Irrational Numbers PYQs
Overview
This page provides comprehensive Class 10 Maths Irrational Numbers PYQs | Real Numbers. Irrational Numbers previous year questions for Class 10 Maths Real Numbers. Practice CBSE board PYQs with step-by-step solutions on SJMaths.
Practice Previous Year Questions Topic-wise
Q2
2025
00:00
$\sqrt{0.4}$ is a/an:
$\sqrt{0.4} = \sqrt{\frac{4}{10}} = \sqrt{\frac{2}{5}} = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{10}}{5}$
Since $\sqrt{10}$ is an irrational number, $\frac{\sqrt{10}}{5}$ is also irrational.
Final Answer: (D) irrational
Q17
2024
00:00
Prove that $\sqrt{5}$ is an irrational number.
Let us assume, to the contrary, that $\sqrt{5}$ is rational.
Then, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{5} = \frac{a}{b}$.
Squaring both sides, we get: $5 = \frac{a^2}{b^2} \implies a^2 = 5b^2$.
This means $a^2$ is divisible by 5. By theorem, $a$ is also divisible by 5.
Let $a = 5c$ for some integer $c$.
Substituting in $a^2 = 5b^2$: $(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2$.
This means $b^2$ is divisible by 5, so $b$ is also divisible by 5.
Therefore, both $a$ and $b$ have 5 as a common factor.
This contradicts our assumption that $a$ and $b$ are co-prime.
This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.
Hence, $\sqrt{5}$ is irrational.
Final Answer: Proved.
Q18
2023
00:00
Prove that $\sqrt{2}$ is irrational.
Let us assume, to the contrary, that $\sqrt{2}$ is rational.
Then, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides: $2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$.
This means $a^2$ is divisible by 2, so $a$ is divisible by 2.
Let $a = 2c$ for some integer $c$.
Substituting: $(2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies b^2 = 2c^2$.
This means $b^2$ is divisible by 2, so $b$ is also divisible by 2.
Both $a$ and $b$ have 2 as a common factor, contradicting that they are co-prime.
Therefore, $\sqrt{2}$ is irrational.
Final Answer: Proved.
Q20
2021
00:00
Show that $(5 - \sqrt{3})$ is irrational, given that $\sqrt{3}$ is irrational.
Let us assume that $(5 - \sqrt{3})$ is rational.
Then, $(5 - \sqrt{3}) = \frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.
Rearranging the terms: $\sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b}$.
Since $a$ and $b$ are integers, $\frac{5b - a}{b}$ is a rational number.
This implies that $\sqrt{3}$ is a rational number.
But this contradicts the given fact that $\sqrt{3}$ is irrational.
This contradiction arose from our incorrect assumption.
Hence, $(5 - \sqrt{3})$ is irrational.
Final Answer: Proved.
Q24
2018
00:00
Prove that $\sqrt{2}$ is irrational.
Let $\sqrt{2} = \frac{a}{b}$ (where $a$ and $b$ are co-prime).
$a^2 = 2b^2 \implies a$ is even. Let $a=2c$.
$4c^2 = 2b^2 \implies b^2 = 2c^2 \implies b$ is even.
Contradiction: $a, b$ are not co-prime. Hence, $\sqrt{2}$ is irrational.
Final Answer: Proved.
Q26
2024
00:00
Prove that $(3 + 2\sqrt{5})$ is irrational, given $\sqrt{5}$ is irrational.
Let $(3 + 2\sqrt{5})$ be a rational number $\frac{a}{b}$ where $a, b$ are integers, $b \neq 0$.
$3 + 2\sqrt{5} = \frac{a}{b}$
$2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$
$\sqrt{5} = \frac{a - 3b}{2b}$
Since $a, b$ are integers, $\frac{a - 3b}{2b}$ is rational.
This implies $\sqrt{5}$ is rational, contradicting the given fact.
Hence, $(3 + 2\sqrt{5})$ is irrational.
Final Answer: Proved.
Q27
2023
00:00
Prove that $(2 + 3\sqrt{5})$ is irrational.
Assume $(2 + 3\sqrt{5})$ is rational, say $\frac{a}{b}$.
$3\sqrt{5} = \frac{a}{b} - 2 = \frac{a - 2b}{b}$
$\sqrt{5} = \frac{a - 2b}{3b}$
This implies $\sqrt{5}$ is rational (as LHS is rational), which is a contradiction.
Therefore, $(2 + 3\sqrt{5})$ is irrational.
Final Answer: Proved.
Q29
2020
00:00
Prove $\sqrt{5}$ is irrational. Hence show $(3 + \sqrt{5})$ is also irrational.
Part 1: Prove $\sqrt{5}$ is irrational (Standard proof: let $\sqrt{5}=a/b$, $a^2=5b^2$, etc. Contradicts co-prime nature).
Part 2: Assume $(3 + \sqrt{5})$ is rational, let it be $r$.
Then $\sqrt{5} = r - 3$.
Since $r$ is rational and 3 is rational, $r-3$ must be rational.
This implies $\sqrt{5}$ is rational, which contradicts Part 1.
Thus, $(3 + \sqrt{5})$ is irrational.
Final Answer: Proved.
Q31
2017
00:00
Show that $\sqrt{2}$ is irrational.
Let $\sqrt{2} = \frac{a}{b}$ (where $a$ and $b$ are co-prime integers, $b \neq 0$).
Squaring both sides: $2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$.
This means $a^2$ is divisible by 2, so $a$ is divisible by 2. Let $a = 2c$.
Substituting: $(2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies b^2 = 2c^2$.
This means $b^2$ is divisible by 2, so $b$ is also divisible by 2.
Both $a$ and $b$ have 2 as a common factor, contradicting that they are co-prime.
Therefore, $\sqrt{2}$ is irrational.
Final Answer: Proved.
$\sqrt{0.4}$ is a/an:
$\sqrt{0.4} = \sqrt{\frac{4}{10}} = \sqrt{\frac{2}{5}} = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{10}}{5}$
Since $\sqrt{10}$ is an irrational number, $\frac{\sqrt{10}}{5}$ is also irrational.
Final Answer: (D) irrational
Q17
2024
00:00
Prove that $\sqrt{5}$ is an irrational number.
Let us assume, to the contrary, that $\sqrt{5}$ is rational.
Then, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{5} = \frac{a}{b}$.
Squaring both sides, we get: $5 = \frac{a^2}{b^2} \implies a^2 = 5b^2$.
This means $a^2$ is divisible by 5. By theorem, $a$ is also divisible by 5.
Let $a = 5c$ for some integer $c$.
Substituting in $a^2 = 5b^2$: $(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2$.
This means $b^2$ is divisible by 5, so $b$ is also divisible by 5.
Therefore, both $a$ and $b$ have 5 as a common factor.
This contradicts our assumption that $a$ and $b$ are co-prime.
This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.
Hence, $\sqrt{5}$ is irrational.
Final Answer: Proved.
Q18
2023
00:00
Prove that $\sqrt{2}$ is irrational.
Let us assume, to the contrary, that $\sqrt{2}$ is rational.
Then, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides: $2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$.
This means $a^2$ is divisible by 2, so $a$ is divisible by 2.
Let $a = 2c$ for some integer $c$.
Substituting: $(2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies b^2 = 2c^2$.
This means $b^2$ is divisible by 2, so $b$ is also divisible by 2.
Both $a$ and $b$ have 2 as a common factor, contradicting that they are co-prime.
Therefore, $\sqrt{2}$ is irrational.
Final Answer: Proved.
Q20
2021
00:00
Show that $(5 - \sqrt{3})$ is irrational, given that $\sqrt{3}$ is irrational.
Let us assume that $(5 - \sqrt{3})$ is rational.
Then, $(5 - \sqrt{3}) = \frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.
Rearranging the terms: $\sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b}$.
Since $a$ and $b$ are integers, $\frac{5b - a}{b}$ is a rational number.
This implies that $\sqrt{3}$ is a rational number.
But this contradicts the given fact that $\sqrt{3}$ is irrational.
This contradiction arose from our incorrect assumption.
Hence, $(5 - \sqrt{3})$ is irrational.
Final Answer: Proved.
Q24
2018
00:00
Prove that $\sqrt{2}$ is irrational.
Let $\sqrt{2} = \frac{a}{b}$ (where $a$ and $b$ are co-prime).
$a^2 = 2b^2 \implies a$ is even. Let $a=2c$.
$4c^2 = 2b^2 \implies b^2 = 2c^2 \implies b$ is even.
Contradiction: $a, b$ are not co-prime. Hence, $\sqrt{2}$ is irrational.
Final Answer: Proved.
Q26
2024
00:00
Prove that $(3 + 2\sqrt{5})$ is irrational, given $\sqrt{5}$ is irrational.
Let $(3 + 2\sqrt{5})$ be a rational number $\frac{a}{b}$ where $a, b$ are integers, $b \neq 0$.
$3 + 2\sqrt{5} = \frac{a}{b}$
$2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$
$\sqrt{5} = \frac{a - 3b}{2b}$
Since $a, b$ are integers, $\frac{a - 3b}{2b}$ is rational.
This implies $\sqrt{5}$ is rational, contradicting the given fact.
Hence, $(3 + 2\sqrt{5})$ is irrational.
Final Answer: Proved.
Q27
2023
00:00
Prove that $(2 + 3\sqrt{5})$ is irrational.
Assume $(2 + 3\sqrt{5})$ is rational, say $\frac{a}{b}$.
$3\sqrt{5} = \frac{a}{b} - 2 = \frac{a - 2b}{b}$
$\sqrt{5} = \frac{a - 2b}{3b}$
This implies $\sqrt{5}$ is rational (as LHS is rational), which is a contradiction.
Therefore, $(2 + 3\sqrt{5})$ is irrational.
Final Answer: Proved.
Q29
2020
00:00
Prove $\sqrt{5}$ is irrational. Hence show $(3 + \sqrt{5})$ is also irrational.
Part 1: Prove $\sqrt{5}$ is irrational (Standard proof: let $\sqrt{5}=a/b$, $a^2=5b^2$, etc. Contradicts co-prime nature).
Part 2: Assume $(3 + \sqrt{5})$ is rational, let it be $r$.
Then $\sqrt{5} = r - 3$.
Since $r$ is rational and 3 is rational, $r-3$ must be rational.
This implies $\sqrt{5}$ is rational, which contradicts Part 1.
Thus, $(3 + \sqrt{5})$ is irrational.
Final Answer: Proved.
Q31
2017
00:00
Show that $\sqrt{2}$ is irrational.
Let $\sqrt{2} = \frac{a}{b}$ (where $a$ and $b$ are co-prime integers, $b \neq 0$).
Squaring both sides: $2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$.
This means $a^2$ is divisible by 2, so $a$ is divisible by 2. Let $a = 2c$.
Substituting: $(2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies b^2 = 2c^2$.
This means $b^2$ is divisible by 2, so $b$ is also divisible by 2.
Both $a$ and $b$ have 2 as a common factor, contradicting that they are co-prime.
Therefore, $\sqrt{2}$ is irrational.
Final Answer: Proved.