.
Hence, 2+√2 is irrational.

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Question 5 D 

Prove that following numbers are irrational :

3 + √5
Sol :
Let us assume 3+√5 is rational
⇒3+√5 can be written in the form $\frac{a}{b}$ where a and b are co-prime.
Hence, 3 + $\sqrt{5}=\frac{a}{b}$
$\sqrt{5}=\frac{a}{b}-3$
$\sqrt{5}=\frac{a-3 b}{b}$
$\sqrt{5}=\left(\frac{a-3 b}{b}\right)$

$\begin{array}{ll}\sqrt{5}&=\left(\frac{a-3 b}{b}\right)\\\text{Irrational}&\text{Rational}\end{array}$

Since, rational ≠ irrational
This is a contradiction.
∴, Our assumption is incorrect.
Hence, 3 + $\sqrt{5}$ is irrational.

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Question 5 E 

Prove that following numbers are irrational :

√3 – √2
Sol :
Let us assume √3-√2 is rational

Let, $\sqrt{3}-\ \sqrt{2}=\frac{a}{b}$
Squaring both sides, we get
$(\sqrt{3}-\sqrt{2}) 2=\frac{a^{2}}{b^{2}}$
$5-2 \sqrt{6}=\frac{a^{2}}{b^{2}}$
$2{\sqrt{6}}=\frac{a^{2}}{b^{2}}-5$
$2 \sqrt{6}=\frac{a^{2}-5 b^{2}}{b^{2}}$
$\sqrt{6}=\left(\frac{a^{2}-5 b^{2}}{2 b^{2}}\right)$

$\begin{array}{ll}\sqrt{6}&=\left(\frac{a^{2}-5 b^{2}}{2 b^{2}}\right)\\\text{Irrational}&\text{Rational}\end{array}$

Since, rational ≠ irrational
This is a contradiction.

∴, Our assumption is incorrect.
Hence,√3-√2 is irrational.

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Question 5 F 

Prove that following numbers are irrational :

√7 – √5
Sol :
Let us assume √7-√5 is rational
Let, $\sqrt{7}-\sqrt{5}=\frac{a}{b}$
Squaring both sides, we get
$(\sqrt{7}-\sqrt{5}) 2=\frac{a^{2}}{b^{2}}$
$12-2 \sqrt{35}=\frac{a^{2}}{b^{2}}$
$2{\sqrt{35}}=\frac{a^{2}}{b^{2}}-12$
$2 \sqrt{35}=\left(\frac{a^{2}-12 b^{2}}{b^{2}}\right)$
$\sqrt{35}=\left(\frac{a^{2}-12 b^{2}}{2 b^{2}}\right)$

$\begin{array}{ll}\sqrt{35}&=\left(\frac{a^{2}-12 b^{2}}{2 b^{2}}\right)\\\text{Irrational}&\text{Rational}\end{array}$

Since, rational ≠ irrational
This is a contradiction.
∴, Our assumption is incorrect.
Hence, √7-√5 is irrational.

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S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1