RBSE Solutions Class 10 Maths Chapter 2 Real Numbers Exercise 2.3

Question 1. Prove that \( 5 - \sqrt{3} \) is irrational.
Answer: Let us assume, for the sake of contradiction, that \( 5 - \sqrt{3} \) is a rational number. If it is rational, we can write it in the form \( \frac{a}{b} \), where 'a' and 'b' are coprime integers and \( b \neq 0 \).
So, \( 5 - \sqrt{3} = \frac{a}{b} \)
Now, we rearrange the equation to isolate \( \sqrt{3} \):
\( 5 - \frac{a}{b} = \sqrt{3} \)
\( \frac{5b - a}{b} = \sqrt{3} \)
Since 'a', 'b', and '5' are all integers, the expression \( \frac{5b - a}{b} \) is a rational number. This means that \( \sqrt{3} \) must also be a rational number.
However, we already know that \( \sqrt{3} \) is an irrational number. This creates a contradiction: a rational number cannot be equal to an irrational number. Therefore, our initial assumption that \( 5 - \sqrt{3} \) is rational must be false. This type of proof is called proof by contradiction.
In simple words: We first pretend that \( 5 - \sqrt{3} \) is a simple fraction. But when we move things around, it shows that \( \sqrt{3} \) would also have to be a simple fraction. We know \( \sqrt{3} \) is not a simple fraction, so our first pretend idea must be wrong. This means \( 5 - \sqrt{3} \) cannot be a simple fraction, so it is irrational.

🎯 Exam Tip: Always start by assuming the opposite of what you need to prove. The contradiction at the end will then prove your original statement.

Question 2. Prove that following numbers are irrational:
(i) \( \frac {1}{\sqrt{2}} \)
(ii) \( 6 + \sqrt{2} \)
(iii) \( 3\sqrt{2} \)
Answer:
(i) To prove \( \frac{1}{\sqrt{2}} \) is irrational:
Let's assume that \( \frac{1}{\sqrt{2}} \) is a rational number. This means we can write it as \( \frac{a}{b} \), where 'a' and 'b' are coprime integers and \( b \neq 0 \).
\( \frac{1}{\sqrt{2}} = \frac{a}{b} \)
To make it easier, we can flip both sides:
\( \sqrt{2} = \frac{b}{a} \)
Since 'a' and 'b' are integers, \( \frac{b}{a} \) is a rational number. This would mean that \( \sqrt{2} \) is also a rational number.
However, we know that \( \sqrt{2} \) is an irrational number. This is a contradiction, as an irrational number cannot be equal to a rational number. Therefore, our initial assumption that \( \frac{1}{\sqrt{2}} \) is rational is incorrect, proving that \( \frac{1}{\sqrt{2}} \) is an irrational number.

(ii) To prove \( 6 + \sqrt{2} \) is irrational:
Let's assume that \( 6 + \sqrt{2} \) is a rational number. So, we can write it as \( \frac{a}{b} \), where 'a' and 'b' are coprime integers and \( b \neq 0 \).
\( 6 + \sqrt{2} = \frac{a}{b} \)
Now, we rearrange the equation to isolate \( \sqrt{2} \):
\( \sqrt{2} = \frac{a}{b} - 6 \)
\( \sqrt{2} = \frac{a - 6b}{b} \)
Since 'a', 'b', and '6' are all integers, the expression \( \frac{a - 6b}{b} \) is a rational number. This means that \( \sqrt{2} \) must also be a rational number. A rational number is any number that can be expressed as a fraction.
But we know that \( \sqrt{2} \) is an irrational number. This contradiction arises from our wrong assumption. Thus, \( 6 + \sqrt{2} \) is an irrational number.

(iii) To prove \( 3\sqrt{2} \) is irrational:
Let's assume that \( 3\sqrt{2} \) is a rational number. So, we can write it as \( \frac{a}{b} \), where 'a' and 'b' are coprime integers and \( b \neq 0 \).
\( 3\sqrt{2} = \frac{a}{b} \)
Now, we rearrange the equation to isolate \( \sqrt{2} \):
\( \sqrt{2} = \frac{a}{3b} \)
Since 'a', 'b', and '3' are all integers, the expression \( \frac{a}{3b} \) is a rational number. This implies that \( \sqrt{2} \) must also be a rational number.
However, we know that \( \sqrt{2} \) is an irrational number. This contradiction proves that our initial assumption was false. Therefore, \( 3\sqrt{2} \) is an irrational number.
In simple words: For all parts, we first pretend the number is a simple fraction. Then we move the numbers around so that only \( \sqrt{2} \) is left on one side. If the other side is a simple fraction, it means \( \sqrt{2} \) must also be a simple fraction. But we know \( \sqrt{2} \) is not, so our pretend idea was wrong, proving the original number is irrational.

🎯 Exam Tip: When proving a sum, difference, product, or quotient involving a known irrational number (like \( \sqrt{2} \) or \( \sqrt{3} \)) is irrational, always aim to isolate the known irrational term. The resulting contradiction will be clear.

Question 3. If p and q are positive prime numbers, then prove that \( \sqrt{p} + \sqrt{q} \) is irrational.
Answer: Let us assume, for contradiction, that \( \sqrt{p} + \sqrt{q} \) is a rational number. This means we can write it as \( \frac{a}{b} \), where 'a' and 'b' are coprime integers and \( b \neq 0 \).
\( \sqrt{p} + \sqrt{q} = \frac{a}{b} \)
Now, we isolate one of the square root terms, for example, \( \sqrt{p} \):
\( \sqrt{p} = \frac{a}{b} - \sqrt{q} \)
Next, we square both sides of the equation:
\( (\sqrt{p})^2 = \left(\frac{a}{b} - \sqrt{q}\right)^2 \)
\( p = \left(\frac{a}{b}\right)^2 - 2 \cdot \frac{a}{b} \cdot \sqrt{q} + (\sqrt{q})^2 \)
\( p = \frac{a^2}{b^2} - \frac{2a}{b}\sqrt{q} + q \)
Now, we rearrange the terms to isolate \( \sqrt{q} \):
\( p - q - \frac{a^2}{b^2} = - \frac{2a}{b}\sqrt{q} \)
Multiply both sides by -1 to make the term positive:
\( \frac{a^2}{b^2} + q - p = \frac{2a}{b}\sqrt{q} \)
Combine the terms on the left side with a common denominator:
\( \frac{a^2 + qb^2 - pb^2}{b^2} = \frac{2a}{b}\sqrt{q} \)
\( \frac{a^2 + (q-p)b^2}{b^2} = \frac{2a}{b}\sqrt{q} \)
Finally, solve for \( \sqrt{q} \):
\( \sqrt{q} = \frac{a^2 + (q-p)b^2}{b^2} \times \frac{b}{2a} \)
\( \sqrt{q} = \frac{a^2 + (q-p)b^2}{2ab} \)
Since 'a' and 'b' are integers, and 'p' and 'q' are prime numbers (which are also integers), the expression \( \frac{a^2 + (q-p)b^2}{2ab} \) is a rational number. This implies that \( \sqrt{q} \) must also be a rational number.
However, we know that the square root of a prime number (like q) is always an irrational number. This is a contradiction, as a rational number cannot be equal to an irrational number. Therefore, our initial assumption that \( \sqrt{p} + \sqrt{q} \) is rational must be false, proving that \( \sqrt{p} + \sqrt{q} \) is an irrational number. The method of squaring both sides is a common technique in these proofs.
In simple words: We pretend that \( \sqrt{p} + \sqrt{q} \) is a simple fraction. We move \( \sqrt{q} \) to the other side and then square both sides to get rid of the square root sign for \( \sqrt{p} \). After doing more math, we find that \( \sqrt{q} \) ends up being equal to a simple fraction. But we know that the square root of a prime number is never a simple fraction. So, our first idea was wrong, and \( \sqrt{p} + \sqrt{q} \) is irrational.

🎯 Exam Tip: For sums of square roots, isolate one root, square both sides, then isolate the other root to demonstrate the contradiction. Remember that the square root of a prime number is always irrational.