| Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
Exercise 1.3
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
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.
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.
Sol :
Suppose 61/3 is rational.
Then, 61/3 = $\frac{n}{m}$ 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
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.
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.
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, $6+\sqrt{2}=\frac{a}{b}$
$\sqrt{2}=\frac{a}{b}-6$
$\sqrt{2}=\frac{a-6 b}{b}$
$\sqrt{2}=\left(\frac{a-6 b}{b}\right)$
This is a contradiction.
∴, Our assumption is incorrect.
Hence, 6+√2 is irrational.
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
-$-\sqrt{3}$ = $\frac{a-5 b}{b}$
$\sqrt{3}$ =- $\left(\frac{a-5 b}{b}\right)$
Hence, $\frac{1}{\sqrt{5}}$ is an irrational number.
Question 4 A
Prove that following numbers are not rational :
(6)1/3Sol :
Suppose 61/3 is rational.
Then, 61/3 = $\frac{n}{m}$ 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√3Sol :
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√3Sol :
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 + √2Sol :
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, $6+\sqrt{2}=\frac{a}{b}$
$\sqrt{2}=\frac{a}{b}-6$
$\sqrt{2}=\frac{a-6 b}{b}$
$\sqrt{2}=\left(\frac{a-6 b}{b}\right)$
$\begin{array}{ll}\sqrt{2}&=\left(\frac{a-6 b}{b}\right)\\\text{Irrational}&\text{Rational}\end{array}$
This is a contradiction.
∴, Our assumption is incorrect.
Hence, 6+√2 is irrational.
Question 5 B
Prove that following numbers are irrational :
5 – √3Sol :
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.
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 + $\sqrt{2}=\frac{a}{b}$
$\sqrt{2}=\frac{a}{b}-2$
$\sqrt{2}=\frac{a-2 b}{b}$
$\sqrt{2}=\left(\frac{a-2 b}{b}\right)$
This is a contradiction.
$\therefore$, Our assumption is incorrect.
Hence, 5-√3 is irrational.
Question 5 C
Prove that following numbers are irrational :
2 + √2Sol :
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.
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 + $\sqrt{5}=\frac{a}{b}$
$\sqrt{5}=\frac{a}{b}-3$
$\sqrt{5}=\frac{a-3 b}{b}$
$\sqrt{5}=\left(\frac{a-3 b}{b}\right)$
This is a contradiction.
∴, Our assumption is incorrect.
Hence, 2+√2 is irrational.
Question 5 D
Prove that following numbers are irrational :
3 + √5Sol :
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.
Sol :
Let us assume √3-√2 is rational
This is a contradiction.
∴, Our assumption is incorrect.
Hence, 3 + $\sqrt{5}$ is irrational.
Question 5 E
Prove that following numbers are irrational :
√3 – √2Sol :
Let us assume √3-√2 is rational
Let, $\sqrt{3}-\ \sqrt{2}=\frac{a}{b}$
Squaring both sides, we get
$(\sqrt{3}-\sqrt{2}) 2=\frac{a^{2}}{b^{2}}$
$5-2 \sqrt{6}=\frac{a^{2}}{b^{2}}$
$2{\sqrt{6}}=\frac{a^{2}}{b^{2}}-5$
$2 \sqrt{6}=\frac{a^{2}-5 b^{2}}{b^{2}}$
$\sqrt{6}=\left(\frac{a^{2}-5 b^{2}}{2 b^{2}}\right)$
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.
This is a contradiction.
∴, Our assumption is incorrect.
Hence,√3-√2 is irrational.
Sol :
Let us assume √7-√5 is rational
Let, $\sqrt{7}-\sqrt{5}=\frac{a}{b}$
Squaring both sides, we get
$(\sqrt{7}-\sqrt{5}) 2=\frac{a^{2}}{b^{2}}$
$12-2 \sqrt{35}=\frac{a^{2}}{b^{2}}$
$2{\sqrt{35}}=\frac{a^{2}}{b^{2}}-12$
$2 \sqrt{35}=\left(\frac{a^{2}-12 b^{2}}{b^{2}}\right)$
$\sqrt{35}=\left(\frac{a^{2}-12 b^{2}}{2 b^{2}}\right)$
Hence,√3-√2 is irrational.
Question 5 F
Prove that following numbers are irrational :
√7 – √5Sol :
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.
This is a contradiction.
∴, Our assumption is incorrect.
Hence, √7-√5 is irrational.
| 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 |
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