Mathematical Reasoning — Statements, Connectives, Proofs

What Mathematical Reasoning Is About

Mathematical reasoning teaches us how to build valid arguments. We learn to distinguish between statements that can be proved true or false, how to combine them using connectives (and, or, if-then), and how to construct proofs. This chapter is about the logic behind mathematics.

For CBSE Class 11, this chapter carries 2-3 marks in boards. In JEE Main, it occasionally appears as a quick 1-mark question. The questions are largely conceptual — no heavy calculations, just clear logical thinking.

graph TD
    A[Need to prove P?] --> B{Which method?}
    B -->|Assume P is true, derive| C[Direct Proof]
    B -->|Assume NOT P, get contradiction| D[Proof by Contradiction]
    B -->|Prove NOT Q implies NOT P| E[Contrapositive]
    B -->|Try specific examples| F[Counterexample to disprove]
    B -->|Prove for n=1, n=k implies n=k+1| G[Mathematical Induction]
    C --> H[Logical chain of implications]
    D --> I[Contradiction found → P must be true]
    E --> J[Equivalent to proving P → Q]
    F --> K[One example enough to disprove]

Key Terms & Definitions

Statement (Proposition) — A sentence that is either true or false, but not both. “5 is prime” is a statement. “Is 5 prime?” is not (it’s a question).

Negation — The opposite of a statement. If $p$ is “It is raining”, then $\sim p$ is “It is not raining”.

Compound Statement — Formed by combining simple statements using connectives.

Connectives:

AND (Conjunction, $\wedge$ ): $p \wedge q$ is true only when both $p$ and $q$ are true.
OR (Disjunction, $\vee$ ): $p \vee q$ is true when at least one is true.
IF-THEN (Implication, $\Rightarrow$ ): $p \Rightarrow q$ is false only when $p$ is true and $q$ is false.
IF AND ONLY IF (Biconditional, $\Leftrightarrow$ ): True when both have the same truth value.

Contrapositive — Of $p \Rightarrow q$ is $\sim q \Rightarrow \sim p$ . They are logically equivalent.

Converse — Of $p \Rightarrow q$ is $q \Rightarrow p$ . NOT logically equivalent to the original.

Truth Tables

$p$	$q$	$p \wedge q$	$p \vee q$	$p \Rightarrow q$	$p \Leftrightarrow q$
T	T	T	T	T	T
T	F	F	T	F	F
F	T	F	T	T	F
F	F	F	F	T	T

The implication $p \Rightarrow q$ is tricky: it’s false ONLY when $p$ is true and $q$ is false. If $p$ is false, the implication is true regardless of $q$ . This is called “vacuous truth” — a promise you never had to keep because the condition was never met.

Types of Proofs

Direct Proof

Start with known truths and use logical steps to reach the conclusion.

Example: Prove that the sum of two even numbers is even.

Let $a = 2m$ and $b = 2n$ . Then $a + b = 2m + 2n = 2(m+n)$ , which is even.

Proof by Contradiction

Assume the negation of what you want to prove. Show this leads to a contradiction.

Example: Prove $\sqrt{2}$ is irrational.

Assume $\sqrt{2} = p/q$ (fully reduced). Then $2 = p^2/q^2$ , so $p^2 = 2q^2$ . This means $p^2$ is even, so $p$ is even. Let $p = 2k$ . Then $4k^2 = 2q^2$ , so $q^2 = 2k^2$ , making $q$ even too. But $p$ and $q$ can’t both be even (we said the fraction was fully reduced). Contradiction. So $\sqrt{2}$ is irrational.

Proof by Contrapositive

To prove $p \Rightarrow q$ , prove $\sim q \Rightarrow \sim p$ instead.

Example: Prove “if $n^2$ is even, then $n$ is even.”

Contrapositive: “if $n$ is odd, then $n^2$ is odd.” If $n = 2k+1$ , then $n^2 = 4k^2+4k+1 = 2(2k^2+2k)+1$ , which is odd. Done.

Counterexample

To disprove a universal statement, one counterexample suffices.

Example: “All prime numbers are odd” — counterexample: 2 is prime and even.

Solved Examples — Easy to Hard

Example 1 (Easy — CBSE)

Identify whether these are statements: (a) “2 + 3 = 6” (b) “Close the door” (c) “Every rectangle is a square”

(a) Statement — it’s false, but it’s declarative. (b) Not a statement — it’s a command. (c) Statement — it’s false (counterexample: a 4x6 rectangle is not a square).

Example 2 (Medium — CBSE)

Write the negation of “For every real number $x$ , $x^2 \geq 0$ .”

Negation: “There exists a real number $x$ such that $x^2 < 0$ .”

Note how “for every” becomes “there exists” and " $\geq$ " becomes " $<$ " in the negation.

Example 3 (Medium — JEE Main)

Find the contrapositive and converse of “If a triangle is equilateral, then it is isosceles.”

Contrapositive: “If a triangle is not isosceles, then it is not equilateral.” (True)

Converse: “If a triangle is isosceles, then it is equilateral.” (False — a triangle with sides 5, 5, 3 is isosceles but not equilateral)

Example 4 (Hard — JEE Main)

Determine the truth value of: “If $3 + 4 = 8$ , then $5 + 6 = 10$ .”

$p$ : " $3+4 = 8$ " is FALSE. $q$ : " $5+6 = 10$ " is FALSE.

$p \Rightarrow q$ : When $p$ is false, the implication is true regardless of $q$ . So this is TRUE.

Exam-Specific Tips

CBSE Class 11: 2-3 marks. Questions on identifying statements, writing negations, truth values of compound statements, and stating the contrapositive. Know the difference between converse and contrapositive — this is asked frequently.

JEE Main: Rarely more than 1 question. Usually about truth values or identifying valid logical equivalences. Quick to solve if you know the truth table rules.

Common Mistakes to Avoid

Mistake 1 — Confusing converse with contrapositive. Converse of $p \Rightarrow q$ is $q \Rightarrow p$ (not equivalent). Contrapositive is $\sim q \Rightarrow \sim p$ (equivalent). Many students mix these up.

Mistake 2 — Thinking “or” means “one or the other but not both.” In maths, “or” is inclusive — $p \vee q$ is true when both are true. Exclusive or is different and rarely used in this chapter.

Mistake 3 — Wrong negation of quantified statements. The negation of “for all $x$ , $P(x)$ ” is “there exists $x$ such that $\sim P(x)$ ” — not “for all $x$ , $\sim P(x)$ ”.

Mistake 4 — Assuming a false hypothesis means the conclusion is false. In $p \Rightarrow q$ , if $p$ is false, the implication is TRUE no matter what $q$ is. Students often mark it as false.

Mistake 5 — Not recognizing open sentences. " $x + 2 = 5$ " is NOT a statement because its truth depends on the value of $x$ . It’s an open sentence. It becomes a statement when $x$ is specified.

Practice Questions

Q1. Is “Every integer is a rational number” a statement? What is its truth value?

Yes, it’s a statement. True — every integer $n$ can be written as $n/1$ .

Q2. Write the negation of “There exists a number $x$ such that $x^2 = -1$ in real numbers.”

“For every real number $x$ , $x^2 \neq -1$ .” (This negation is true.)

Q3. Find the truth value of $(p \wedge q) \vee (\sim p)$ when $p$ is true and $q$ is false.

$p \wedge q = T \wedge F = F$ . $\sim p = F$ . $F \vee F = F$ .

Q4. State the contrapositive: “If $x$ is divisible by 6, then $x$ is divisible by 2.”

“If $x$ is not divisible by 2, then $x$ is not divisible by 6.”

Q5. Disprove: “For all positive integers $n$ , $n^2 + n + 41$ is prime.”

Try $n = 40$ : $40^2 + 40 + 41 = 1600 + 40 + 41 = 1681 = 41^2$ . Not prime. Disproved.

Q6. Using truth table, verify $\sim(p \wedge q) = (\sim p) \vee (\sim q)$ (De Morgan’s Law).

Check all four combinations: TT → F vs F∨F=F. TF → T vs F∨T=T. FT → T vs T∨F=T. FF → T vs T∨T=T. All match. Verified.

Q7. Is the following valid? “If it rains, the match is cancelled. The match is not cancelled. Therefore, it did not rain.”

Yes. This is the contrapositive in action: $p \Rightarrow q$ and $\sim q$ together imply $\sim p$ . This is called modus tollens.

Q8. Write the negation of " $p \Rightarrow q$ " in terms of $p$ and $q$ .

$\sim(p \Rightarrow q) = p \wedge (\sim q)$ . The implication is false only when $p$ is true and $q$ is false.

Mathematical Induction — The Bridge from Pattern to Proof

Mathematical induction is the formal method for proving statements about all natural numbers. It connects to this chapter because induction IS mathematical reasoning in action.

The Two Steps of Induction

Prove the statement is true for $n = 1$ (or the starting value).

Assume the statement is true for $n = k$ (the inductive hypothesis). Then prove it must be true for $n = k + 1$ .

If both steps succeed, the statement is true for all natural numbers. Think of it as a chain of dominoes — the base case knocks over the first, and the inductive step ensures each domino knocks over the next.

Worked Example — Induction

Prove that $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$ for all $n \geq 1$ .

Base case: $n = 1$ : LHS $= 1$ . RHS $= \frac{1 \times 2}{2} = 1$ . True.

Inductive step: Assume true for $n = k$ : $1 + 2 + \cdots + k = \frac{k(k+1)}{2}$ .

Now prove for $n = k+1$ :

1 + 2 + \cdots + k + (k+1) = \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}

This is exactly $\frac{(k+1)((k+1)+1)}{2}$ — the formula with $n = k+1$ . Proved.

\sim(p \wedge q) \equiv (\sim p) \vee (\sim q)

\sim(p \vee q) \equiv (\sim p) \wedge (\sim q)

The negation of AND is OR (with negated components), and vice versa. These are essential for writing negations of compound statements correctly.

CBSE Class 11 boards often ask: “Write the negation of the compound statement: It is raining AND it is cold.” Using De Morgan’s Law: “It is NOT raining OR it is NOT cold.” Students who don’t know De Morgan’s Law write “It is NOT raining AND it is NOT cold” — which is the negation of the OR, not the AND.

Q9. Prove by mathematical induction that $n^2 + n$ is even for all $n \geq 1$ .

Base case: $n = 1$ : $1 + 1 = 2$ (even). True.

Inductive step: Assume $k^2 + k$ is even. Then $(k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = (k^2 + k) + 2(k+1)$ .

By hypothesis, $k^2 + k$ is even. $2(k+1)$ is always even. Sum of two even numbers is even. Proved.

Alternatively, $n^2 + n = n(n+1)$ , which is the product of two consecutive integers — one of them is always even, making the product even.

Q10. Determine the truth value of: ” $2 + 3 = 5$ if and only if $3 + 2 = 6$ .”

$p$ : " $2 + 3 = 5$ " is TRUE. $q$ : " $3 + 2 = 6$ " is FALSE.

$p \Leftrightarrow q$ is true only when both have the same truth value. Here $p$ is true and $q$ is false, so $p \Leftrightarrow q$ is FALSE.

FAQs

What is the difference between a statement and a sentence?

All statements are sentences, but not all sentences are statements. Commands (“shut the door”), questions (“is it raining?”), and exclamations (“wow!”) are sentences but not statements. A statement must be declarative and have a definite truth value.

Why is implication true when the hypothesis is false?

Think of it as a promise: “If it rains, I’ll bring an umbrella.” If it doesn’t rain, the promise isn’t broken regardless of whether I bring an umbrella. The implication is only falsified when the hypothesis holds but the conclusion doesn’t.

What is a tautology?

A compound statement that is true for ALL possible truth values of its components. Example: $p \vee (\sim p)$ is always true.

What is a contradiction?

A compound statement that is false for ALL truth values. Example: $p \wedge (\sim p)$ is always false.

How many rows does a truth table need for $n$ variables?

$2^n$ rows. For 2 variables: 4 rows. For 3 variables: 8 rows.