OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Exercise 27 (G)

Question 1. Show that the statement p: 'If x is a real number such that \( x^3 + 4x = 0 \), then \( x = 0 \)' is true by
(i) Direct method,
(ii) Method of contradiction,
(iii) Method of contrapositive.
Answer:
Let \( q: x \in R \) such that \( x^3 + 4x = 0 \).
Let \( r: x = 0 \).
The statement p can be written as \( q \rightarrow r \).

(i) **Direct Method:**
Assume \( q \) is true, which means \( x^3 + 4x = 0 \).
We can factor this as \( x(x^2 + 4) = 0 \).
This means either \( x = 0 \) or \( x^2 + 4 = 0 \).
Since \( x \) is a real number, \( x^2 \) must be greater than or equal to 0. So, \( x^2 + 4 \) will always be greater than or equal to 4, which means \( x^2 + 4 \) cannot be 0.
Therefore, the only possibility is \( x = 0 \).
This shows that \( r \) is true.
Since \( q \rightarrow r \) is true, the statement \( p \) is true.

(ii) **Method of Contradiction:**
Assume the statement \( p \) is false. This means \( q \) is true and \( r \) is false.
So, \( x^3 + 4x = 0 \) (q is true) and \( x \neq 0 \) (r is false).
If \( x \neq 0 \), then we can divide \( x^3 + 4x = 0 \) by \( x \).
\( x^3 + 4x = 0 \)
\( \implies x(x^2 + 4) = 0 \)
Since we assumed \( x \neq 0 \), then it must be that \( x^2 + 4 = 0 \).
However, for any real number \( x \), \( x^2 \ge 0 \), so \( x^2 + 4 \ge 4 \).
Thus, \( x^2 + 4 \) can never be 0 for a real number \( x \).
This creates a contradiction.
Therefore, our initial assumption that \( p \) is false must be wrong. So, \( p \) is true.

(iii) **Method of Contrapositive:**
The contrapositive of \( p: q \rightarrow r \) is \( \sim r \rightarrow \sim q \).
Here, \( \sim r \) means \( x \neq 0 \).
And \( \sim q \) means \( x^3 + 4x \neq 0 \).
Let's assume \( \sim r \) is true, so \( x \neq 0 \).
If \( x \neq 0 \), then \( x^2 \) is a positive number. So, \( x^2 + 4 \) will always be greater than or equal to 4.
Since \( x \neq 0 \) and \( x^2 + 4 \neq 0 \), their product \( x(x^2 + 4) \) will also not be 0.
So, \( x^3 + 4x \neq 0 \).
This means \( \sim q \) is true.
Since \( \sim r \rightarrow \sim q \) is true, the original statement \( p: q \rightarrow r \) is also true. These methods are fundamental for proving mathematical statements.
In simple words: We showed the statement is true using three ways: first, by directly showing the result; second, by assuming the opposite and finding a problem; and third, by checking the reverse-negative version of the statement. All three methods led to the same conclusion, proving the statement is true.

🎯 Exam Tip: When asked to prove a statement using multiple methods, ensure each method is clearly separated and follows its specific logical structure. Pay close attention to definitions like 'real number' as they affect possible values for variables.

Question 2. Show that the statement 'For any real numbers a and b, \( a^2 = b^2 \) implies that \( a = b \)' is not true by giving counter examples.
Answer:
The given statement is: "For any real numbers \( a \) and \( b \), if \( a^2 = b^2 \), then \( a = b \)".
To show that this statement is not true, we need to find just one example where \( a^2 = b^2 \) is true, but \( a = b \) is false. This is called a counterexample.

Let's take \( a = 2 \) and \( b = -2 \).
First, let's check if \( a^2 = b^2 \):
\( a^2 = (2)^2 = 4 \)
\( b^2 = (-2)^2 = 4 \)
So, \( a^2 = b^2 \) is true in this case.

Next, let's check if \( a = b \):
Here, \( a = 2 \) and \( b = -2 \).
Clearly, \( 2 \neq -2 \).
So, \( a = b \) is false in this case. Finding even one such example disproves the general statement.
Since we found a case where \( a^2 = b^2 \) is true but \( a = b \) is false, the original statement "if \( a^2 = b^2 \), then \( a = b \)" is not true for all real numbers \( a \) and \( b \).
In simple words: The statement says if two numbers squared are the same, then the numbers themselves must be the same. But this is not always true. For example, 2 squared is 4, and -2 squared is also 4. So \( a^2 = b^2 \) is true for \( a=2 \) and \( b=-2 \), but \( a \) is not equal to \( b \). This one example proves the statement is false.

🎯 Exam Tip: To disprove a universal statement (one that claims something is true "for all"), you only need to provide a single counterexample that clearly violates the statement. No complex proofs are needed.

Question 3. Show that the following statement is true by the method of contrapositive. p : If x is an integer and \( x^2 \) is even, then x is also even.
Answer:
Let the given statement be \( p \).
Let \( q \) be "x is an integer and \( x^2 \) is even".
Let \( r \) be "x is also even".
So, the statement \( p \) is \( q \rightarrow r \).

To prove \( q \rightarrow r \) using the method of contrapositive, we need to prove its contrapositive statement, which is \( \sim r \rightarrow \sim q \).
Here, \( \sim r \) means "x is not even", which implies "x is odd".
And \( \sim q \) means "x is an integer and \( x^2 \) is not even", which implies "x is an integer and \( x^2 \) is odd".

So, we need to prove: "If x is an integer and x is odd, then \( x^2 \) is odd."

Let's assume \( \sim r \) is true, meaning x is an odd integer.
An odd integer can always be written in the form \( 2k + 1 \), where \( k \) is some integer.
So, let \( x = 2k + 1 \).

Now, let's find \( x^2 \):
\( x^2 = (2k + 1)^2 \)
\( \implies x^2 = (2k)^2 + 2(2k)(1) + (1)^2 \)
\( \implies x^2 = 4k^2 + 4k + 1 \)
We can factor out 2 from the first two terms:
\( \implies x^2 = 2(2k^2 + 2k) + 1 \)
Let \( m = 2k^2 + 2k \). Since \( k \) is an integer, \( m \) will also be an integer.
\( \implies x^2 = 2m + 1 \)
This form \( 2m + 1 \) clearly shows that \( x^2 \) is an odd number. The property of odd numbers \( 2m+1 \) is quite useful here.
Thus, if x is odd, then \( x^2 \) is odd. This is \( \sim q \).

Since we have proven that \( \sim r \rightarrow \sim q \) is true, the original statement \( p: q \rightarrow r \) is also true.
In simple words: We want to show that if a number squared is even, then the number itself must be even. Instead of proving this directly, we proved the opposite: if a number is NOT even (meaning it's odd), then its square is also NOT even (meaning its square is odd). We showed that if you square any odd number, the result is always odd. Since this opposite idea is true, the original statement must also be true.

🎯 Exam Tip: The contrapositive method is often easier when dealing with proofs involving "even" or "odd" numbers. Remember to correctly form the negation (\( \sim \)) of both the hypothesis and the conclusion before starting the proof.

Question 4. Given below are two statements : p : 30 is a multiple of 5. q : 30 is a multiple of 9. Write the compound statement, connecting these two statements with 'and' and 'or'. In both cases, check the validity of the compound statement.
Answer:
Given statements:
\( p \): "30 is a multiple of 5."
\( q \): "30 is a multiple of 9."

First, let's determine the truth value of each simple statement:
For \( p \): 30 is \( 5 \times 6 \), so 30 is a multiple of 5. Therefore, \( p \) is **True (T)**.
For \( q \): 30 cannot be divided evenly by 9 ( \( 30 \div 9 = 3 \) with a remainder of 3). So, 30 is not a multiple of 9. Therefore, \( q \) is **False (F)**.

Now, let's form the compound statements:

**1. Compound statement with 'and' (conjunction): \( p \wedge q \)**
Statement: "30 is a multiple of 5 and 30 is a multiple of 9."
Truth value: A conjunction \( p \wedge q \) is true only if *both* \( p \) and \( q \) are true. Since \( p \) is true (T) and \( q \) is false (F), the compound statement \( p \wedge q \) is **False (F)**.
Validity: The statement is not valid because one of its components is false.

**2. Compound statement with 'or' (disjunction): \( p \vee q \)**
Statement: "30 is a multiple of 5 or 30 is a multiple of 9."
Truth value: A disjunction \( p \vee q \) is true if *at least one* of \( p \) or \( q \) is true. Since \( p \) is true (T) and \( q \) is false (F), the compound statement \( p \vee q \) is **True (T)**.
Validity: The statement is valid because at least one of its components is true. This illustrates the fundamental difference between 'and' and 'or' in logic.
In simple words: We have two facts: "30 can be divided by 5" (which is true) and "30 can be divided by 9" (which is false). When we join them with "and", the whole sentence becomes false because one part is false. But when we join them with "or", the whole sentence becomes true because at least one part is true.

🎯 Exam Tip: Remember the truth tables for 'and' (\( \wedge \)) and 'or' (\( \vee \)): for 'and', both parts must be true for the whole statement to be true; for 'or', only one part needs to be true for the whole statement to be true. Clearly stating the truth value of individual simple statements is key.

Question 5. Verify by the method of contradiction that \( \sqrt{7} \) is irrational.
Answer:
Let the given statement \( p \) be " \( \sqrt{7} \) is irrational".
To prove this by the method of contradiction, we first assume the opposite of the statement is true.
So, let's assume \( \sim p \) is true, meaning " \( \sqrt{7} \) is rational".

If \( \sqrt{7} \) is rational, then it can be written as a fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers, \( b \neq 0 \), and \( a \) and \( b \) have no common factors other than 1 (i.e., they are in their simplest form, or coprime).
So, \( \sqrt{7} = \frac{a}{b} \).

Square both sides of the equation:
\( (\sqrt{7})^2 = \left(\frac{a}{b}\right)^2 \)
\( \implies 7 = \frac{a^2}{b^2} \)
\( \implies 7b^2 = a^2 \)

This equation tells us that \( a^2 \) is a multiple of 7. If \( a^2 \) is a multiple of 7, it means that \( a \) itself must also be a multiple of 7. This is a property of prime numbers (7 is a prime number).
So, we can write \( a = 7k \) for some integer \( k \).

Now substitute \( a = 7k \) back into the equation \( 7b^2 = a^2 \):
\( 7b^2 = (7k)^2 \)
\( \implies 7b^2 = 49k^2 \)
Divide both sides by 7:
\( \implies b^2 = 7k^2 \)

This equation shows that \( b^2 \) is a multiple of 7. Similar to \( a \), if \( b^2 \) is a multiple of 7, then \( b \) itself must also be a multiple of 7.

So, we have found that both \( a \) and \( b \) are multiples of 7.
This means that \( a \) and \( b \) have a common factor of 7.
However, earlier we defined \( a \) and \( b \) as having no common factors other than 1. This new finding that they share a common factor of 7 directly contradicts our initial assumption.
This contradiction means our original assumption that \( \sqrt{7} \) is rational must be false.
Therefore, \( \sqrt{7} \) must be irrational. The method of contradiction is a powerful technique in mathematics.
In simple words: We wanted to prove that \( \sqrt{7} \) cannot be written as a simple fraction. So, we pretended it *could* be written as a simple fraction \( \frac{a}{b} \). By doing some math steps, we found that both \( a \) and \( b \) had to be divisible by 7. But if they are both divisible by 7, then the fraction \( \frac{a}{b} \) wasn't in its simplest form, which goes against our first rule for a simple fraction. This problem means our original guess was wrong, so \( \sqrt{7} \) must be irrational.

🎯 Exam Tip: When using the method of contradiction for irrationality proofs, always clearly state your initial assumption (that the number is rational), perform algebraic manipulations to show a common factor, and then clearly state how this contradicts the initial definition of coprime integers.