CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
Understanding the decimal representation of real numbers becomes essential for mathematics because real numbers contain all rational and irrational elements. The behavior of these numbers in decimal format is addressed in this exercise. We use this method to identify non-terminating decimals while understanding how rational numbers form repeating patterns. Learning these concepts establishes our ability to correctly identify numbers while linking theoretical number concepts to their actual decimal expressions.
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Students obtain maximum benefit from these concepts when they consult with NCERT Solutions. The NCERT Books related solutions provide step-by-step explanations to simplify difficult proofs. Students who refer to these educational resources can develop fundamental number theory knowledge that helps them succeed in advanced mathematics studies as well as competitive exams.
Q1 Prove that $\sqrt 5$ is irrational.
Answer:
Let us assume $\sqrt{5}$ is rational, which means it can be written in the form $\frac{p}{q}$ where p and q are co-primes and $q\neq 0$
$\\\sqrt{5}=\frac{p}{q}$
Squaring both sides, we obtain
$\\\left ( \sqrt{5} \right )^{2}=\left (\frac{p}{q} \right )^{2}\\$
$5=\frac{p^{2}}{q^{2}}\\$
$p^{2}=5q^{2}$
From the above equation, we can see that p2 is divisible by 5, therefore, p will also be divisible by 5, as 5 is a prime number. $(i)$
Therefore, p can be written as 5r
p = 5r
p2 = (5r)2
5q2 = 25r2
q2 = 5r2
From the above equation, we can see that q2 is divisible by 5, Therefore, q will also be divisible by 5 as 5 is a prime number. $(ii)$
From (i) and (ii), we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that $\sqrt{5}$ is rational was wrong. Hence proved that $\sqrt{5}$ is irrational.
Q2 Prove that $3 + 2 \sqrt 5$ is irrational.
Answer:
Let us assume $3 + 2 \sqrt 5$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.
$\\3+2\sqrt{5}=\frac{p}{q}$
$2\sqrt{5}=\frac{p}{q}-3$
$\sqrt{5}=\frac{p-3q}{2q}$
As p and q are integers $\frac{p-3q}{2q}\\$ would be rational, which contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $3 + 2 \sqrt 5$ is rational was wrong. Therefore $3 + 2 \sqrt 5$ is irrational.
Q3 Prove that the following are irrationals :
(i) $\frac{1}{\sqrt 2}$
Answer:
Let us assume $\frac{1}{\sqrt{2}}$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.
$\frac{1}{\sqrt{2}}=\frac{p}{q}$
$\sqrt{2}=\frac{q}{p}$
Since p and q are co-prime integers $\frac{q}{p}$ will be rational, which contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $\frac{1}{\sqrt{2}}$ is rational was wrong. Therefore $\frac{1}{\sqrt{2}}$ is irrational.
Q3 (2) Prove that the following are irrationals :
(ii) $7 \sqrt 5$
Answer:
Let us assume $7 \sqrt 5$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.
$7\sqrt{5}=\frac{p}{q}$
$\sqrt{5}=\frac{p}{7q}$
As p and q are integers $\frac{p}{7q}\\$ would be rational, which contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $7 \sqrt 5$ is rational was wrong. Therefore $7 \sqrt 5$ is irrational.
Q3 (3) Prove that the following are irrationals : $6 + \sqrt 2$
Answer:
Let us assume $6 + \sqrt 2$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.
$6+\sqrt{2}=\frac{p}{q}$
$\sqrt{2}=\frac{p}{q}-6$
$\sqrt{2}=\frac{p-6q}{q}$
As p and q are integers $\frac{p-6q}{q}$ would be rational, which contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $6 + \sqrt 2$ is rational was wrong. Therefore $6 + \sqrt 2$ is irrational.
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Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.
Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.
Frequently Asked Questions (FAQs)
Sum and difference of a rational and irrational number is irrational.
“Fundamental theorem of Arithmetic” given in the Class 10 Maths chapter 1 states that “Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur”.
The proof is based on a most common technique called ‘proof by contradiction.
important in board exams, you can check previous year papers for better understanding.
On Question asked by student community
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