NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3

The Core Concept: Theorem 1.3 & Proof by Contradiction

To prove that a number is irrational, we rely on a fundamental mathematical theorem. Theorem 1.3 states: Let $p$ be a prime number. If $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer. For example, if the prime number $3$ divides $9^2$ ($81$), it also divides $9$. This simple rule is the key to unlocking proofs of irrationality.

The strategy we use is called the Method of Contradiction. Instead of proving a number is irrational directly, we assume the opposite: that the number is rational. A rational number can be written in the simplest fraction form $a/b$, where $a$ and $b$ are integers and coprime (meaning they share no common factors other than 1). If we can logically demonstrate that $a$ and $b$ actually do share a common factor, our starting assumption collapses. This forces us to conclude that the number cannot be rational, meaning it must be irrational.

The Step-by-Step Template for Proving Square Roots are Irrational

1. Step 1: State the Opposite Assumption — Assume to the contrary that the given square root (e.g., $\sqrt{5}$) is a rational number. This means we can write $\sqrt{5} = a/b$, where $a$ and $b$ are coprime integers and $b \neq 0$.
2. Step 2: Square and Rearrange — Square both sides of the equation to eliminate the radical sign: $5 = a^2 / b^2$. Multiply both sides by $b^2$ to get $5b^2 = a^2$.
3. Step 3: Apply Theorem 1.3 — Since $5b^2 = a^2$, it means $5$ divides $a^2$. By Theorem 1.3, since $5$ is a prime dividing $a^2$, $5$ must also divide $a$. Therefore, we can write $a = 5c$ for some integer $c$.
4. Step 4: Substitute and Find the Factor for b — Substitute $a = 5c$ back into our original squared equation: $5b^2 = (5c)^2 \implies 5b^2 = 25c^2$. Simplify this by dividing by 5 to get $b^2 = 5c^2$. This means $5$ divides $b^2$, and consequently, $5$ divides $b$.
5. Step 5: Declare the Contradiction — Now, we have shown that $5$ divides both $a$ and $b$. This means $a$ and $b$ share a common factor of $5$, contradicting our initial assumption that $a$ and $b$ are coprime. Thus, our assumption is wrong, proving that $\sqrt{5}$ is irrational.

Worked Examples: Proving Composite Irrational Numbers

• Example 1: Prove that $3 + 2\sqrt{5}$ is irrational, given that $\sqrt{5}$ is irrational. Step 1: Assume $3 + 2\sqrt{5}$ is rational. Write it as $3 + 2\sqrt{5} = a/b$, where $a$ and $b$ are coprime integers and $b \neq 0$. Step 2: Isolate the radical term on one side. $2\sqrt{5} = (a/b) - 3$ $2\sqrt{5} = (a - 3b)/b$ $\sqrt{5} = (a - 3b)/(2b)$ Step 3: Analyze both sides of the equation. Since $a$ and $b$ are integers, $(a - 3b)/(2b)$ is a rational number. However, we are given that $\sqrt{5}$ is irrational. Step 4: State the contradiction. An irrational number cannot equal a rational number. Therefore, our assumption was incorrect, and $3 + 2\sqrt{5}$ is irrational.
• Example 2: Prove that $7\sqrt{5}$ is irrational. Step 1: Assume $7\sqrt{5}$ is rational. Let $7\sqrt{5} = a/b$, where $a$ and $b$ are coprime integers and $b \neq 0$. Step 2: Rearrange to isolate the radical. $\sqrt{5} = a/(7b)$ Step 3: Since $a$, $b$, and $7$ are integers, $a/(7b)$ is rational. This implies that $\sqrt{5}$ is rational, which contradicts the known fact that $\sqrt{5}$ is irrational. Hence, $7\sqrt{5}$ is irrational.

CBSE Board Exam Tips & Common Student Pitfalls

• Never omit the word 'Coprime': When assuming $a/b$, you must state that $a$ and $b$ are coprime. If you miss this term, you will lose marks because the whole contradiction depends on it.
• Write down the theorems explicitly: Don't just say 'so 5 divides a'. Write 'Since 5 divides $a^2$, by Theorem 1.3, 5 also divides $a$.' Explaining the logic earns full process points.
• Do not reprove standard roots in composite questions: In questions like proving $6 + \sqrt{2}$ is irrational, you do not need to prove $\sqrt{2}$ is irrational from scratch unless the question asks you to do so in a separate sub-question.

Practice Questions with Solutions

• Q: Prove that $\sqrt{3}$ is irrational. A: Step 1: Assume to the contrary that $\sqrt{3}$ is rational. Therefore, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{3} = a/b$. Step 2: Squaring both sides, we get $3 = a^2 / b^2 \implies 3b^2 = a^2$. This means $3$ divides $a^2$, so by Theorem 1.3, $3$ divides $a$. Step 3: Write $a = 3c$ for some integer $c$. Substituting this in $3b^2 = a^2$ gives $3b^2 = (3c)^2 \implies 3b^2 = 9c^2 \implies b^2 = 3c^2$. This means $3$ divides $b^2$, so $3$ divides $b$. Step 4: Therefore, $a$ and $b$ have at least $3$ as a common factor. But this contradicts our assumption that $a$ and $b$ are coprime. Final answer: Hence, $\sqrt{3}$ is irrational.
• Q: Prove that $6 + \sqrt{2}$ is irrational. A: Step 1: Assume to the contrary that $6 + \sqrt{2}$ is rational. Let $6 + \sqrt{2} = a/b$, where $a$ and $b$ are coprime integers ($b \neq 0$). Step 2: Rearranging the equation to isolate $\sqrt{2}$: $\sqrt{2} = a/b - 6 \implies \sqrt{2} = (a - 6b)/b$. Step 3: Since $a$ and $b$ are integers, $(a - 6b)/b$ is rational. This implies $\sqrt{2}$ is rational. Step 4: But this contradicts the fact that $\sqrt{2}$ is irrational. Final answer: Therefore, our assumption is false, and $6 + \sqrt{2}$ is irrational.
• Q: Prove that $1/\sqrt{2}$ is irrational. A: Step 1: Assume $1/\sqrt{2}$ is rational. Therefore, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $1/\sqrt{2} = a/b$. Step 2: Taking the reciprocal on both sides gives $\sqrt{2} = b/a$. Step 3: Since $a$ and $b$ are non-zero integers, $b/a$ is a rational number. This implies that $\sqrt{2}$ is rational. Step 4: This contradicts the well-established fact that $\sqrt{2}$ is irrational. Final answer: Hence, $1/\sqrt{2}$ is irrational.
• Q: Show that $3\sqrt{2}$ is irrational. A: Step 1: Assume $3\sqrt{2}$ is rational. Let $3\sqrt{2} = a/b$, where $a$ and $b$ are coprime integers and $b \neq 0$. Step 2: Isolate the square root: $\sqrt{2} = a/(3b)$. Step 3: Since $a$, $b$, and $3$ are integers, $a/(3b)$ is rational. This means $\sqrt{2}$ must be rational. Step 4: But this contradicts the fact that $\sqrt{2}$ is irrational. Final answer: Consequently, our assumption is incorrect, and $3\sqrt{2}$ is irrational.

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