JEE Main & Advanced

Mathematical Reasoning

Mathematical Reasoning for JEE Main & Advanced

1
Module 1

Statements and Logical Connectives

Mathematical Statements and Truth ValuesTopic 1

A mathematical statement (or proposition) is a declarative sentence that is either definitively true or definitively false, but not both simultaneously. The truth or falsity of a statement is called its truth value, denoted by $T$ (true) or $F$ (false).

Examples of statements:
  • "The sum of two odd integers is even." (True statement)
  • "$\sqrt{2}$ is a rational number." (False statement)
  • "$7 > 5$." (True statement)
Examples that are NOT statements:
  • "Close the door." (command, not declarative)
  • "What time is it?" (question)
  • "$x + 5 = 8$." (open sentence — truth value depends on $x$)
  • "May you have a great day." (wish/exclamation)

Open Sentences: A declarative sentence containing one or more variables whose truth value depends on the value(s) assigned to the variable(s). For example, "$x$ is a prime number" is an open sentence — it becomes a statement once $x$ is specified.

Simple vs. Compound Statements:
  • Simple Statement: Cannot be broken down further into simpler component statements. (e.g., "It is raining.")
  • Compound Statement: Formed by combining two or more simple statements using logical connectives (e.g., "It is raining AND the road is wet.").

Component Statements: The simple statements making up a compound statement. To analyze a compound statement, identify its component statements and the connectives used.

Notation: Statements are typically denoted by lowercase letters $p, q, r, s, \ldots$

Worked Examples
1

Determine which of the following are statements. For those that are, identify the truth value. (i) "The earth has one moon." (ii) "$\sqrt{3}$ is irrational." (iii) "How beautiful!" (iv) "$x^2 = 4$." (v) "5 is a perfect square."

Show solution

(i) Statement, True. (It is declarative and definitely true.)
(ii) Statement, True. (Famous theorem: $\sqrt{3}$ cannot be expressed as $p/q$.)
(iii) NOT a statement. (Exclamation, not declarative.)
(iv) NOT a statement. (Open sentence — truth depends on $x$.)
(v) Statement, False. (Since 5 is not the square of any integer.)

2

Identify the component statements of the following compound statement:
"The number 36 is divisible by 4 and by 6."

Show solution
Component statements:
  • $p$: "The number 36 is divisible by 4." (True)
  • $q$: "The number 36 is divisible by 6." (True)
The connective used is "and". Since both $p$ and $q$ are true, the compound statement is true.
✎ Self-Check — 5 questions0 / 5
Q1.Which of the following is a mathematical statement?
Q2.The sentence "$x - 5 = 7$" is:
Q3.The compound statement "2 is even and 3 is odd" has truth value:
Q4.Which of the following statements is FALSE?
Q5.The statement "All multiples of 4 are even" is:

Logical Connectives and Truth TablesTopic 2

Compound statements are formed using logical connectives (also called logical operators). The five fundamental connectives in propositional logic are:

1. Negation ($\sim p$ or $\neg p$): The opposite truth value of $p$.
$\sim p$ is read as "not $p$" or "it is not the case that $p$". \[ \begin{array}{|c|c|} p & \sim p \\ T & F \\ F & T \\ \end{array} \]

2. Conjunction ($p \wedge q$): "$p$ AND $q$". True only when both $p$ and $q$ are true. \[ \begin{array}{|c|c|c|} p & q & p \wedge q \\ T & T & T \\ T & F & F \\ F & T & F \\ F & F & F \\ \end{array} \]

3. Disjunction ($p \vee q$): "$p$ OR $q$" (inclusive or). False only when both $p$ and $q$ are false. \[ \begin{array}{|c|c|c|} p & q & p \vee q \\ T & T & T \\ T & F & T \\ F & T & T \\ F & F & F \\ \end{array} \]

4. Conditional / Implication ($p \Rightarrow q$): "If $p$ then $q$". False only when $p$ is true and $q$ is false. \[ \begin{array}{|c|c|c|} p & q & p \Rightarrow q \\ T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \\ \end{array} \] $p$ is called the antecedent (or hypothesis); $q$ is the consequent (or conclusion). The rows with $p$ false are called vacuously true.

5. Biconditional ($p \Leftrightarrow q$): "$p$ if and only if $q$" (often abbreviated "$p$ iff $q$"). True when $p$ and $q$ have the same truth value. \[ \begin{array}{|c|c|c|} p & q & p \Leftrightarrow q \\ T & T & T \\ T & F & F \\ F & T & F \\ F & F & T \\ \end{array} \]

Relationships and Equivalences:
  • $p \Leftrightarrow q \equiv (p \Rightarrow q) \wedge (q \Rightarrow p)$
  • $p \Rightarrow q \equiv \sim p \vee q$ (Material implication)
  • Double negation: $\sim(\sim p) \equiv p$
Tautology, Contradiction, Contingency:
  • Tautology: A compound statement that is always true regardless of truth values of components. E.g., $p \vee \sim p$.
  • Contradiction (Fallacy): Always false. E.g., $p \wedge \sim p$.
  • Contingency: Sometimes true and sometimes false. Most compound statements are contingencies.
Worked Examples
1

Construct the truth table for $(p \Rightarrow q) \wedge (q \Rightarrow p)$ and verify that it is logically equivalent to $p \Leftrightarrow q$.

Show solution

\[ \begin{array}{|c|c|c|c|c|c|} p & q & p \Rightarrow q & q \Rightarrow p & (p \Rightarrow q) \wedge (q \Rightarrow p) & p \Leftrightarrow q \\ T & T & T & T & T & T \\ T & F & F & T & F & F \\ F & T & T & F & F & F \\ F & F & T & T & T & T \\ \end{array} \] Columns 5 and 6 are identical, so $(p \Rightarrow q) \wedge (q \Rightarrow p) \equiv p \Leftrightarrow q$.
Final Answer: The two compound statements are logically equivalent.

2

Show that $p \vee \sim p$ is a tautology and $p \wedge \sim p$ is a contradiction.

Show solution

\[ \begin{array}{|c|c|c|c|} p & \sim p & p \vee \sim p & p \wedge \sim p \\ T & F & T & F \\ F & T & T & F \\ \end{array} \] The column for $p \vee \sim p$ is all $T$, so it is a tautology.
The column for $p \wedge \sim p$ is all $F$, so it is a contradiction.
(These are known as the law of excluded middle and the law of non-contradiction.)

3

Determine whether the compound statement $[(p \Rightarrow q) \wedge p] \Rightarrow q$ is a tautology, contradiction, or contingency.

Show solution

\[ \begin{array}{|c|c|c|c|c|} p & q & p \Rightarrow q & (p \Rightarrow q) \wedge p & [(p \Rightarrow q) \wedge p] \Rightarrow q \\ T & T & T & T & T \\ T & F & F & F & T \\ F & T & T & F & T \\ F & F & T & F & T \\ \end{array} \] The last column is all $T$, so this statement is a tautology.
This is the classical rule of inference called Modus Ponens: "If $p$ implies $q$, and $p$ is true, then $q$ must be true."

✎ Self-Check — 5 questions0 / 5
Q1.The negation of "$2 + 3 = 5$" is:
Q2.If $p$ is true and $q$ is false, then which of the following compound statements is true?
Q3.The statement $p \Rightarrow (q \Rightarrow p)$ is:
Q4.Which of the following is logically equivalent to $p \Rightarrow q$?
Q5.The truth value of $(\sim p \vee q) \wedge (\sim q \vee p)$ when $p$ is true and $q$ is false is:
2
Module 2

Logical Equivalences & Conditional Statements

Logical Equivalences and De Morgan's LawsTopic 1

Two compound statements $P$ and $Q$ are logically equivalent, written $P \equiv Q$, if they have the same truth value for every assignment of truth values to their atomic components. Equivalently, $P \Leftrightarrow Q$ is a tautology.

Important Logical Equivalences (Laws of Logic):

Idempotent Laws: \[ p \vee p \equiv p, \quad p \wedge p \equiv p \]

Commutative Laws: \[ p \vee q \equiv q \vee p, \quad p \wedge q \equiv q \wedge p \]

Associative Laws: \[ (p \vee q) \vee r \equiv p \vee (q \vee r), \quad (p \wedge q) \wedge r \equiv p \wedge (q \wedge r) \]

Distributive Laws: \[ p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r) \] \[ p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) \]

Identity Laws (where $\mathbf{T}$ is any tautology and $\mathbf{F}$ is any contradiction): \[ p \vee \mathbf{F} \equiv p, \quad p \wedge \mathbf{T} \equiv p \] \[ p \vee \mathbf{T} \equiv \mathbf{T}, \quad p \wedge \mathbf{F} \equiv \mathbf{F} \]

Complement (Negation) Laws: \[ p \vee \sim p \equiv \mathbf{T}, \quad p \wedge \sim p \equiv \mathbf{F} \] \[ \sim(\sim p) \equiv p \quad \text{(Double negation)} \]

De Morgan's Laws (Critical for JEE): \[ \sim(p \wedge q) \equiv \sim p \vee \sim q \] \[ \sim(p \vee q) \equiv \sim p \wedge \sim q \]

Negation of Conditional: \[ \sim(p \Rightarrow q) \equiv p \wedge \sim q \] This is derived because $p \Rightarrow q \equiv \sim p \vee q$, and by De Morgan's: $\sim(\sim p \vee q) \equiv p \wedge \sim q$.

Negation of Biconditional: \[ \sim(p \Leftrightarrow q) \equiv (p \wedge \sim q) \vee (\sim p \wedge q) \] (Which means "exactly one of $p, q$ is true" — the XOR operation.)

Absorption Laws: \[ p \vee (p \wedge q) \equiv p, \quad p \wedge (p \vee q) \equiv p \]

Worked Examples
1

Use logical equivalences (without truth tables) to simplify the expression $\sim[\sim p \vee (p \wedge q)]$.

Show solution

Apply De Morgan's law: \[ \sim[\sim p \vee (p \wedge q)] \equiv \sim(\sim p) \wedge \sim(p \wedge q) \] Apply double negation and De Morgan's again: \[ \equiv p \wedge (\sim p \vee \sim q) \] Apply distributive law: \[ \equiv (p \wedge \sim p) \vee (p \wedge \sim q) \] Apply complement law $p \wedge \sim p \equiv \mathbf{F}$: \[ \equiv \mathbf{F} \vee (p \wedge \sim q) \] Apply identity law: \[ \equiv p \wedge \sim q \] Final Answer: $p \wedge \sim q$.

2

Write the negation of the following statement: "If it rains, then the streets are wet."

Show solution

Let $p$: "It rains." and $q$: "The streets are wet."
The statement is $p \Rightarrow q$.
Its negation is: \[ \sim(p \Rightarrow q) \equiv p \wedge \sim q \] In words: "It rains AND the streets are NOT wet."
Final Answer: "It rains and the streets are not wet."

3

Write the negation of "$x = 5$ AND $y \geq 7$" using De Morgan's law.

Show solution

Let $p$: "$x = 5$" and $q$: "$y \geq 7$".
The given statement is $p \wedge q$. Its negation is: \[ \sim(p \wedge q) \equiv \sim p \vee \sim q \] In words: "$x \neq 5$ OR $y < 7$".
Final Answer: "$x \neq 5$ or $y < 7$".

✎ Self-Check — 5 questions0 / 5
Q1.The negation of $p \wedge (q \vee \sim r)$ is logically equivalent to:
Q2.The negation of "Either $x > 5$ or $y < 3$" is:
Q3.$\sim(p \Leftrightarrow q)$ is equivalent to:
Q4.Which of the following is a tautology?
Q5.The simplification of $(p \wedge q) \vee (p \wedge \sim q)$ is:

Converse, Inverse, and ContrapositiveTopic 2

Given a conditional statement $p \Rightarrow q$, we can form three related conditional statements:

Converse: $q \Rightarrow p$
(Switches the antecedent and consequent.)

Inverse: $\sim p \Rightarrow \sim q$
(Negates both the antecedent and consequent, keeping the order.)

Contrapositive: $\sim q \Rightarrow \sim p$
(Negates both AND switches them.)

Crucial Logical Equivalences: \[ p \Rightarrow q \equiv \sim q \Rightarrow \sim p \quad \text{(Original is equivalent to its contrapositive)} \] \[ q \Rightarrow p \equiv \sim p \Rightarrow \sim q \quad \text{(Converse is equivalent to inverse)} \]

However: \[ p \Rightarrow q \not\equiv q \Rightarrow p \quad \text{(Original NOT equivalent to converse, in general)} \] \[ p \Rightarrow q \not\equiv \sim p \Rightarrow \sim q \quad \text{(Original NOT equivalent to inverse, in general)} \]

Verification table: \[ \begin{array}{|c|c|c|c|c|c|c|c|} p & q & \sim p & \sim q & p \Rightarrow q & q \Rightarrow p & \sim p \Rightarrow \sim q & \sim q \Rightarrow \sim p \\ T & T & F & F & T & T & T & T \\ T & F & F & T & F & T & T & F \\ F & T & T & F & T & F & F & T \\ F & F & T & T & T & T & T & T \\ \end{array} \] Columns 5 and 8 match (original $\equiv$ contrapositive). Columns 6 and 7 match (converse $\equiv$ inverse).

Necessary and Sufficient Conditions: In "$p \Rightarrow q$":
  • $p$ is a sufficient condition for $q$. (If you have $p$, you're guaranteed $q$.)
  • $q$ is a necessary condition for $p$. (You can't have $p$ without $q$.)
If $p \Leftrightarrow q$, then $p$ and $q$ are each both necessary and sufficient for each other ("if and only if").

Importance in Theorems: Many mathematical theorems are biconditionals. To prove $p \Leftrightarrow q$, one must prove both $p \Rightarrow q$ and $q \Rightarrow p$.
An alternative to proving $p \Rightarrow q$ directly is to prove its contrapositive $\sim q \Rightarrow \sim p$. This is the basis of proof by contrapositive.

Worked Examples
1

Write the converse, inverse, and contrapositive of the following statement:
"If $n$ is an even integer, then $n^2$ is even."

Show solution
Let $p$: "$n$ is an even integer." and $q$: "$n^2$ is even."
The given statement is $p \Rightarrow q$.
  • Converse ($q \Rightarrow p$): "If $n^2$ is even, then $n$ is an even integer."
  • Inverse ($\sim p \Rightarrow \sim q$): "If $n$ is not an even integer, then $n^2$ is not even."
  • Contrapositive ($\sim q \Rightarrow \sim p$): "If $n^2$ is not even, then $n$ is not an even integer."
(All four — the original, converse, inverse, and contrapositive — happen to be true in this case, but in general only the original and contrapositive are guaranteed to have the same truth value.)
2

State the contrapositive of: "If $x^2 - 5x + 6 = 0$, then $x = 2$ or $x = 3$."

Show solution

Let $p$: "$x^2 - 5x + 6 = 0$" and $q$: "$x = 2$ or $x = 3$".
$\sim p$: "$x^2 - 5x + 6 \neq 0$".
$\sim q$ (by De Morgan): "$x \neq 2$ AND $x \neq 3$".
Contrapositive: "If $x \neq 2$ and $x \neq 3$, then $x^2 - 5x + 6 \neq 0$."

3

Using proof by contrapositive, show that if $n^2$ is odd, then $n$ is odd (for $n \in \mathbb{Z}$).

Show solution

Statement to prove: $p \Rightarrow q$ where $p$: "$n^2$ is odd" and $q$: "$n$ is odd".
Contrapositive: $\sim q \Rightarrow \sim p$: "If $n$ is even, then $n^2$ is even."
Proof of contrapositive: Assume $n$ is even, so $n = 2k$ for some $k \in \mathbb{Z}$. Then $n^2 = 4k^2 = 2(2k^2)$, which is even.
Since the contrapositive is true, the original statement is also true.
Final Conclusion: If $n^2$ is odd, then $n$ is odd.

✎ Self-Check — 5 questions0 / 5
Q1.The contrapositive of "If $x$ is rational, then $x^2$ is rational" is:
Q2.Which of the following pairs is NOT logically equivalent?
Q3.The inverse of "If a triangle is equilateral, then it is isosceles" is:
Q4.In the statement "$x$ is a multiple of 6 $\Rightarrow$ $x$ is a multiple of 3", being a multiple of 6 is:
Q5.The contrapositive of "$p \vee q \Rightarrow r$" is:
3
Module 3

Quantifiers and Methods of Proof

Quantifiers and Their NegationsTopic 1

A quantifier is a word or phrase that indicates the quantity or range of objects to which a statement applies. In mathematical logic, there are two fundamental quantifiers.

Universal Quantifier ($\forall$): "for all", "for every", "for each".
$\forall x \in S, P(x)$ means "for every element $x$ in $S$, the property $P(x)$ holds."
Example: $\forall n \in \mathbb{N}, n + 1 > n$.

Existential Quantifier ($\exists$): "there exists", "for some", "for at least one".
$\exists x \in S, P(x)$ means "there exists at least one $x$ in $S$ such that $P(x)$ holds."
Example: $\exists n \in \mathbb{N}, n^2 = 25$ (namely $n = 5$).

Negation of Quantified Statements (Critical Rules): \[ \sim(\forall x, P(x)) \equiv \exists x, \sim P(x) \] \[ \sim(\exists x, P(x)) \equiv \forall x, \sim P(x) \]

In words:
  • The negation of "All $x$ satisfy $P$" is "There exists an $x$ that does not satisfy $P$".
  • The negation of "Some $x$ satisfies $P$" is "Every $x$ fails to satisfy $P$".

Combining Quantifiers: A statement may have multiple quantifiers, and the order matters!
$\forall x \exists y P(x, y)$ is generally NOT equivalent to $\exists y \forall x P(x, y)$.

Example: "For every real number $x$, there exists a real number $y$ such that $y > x$." (True)
But: "There exists a real number $y$ such that for every real number $x$, $y > x$." (False — no universal upper bound exists.)

Negation of Multiple Quantifiers:
To negate, flip each quantifier and negate the innermost statement: \[ \sim(\forall x \exists y P(x, y)) \equiv \exists x \forall y \sim P(x, y) \] \[ \sim(\exists x \forall y P(x, y)) \equiv \forall x \exists y \sim P(x, y) \]

Common Language Patterns:
  • "All triangles have three sides" $= \forall t \in T, \text{sides}(t) = 3$.
  • "Some real numbers are negative" $= \exists r \in \mathbb{R}, r < 0$.
  • "No prime is even except 2" $= \forall p \in \text{Primes}, (p = 2 \vee p \neq \text{even})$.
Worked Examples
1

Write the negation of the following statements:
(i) "All students passed the exam."
(ii) "There exists an integer $n$ such that $n^2 = 17$."
(iii) "For every real number $x$, $x^2 \geq 0$."

Show solution

(i) Symbolically: $\forall s, P(s)$ where $P(s)$ = "$s$ passed".
Negation: $\exists s, \sim P(s)$, i.e., "There exists a student who did not pass."
(ii) Symbolically: $\exists n \in \mathbb{Z}, n^2 = 17$.
Negation: $\forall n \in \mathbb{Z}, n^2 \neq 17$, i.e., "For every integer $n$, $n^2 \neq 17$."
(iii) Symbolically: $\forall x \in \mathbb{R}, x^2 \geq 0$.
Negation: $\exists x \in \mathbb{R}, x^2 < 0$, i.e., "There exists a real number $x$ such that $x^2 < 0$."
(Note: The original (iii) is true, so the negation must be false, which it is.)

2

Negate the statement: "For every real number $x$, there exists a real number $y$ such that $x + y = 0$."

Show solution

Symbolically: $\forall x \in \mathbb{R}, \exists y \in \mathbb{R}, x + y = 0$.
Negation: $\exists x \in \mathbb{R}, \forall y \in \mathbb{R}, x + y \neq 0$.
In words: "There exists a real number $x$ such that for every real number $y$, $x + y \neq 0$."
(The original is true with $y = -x$; the negation is false.)

3

Determine the truth value of the statement: "$\exists x \in \mathbb{R}, \forall y \in \mathbb{R}, x \cdot y = y$." Provide a justification.

Show solution

We need a single $x$ that works for ALL $y$. Try $x = 1$: $1 \cdot y = y$ for every real $y$. \checkmark
So the statement is True, with witness $x = 1$.

✎ Self-Check — 5 questions0 / 5
Q1.The negation of "All students are intelligent" is:
Q2.The negation of $\forall x \in \mathbb{R}, x^2 + 1 > 0$ is:
Q3.Which of the following is the symbolic form of "There exists a real number that is its own square root"?
Q4.The negation of "For some prime $p$, $p$ is even" is:
Q5.The statement "$\forall x \exists y$ such that $xy = 1$" is true when the domain is:

Methods of ProofTopic 2

A proof is a logical argument that establishes the truth of a mathematical statement using axioms, definitions, and previously proven results. Several proof techniques are essential for JEE:

1. Direct Proof: To prove $p \Rightarrow q$, assume $p$ is true, then deduce $q$ through a chain of logical implications.
Example: "If $n$ is odd, then $n^2$ is odd."
Assume $n = 2k + 1$. Then $n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$, which is odd. \checkmark

2. Proof by Contrapositive: To prove $p \Rightarrow q$, instead prove $\sim q \Rightarrow \sim p$, which is logically equivalent.
Useful when: The contrapositive is easier to manipulate. (E.g., when negating $q$ produces a more concrete algebraic condition.)

3. Proof by Contradiction (Reductio ad Absurdum): To prove $p$, assume $\sim p$ is true and derive a contradiction. Since $\sim p$ leads to a contradiction, $\sim p$ must be false, so $p$ is true.
Classic Example: Prove $\sqrt{2}$ is irrational. Assume $\sqrt{2} = p/q$ in lowest terms. Squaring: $2q^2 = p^2$, so $p^2$ is even, hence $p$ is even. Write $p = 2m$; then $2q^2 = 4m^2 \implies q^2 = 2m^2$, so $q$ is also even — contradicting the assumption that $p/q$ was in lowest terms. Therefore $\sqrt{2}$ is irrational.

4. Proof by Exhaustion (Case Analysis): Break the problem into a finite number of cases and prove each case separately.
Example: Show that for any integer $n$, $n^2 + n$ is even. Consider two cases:
  • Case 1: $n$ is even. Then $n^2$ is even, and even + even = even.
  • Case 2: $n$ is odd. Then $n^2$ is odd, and odd + odd = even.
5. Proof by Mathematical Induction: To prove $P(n)$ for all $n \geq n_0$:
  • Base Case: Prove $P(n_0)$.
  • Inductive Step: Assume $P(k)$ holds for some arbitrary $k \geq n_0$ (Inductive Hypothesis). Prove $P(k + 1)$.
Variants include strong induction (assume $P(j)$ for all $n_0 \leq j \leq k$) and infinite descent.

6. Counterexample: To disprove a universal statement $\forall x, P(x)$, provide a single $x$ for which $P(x)$ fails.
Example: "Every prime is odd" is disproved by the counterexample $p = 2$.

Validity of Arguments: An argument with premises $P_1, P_2, \ldots, P_n$ and conclusion $C$ is valid iff $(P_1 \wedge P_2 \wedge \cdots \wedge P_n) \Rightarrow C$ is a tautology. Key rules of inference:
  • Modus Ponens: From $p \Rightarrow q$ and $p$, infer $q$.
  • Modus Tollens: From $p \Rightarrow q$ and $\sim q$, infer $\sim p$.
  • Hypothetical Syllogism: From $p \Rightarrow q$ and $q \Rightarrow r$, infer $p \Rightarrow r$.
  • Disjunctive Syllogism: From $p \vee q$ and $\sim p$, infer $q$.
Worked Examples
1

Prove by mathematical induction that $1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$ for all positive integers $n$.

Show solution

Base Case ($n = 1$): LHS $= 1$, RHS $= \frac{1 \cdot 2}{2} = 1$. \checkmark
Inductive Step: Assume the formula holds for $n = k$: \[ 1 + 2 + \cdots + k = \frac{k(k+1)}{2} \quad \text{(Inductive Hypothesis)} \] We must show it holds for $n = k + 1$: \[ 1 + 2 + \cdots + k + (k + 1) = \frac{k(k+1)}{2} + (k+1) = (k+1)\left[\frac{k}{2} + 1\right] = (k+1) \cdot \frac{k+2}{2} = \frac{(k+1)(k+2)}{2} \] This matches the formula with $n = k + 1$. By the principle of mathematical induction, the formula holds for all $n \geq 1$. \qed

2

Prove by contradiction: $\log_2 3$ is irrational.

Show solution

Assume for contradiction that $\log_2 3$ is rational, say $\log_2 3 = \frac{p}{q}$ where $p, q$ are positive integers (since $\log_2 3 > 0$) with $\gcd(p, q) = 1$.
Then $2^{p/q} = 3 \implies 2^p = 3^q$.
But $2^p$ is even (a power of 2) while $3^q$ is odd (a power of 3). This is impossible for positive integers $p, q$.
Hence our assumption is false; $\log_2 3$ is irrational. \qed

3

Determine whether the following argument is valid: "If it rains, then the grass is wet. The grass is wet. Therefore, it rained."

Show solution

Let $p$: "it rains" and $q$: "the grass is wet."
Premises: $p \Rightarrow q$ and $q$. Conclusion: $p$.
This is the fallacy of affirming the consequent. We can construct a counterexample: it could be that the grass is wet because of a sprinkler, not rain. In truth-table terms, the conjunction $(p \Rightarrow q) \wedge q$ is true when $p = F, q = T$, but the conclusion $p$ would be false.
Conclusion: The argument is INVALID.

✎ Self-Check — 5 questions0 / 5
Q1.To prove "If $n^2$ is even, then $n$ is even" by contrapositive, we should prove:
Q2.The standard proof of "$\sqrt{2}$ is irrational" uses the method of:
Q3.The argument "$p \Rightarrow q$, $\sim q$, therefore $\sim p$" is an example of:
Q4.A counterexample to "Every continuous function is differentiable" is:
Q5.Which proof method is BEST suited to proving "For all $n \geq 1$, $n!$ exceeds $2^{n-2}$ when $n \geq 5$"?
4
Module 4

Validity of Arguments & Advanced Reasoning

Validity, Rules of Inference, and Argument FormsTopic 1

An argument is a sequence of statements (premises) followed by a statement (conclusion). The argument is valid if the truth of the premises guarantees the truth of the conclusion. Symbolically, if premises are $P_1, P_2, \ldots, P_n$ and conclusion is $C$, then validity requires: \[ (P_1 \wedge P_2 \wedge \cdots \wedge P_n) \Rightarrow C \quad \text{is a tautology} \]

Key Rules of Inference:

1. Modus Ponens (Affirming the Antecedent): \[ \frac{p \Rightarrow q, \; p}{\therefore q} \]

2. Modus Tollens (Denying the Consequent): \[ \frac{p \Rightarrow q, \; \sim q}{\therefore \sim p} \]

3. Hypothetical Syllogism (Chain Rule): \[ \frac{p \Rightarrow q, \; q \Rightarrow r}{\therefore p \Rightarrow r} \]

4. Disjunctive Syllogism: \[ \frac{p \vee q, \; \sim p}{\therefore q} \]

5. Constructive Dilemma: \[ \frac{(p \Rightarrow q) \wedge (r \Rightarrow s), \; p \vee r}{\therefore q \vee s} \]

6. Simplification: \[ \frac{p \wedge q}{\therefore p} \]

7. Conjunction: \[ \frac{p, \; q}{\therefore p \wedge q} \]

8. Addition: \[ \frac{p}{\therefore p \vee q} \]

Common Fallacies (Invalid Argument Forms):

Fallacy of Affirming the Consequent (Converse Fallacy): \[ \frac{p \Rightarrow q, \; q}{\therefore p} \quad \text{(INVALID)} \]

Fallacy of Denying the Antecedent (Inverse Fallacy): \[ \frac{p \Rightarrow q, \; \sim p}{\therefore \sim q} \quad \text{(INVALID)} \]

Checking Argument Validity Using Truth Tables: Construct a truth table for all premises and the conclusion. Look for rows where all premises are true. If the conclusion is also true in every such row, the argument is valid.

Validity vs. Soundness: An argument is valid if the conclusion logically follows from the premises (regardless of whether premises are true). It is sound if it is valid AND all premises are actually true.

Worked Examples
1

Determine if the following argument is valid:
"If I study hard, then I will pass. I did not pass. Therefore, I did not study hard."

Show solution

Let $p$: "I study hard." and $q$: "I will pass."
Premises: $p \Rightarrow q$ and $\sim q$. Conclusion: $\sim p$.
This is precisely Modus Tollens, a valid rule of inference. The argument is VALID.

2

Test the validity of the following argument using a truth table:
Premises: $p \vee q$, $p \Rightarrow r$, $q \Rightarrow r$. Conclusion: $r$.

Show solution

Construct the truth table: \[ \begin{array}{|c|c|c|c|c|c|c|} p & q & r & p \vee q & p \Rightarrow r & q \Rightarrow r & r \\ T & T & T & T & T & T & T \\ T & T & F & T & F & F & F \\ T & F & T & T & T & T & T \\ T & F & F & T & F & T & F \\ F & T & T & T & T & T & T \\ F & T & F & T & T & F & F \\ F & F & T & F & T & T & T \\ F & F & F & F & T & T & F \\ \end{array} \] Find rows where all three premises are TRUE: rows 1, 3, 5. In all these rows, the conclusion $r$ is also TRUE. Therefore the argument is VALID (this is the rule of constructive dilemma).

3

Show that the following argument is INVALID by giving a counterexample:
"If $x$ is divisible by 4, then $x$ is divisible by 2. The number $x = 6$ is divisible by 2. Therefore, $x = 6$ is divisible by 4."

Show solution

Let $p$: "$x$ is divisible by 4" and $q$: "$x$ is divisible by 2".
Premises: $p \Rightarrow q$ (true mathematically) and $q$ (true for $x = 6$). Conclusion: $p$.
This is the fallacy of affirming the consequent. The argument is INVALID. The number $x = 6$ is a counterexample: $q$ is true ($6$ is divisible by $2$), but $p$ is false ($6$ is not divisible by $4$).

✎ Self-Check — 5 questions0 / 5
Q1.The argument "$p \Rightarrow q$, $\sim p$, therefore $\sim q$" is an example of:
Q2.Which of the following is a valid argument form?
Q3.From the premises "$p \vee q$" and "$\sim q$", we can validly conclude:
Q4.Consider the argument: "If it is sunny, I will go for a walk. It is not sunny. Therefore, I will not go for a walk." This argument is:
Q5.Which fallacy is committed in: "If a student is in this class, then they are taking math. John is taking math. Therefore, John is in this class"?

Boolean Algebra Connections & Switching Circuits (Enrichment)Topic 2

Logic and Boolean algebra are equivalent algebraic systems. Each logical statement corresponds to a Boolean expression, where: \[ \text{TRUE} \longleftrightarrow 1, \quad \text{FALSE} \longleftrightarrow 0 \] \[ p \wedge q \longleftrightarrow p \cdot q \text{ (AND/multiplication)} \] \[ p \vee q \longleftrightarrow p + q \text{ (OR/addition with } 1 + 1 = 1\text{)} \] \[ \sim p \longleftrightarrow p' \text{ or } \bar{p} \text{ (NOT/complementation)} \]

Boolean Identities: Direct translations of the logical equivalences:
  • Idempotent: $p \cdot p = p$, $p + p = p$
  • Absorption: $p + (p \cdot q) = p$, $p \cdot (p + q) = p$
  • De Morgan: $(p \cdot q)' = p' + q'$, $(p + q)' = p' \cdot q'$
  • Complement: $p \cdot p' = 0$, $p + p' = 1$
Switching Circuits: A switch can be ON (closed, 1) or OFF (open, 0). Two basic configurations:
  • Series Configuration (current flows iff BOTH switches are closed): Models AND ($p \wedge q$ or $p \cdot q$).
  • Parallel Configuration (current flows iff AT LEAST ONE switch is closed): Models OR ($p \vee q$ or $p + q$).

A complex switching circuit is equivalent to a Boolean expression, and minimizing the circuit corresponds to simplifying the Boolean expression using algebraic identities (or methods like Karnaugh maps in computer engineering).

Practical Applications:
  • Logic gates in computers (AND, OR, NOT, NAND, NOR, XOR).
  • Truth tables for combinational circuits.
  • Simplifying networks of relays.

Exclusive OR (XOR): The Boolean function $p \oplus q$ outputs 1 iff exactly one of $p, q$ is 1. \[ p \oplus q = (p \cdot q') + (p' \cdot q) = \sim(p \Leftrightarrow q) \]

NAND and NOR Gates: The functions NAND $= \sim(p \wedge q)$ and NOR $= \sim(p \vee q)$ are called functionally complete — any logical function can be expressed using only NAND gates or only NOR gates.

Worked Examples
1

Simplify the Boolean expression $x \cdot y + x \cdot y' + x' \cdot y$.

Show solution

Factor $x$ from the first two terms: \[ x \cdot y + x \cdot y' + x' \cdot y = x(y + y') + x' \cdot y \] Apply the complement law $y + y' = 1$: \[ = x \cdot 1 + x' \cdot y = x + x' \cdot y \] Apply the absorption-related identity $x + x' \cdot y = x + y$ (proof: $x + x'y = (x + x')(x + y) = 1 \cdot (x + y) = x + y$): \[ = x + y \] Final Answer: $x + y$.

2

Express the truth table of the XOR function and write it as a Boolean expression.

Show solution

\[ \begin{array}{|c|c|c|} p & q & p \oplus q \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \] XOR outputs 1 in rows 2 and 3. By sum-of-products: \[ p \oplus q = p' \cdot q + p \cdot q' \] This is equivalent to $\sim(p \Leftrightarrow q)$.

3

Design a switching circuit corresponding to the Boolean expression $(p + q) \cdot (p' + r)$ and simplify if possible.

Show solution
Circuit: The expression $(p + q) \cdot (p' + r)$ consists of two parallel sub-circuits connected in series:
  • Sub-circuit 1 (parallel): switches $p$ and $q$ in parallel.
  • Sub-circuit 2 (parallel): switches $p'$ and $r$ in parallel.
  • These two are then connected in series.

Simplification by distribution: \[ (p + q)(p' + r) = p \cdot p' + p \cdot r + q \cdot p' + q \cdot r \] Since $p \cdot p' = 0$: \[ = 0 + p \cdot r + p' \cdot q + q \cdot r = pr + p'q + qr \] Use consensus: $pr + p'q + qr = pr + p'q$ (the consensus term $qr$ is absorbed).
Final Simplified Form: $pr + p'q$.

✎ Self-Check — 5 questions0 / 5
Q1.The Boolean expression $x + x \cdot y$ simplifies to:
Q2.Using De Morgan's law, $(x + y)'$ equals:
Q3.In a series-parallel switching circuit, two switches in parallel correspond to which logical operation?
Q4.The XOR function $p \oplus q$ is logically equivalent to:
Q5.Which gate is functionally complete (can implement any Boolean function on its own)?

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