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

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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.

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

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.