Mathematical Reasoning Miscellaneous Ex Class 11 NCERT

Deep Understanding of Logic & Proof Methods

In CBSE Class 11, the miscellaneous exercise brings together multiple logical domains. First, you must master the logic of conditional statements ($p \implies q$). The contrapositive of this conditional statement is written as $\sim q \implies \sim p$, which holds the exact same truth value as the original implication. On the other hand, the converse is $q \implies p$, which is not logically equivalent to the original. Second, negating compound statements requires precise application of De Morgan's Laws: negating an 'AND' statement converts it into an 'OR' statement with negated components, while negating an 'OR' statement converts it into an 'AND' statement. Third, when validating statements, you will learn to use the proof by contrapositive and proof by contradiction methods. A proof by contradiction starts by assuming the negation of your statement is true, and logically working your way to an absurd or contradictory mathematical fact.

Key Logic Definitions

Contrapositive

The statement 'If not $q$, then not $p

derived from the conditional statement 'If $p$, then $q

. It is logically equivalent to the original statement.

Converse

The statement 'If $q$, then $p

derived from the conditional statement 'If $p$, then $q

. It is not logically equivalent to the original statement.

Negation

The assertion of the opposite truth value of a given statement. If statement $p$ is true, its negation $\sim p$ is false.

Counterexample

A single specific case or example which disproves a universal mathematical assertion.

Step-by-Step Worked Solutions

• Example 1: Write the converse and contrapositive of the statement: 'If a number $n$ is divisible by 9, then it is divisible by 3.' Step 1: Identify the component statements. Let $p$: 'A number $n$ is divisible by 9.' Let $q$: 'A number $n$ is divisible by 3.' Step 2: Formulate the converse ($q \implies p$). 'If a number $n$ is divisible by 3, then it is divisible by 9.' (Note: This is mathematically false, e.g., for $n = 6$). Step 3: Formulate the contrapositive ($\sim q \implies \sim p$). 'If a number $n$ is not divisible by 3, then it is not divisible by 9.' (This is mathematically true).
• Example 2: Prove by the method of contradiction that $\sqrt{2}$ is irrational. Step 1: Assume the negation of the given statement is true. Let $\sqrt{2}$ be a rational number. Step 2: Express in the form of a fraction. $\sqrt{2} = a/b$, where $a$ and $b$ are co-prime integers ($b \neq 0$). Step 3: Square both sides and rearrange. $2 = a^2 / b^2 \implies a^2 = 2b^2$. This implies $a^2$ is even, which means $a$ is even (let $a = 2c$ for some integer $c$). Step 4: Substitute $a = 2c$ back into the equation. $(2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies b^2 = 2c^2$. This implies $b^2$ is even, which means $b$ is even. Step 5: Identify the contradiction. Since both $a$ and $b$ are even, they have a common factor of 2. This contradicts our assumption that $a$ and $b$ are co-prime. Final Conclusion: Our assumption was wrong. Therefore, $\sqrt{2}$ is irrational.

Common Board Exam Mistakes & Tips

1. Watch the Negation of 'AND'/'OR': When negating a statement like '$p$ and $q
   , students often write '$\sim p$ and $\sim q
   . The correct negation is '$\sim p$ or $\sim q
   (using De Morgan's Law).
2. Converse vs. Contrapositive: Remember that the converse ($q \implies p$) is NOT logically equivalent to the original statement, but the contrapositive ($\sim q \implies \sim p$) is ALWAYS logically equivalent. Do not confuse the two during your Class 11 exams!
3. Universal Quantifiers: To negate 'For all $x
   , change it to 'There exists at least one $x$ for which the condition does not hold'.

Practice Questions with Solutions

• Q: Write the negation of the statement: 'Every natural number is an integer and some real numbers are not complex.' A: Step 1: Identify the compound structure. The statement is of the form $P \land Q$. Step 2: Apply De Morgan's Law for negation: $\sim(P \land Q) \equiv \sim P \lor \sim Q$. Step 3: Negate component $P$ ('Every natural number is an integer'). The negation is 'There exists a natural number which is not an integer'. Step 4: Negate component $Q$ ('some real numbers are not complex'). The negation is 'All real numbers are complex'. Final answer: 'There exists a natural number which is not an integer or all real numbers are complex.'
• Q: Check the validity of the statement by the contrapositive method: 'If $n$ is an integer and $n^2$ is even, then $n$ is even.' A: Step 1: Identify $p$ and $q$. Here, $p$: '$n^2$ is even' and $q$: '$n$ is even'. Step 2: State the contrapositive ($\sim q \implies \sim p$). 'If $n$ is not even (odd), then $n^2$ is not even (odd)'. Step 3: Prove the contrapositive statement. Let $n$ be an odd integer. Then, $n = 2k + 1$ for some integer $k$. Step 4: Compute $n^2$. $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$, which is of the form $2m + 1$ (odd). Final answer: Since the contrapositive statement 'If $n$ is odd, then $n^2$ is odd' is true, the original statement is valid.
• Q: Write the converse and contrapositive of: 'If two triangles are congruent, then they are similar.' A: Step 1: Identify components. $p$: 'Two triangles are congruent.' $q$: 'They are similar.' Step 2: Write converse ($q \implies p$). 'If two triangles are similar, then they are congruent.' Step 3: Write contrapositive ($\sim q \implies \sim p$). 'If two triangles are not similar, then they are not congruent.' Final answer: Converse: 'If two triangles are similar, then they are congruent.' Contrapositive: 'If two triangles are not similar, then they are not congruent.'
• Q: Show that the statement 'For any real numbers $a$ and $b$, $a^2 = b^2 \implies a = b
  is false by providing a counterexample. A: Step 1: Understand the claim. The claim states that if the squares of two real numbers are equal, the numbers themselves must be equal. Step 2: Search for a counterexample where $a^2 = b^2$ but $a \neq b$. Step 3: Let $a = 5$ and $b = -5$. Step 4: Verify the hypothesis. $a^2 = (5)^2 = 25$ and $b^2 = (-5)^2 = 25$. So, $a^2 = b^2$ is true. Step 5: Check the conclusion. Here, $5 \neq -5$, so $a = b$ is false. Final answer: Since $a = 5$ and $b = -5$ satisfies $a^2 = b^2$ but $a \neq b$, the statement is false.

Frequently Asked Questions