Mathematical Reasoning Formulas for JEE 2026, Complete PDF

Worked Example: Negation

Write the negation of: "Every natural number is greater than zero."

Negation: "There exists a natural number that is not greater than zero."

Note how "every" changed to "there exists" and "greater than" changed to "not greater than." This is a general pattern we'll see with quantifiers.

Worked Example: Conjunction

Let $$p$$: "$$2$$ is even" (T) and $$q$$: "$$3$$ is even" (F). Find the truth value of $$p \land q$$.

$$p \land q$$: "$$2$$ is even AND $$3$$ is even."

Since $$q$$ is false, $$p \land q$$ is false.

Worked Example: Conditional

Determine the truth value of: "If $$4 > 5$$, then $$10 > 9$$."

Here $$p$$: "$$4 > 5$$" is false. Since the hypothesis is false, $$p \Rightarrow q$$ is true (vacuously true), regardless of $$q$$.

Worked Example: Converse, Inverse, and Contrapositive

Given: "If a number is divisible by 6, then it is divisible by 3."

Converse: "If a number is divisible by 3, then it is divisible by 6." (False: 9 is divisible by 3 but not 6.)

Inverse: "If a number is not divisible by 6, then it is not divisible by 3." (False: same counterexample.)

Contrapositive: "If a number is not divisible by 3, then it is not divisible by 6." (True — same truth value as the original.)

Worked Example: Verifying a Tautology

Show that $$p \lor (\sim p)$$ is a tautology using a truth table.

The last column is always T, so $$p \lor (\sim p)$$ is a tautology.

Worked Example: Is $$(p \Rightarrow q) \lor (q \Rightarrow p)$$ a Tautology?

Always T — so yes, it is a tautology.

Worked Example: Negating a Quantified Statement

Negate: "All prime numbers are odd."

The statement is: $$\forall x$$ (if $$x$$ is prime, then $$x$$ is odd).

Negation: $$\exists x$$ such that $$x$$ is prime and $$x$$ is not odd.

In plain English: "There exists a prime number that is not odd." (This is actually true — the number 2.)

Worked Example: Direct Proof

Prove: "If $$n$$ is even, then $$n^2$$ is even."

Assume $$n$$ is even. By definition, $$n = 2k$$ for some integer $$k$$.

Then $$n^2 = (2k)^2 = 4k^2 = 2(2k^2)$$.

Since $$2k^2$$ is an integer, $$n^2 = 2 \times (\text{integer})$$, so $$n^2$$ is even. $$\square$$

Worked Example: Proof by Contradiction

Prove that $$\sqrt{2}$$ is irrational.

Assume $$\sqrt{2}$$ is rational. Then $$\sqrt{2} = \dfrac{a}{b}$$ where $$a, b$$ are integers with no common factor (i.e., the fraction is in lowest terms) and $$b \neq 0$$.

Squaring: $$2 = \dfrac{a^2}{b^2}$$, so $$a^2 = 2b^2$$.

This means $$a^2$$ is even, so $$a$$ must be even. Write $$a = 2k$$.

Then $$(2k)^2 = 2b^2 \Rightarrow 4k^2 = 2b^2 \Rightarrow b^2 = 2k^2$$.

So $$b^2$$ is even, meaning $$b$$ is also even.

Contradiction: Both $$a$$ and $$b$$ are even, so they share the common factor 2. But we assumed $$a/b$$ was in lowest terms.

Therefore, $$\sqrt{2}$$ is irrational. $$\square$$

Worked Example: Proof by Contrapositive

Prove: "If $$n^2$$ is even, then $$n$$ is even."

Contrapositive: "If $$n$$ is not even (i.e., $$n$$ is odd), then $$n^2$$ is not even (i.e., $$n^2$$ is odd)."

Assume $$n$$ is odd: $$n = 2k + 1$$ for some integer $$k$$.

$$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$$

Since $$2k^2 + 2k$$ is an integer, $$n^2$$ is odd. $$\square$$

Worked Example: Sum of First $$n$$ Natural Numbers

Prove by induction: $$1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$$ for all $$n \in \mathbb{N}$$.

Let $$P(n)$$: $$\displaystyle\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}$$.

Base case ($$n = 1$$): LHS $$= 1$$. RHS $$= \dfrac{1 \times 2}{2} = 1$$. LHS $$=$$ RHS. ✓

Inductive hypothesis: Assume $$P(k)$$ is true: $$1 + 2 + \cdots + k = \dfrac{k(k+1)}{2}$$.

Inductive step: Prove $$P(k+1)$$: $$1 + 2 + \cdots + k + (k+1) = \dfrac{(k+1)(k+2)}{2}$$.

LHS $$= \underbrace{1 + 2 + \cdots + k}_{\text{use } P(k)} + (k+1) = \dfrac{k(k+1)}{2} + (k+1)$$

$$= (k+1)\left(\dfrac{k}{2} + 1\right) = (k+1) \cdot \dfrac{k+2}{2} = \dfrac{(k+1)(k+2)}{2}$$ = RHS. ✓

By PMI, $$P(n)$$ is true for all $$n \in \mathbb{N}$$. $$\square$$

Worked Example: Sum of Squares

Prove by induction: $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$$.

Let $$P(n)$$: $$\displaystyle\sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}$$.

Base case ($$n = 1$$): LHS $$= 1$$. RHS $$= \dfrac{1 \times 2 \times 3}{6} = 1$$. ✓

Inductive hypothesis: Assume $$P(k)$$: $$1^2 + 2^2 + \cdots + k^2 = \dfrac{k(k+1)(2k+1)}{6}$$.

Inductive step: Prove $$P(k+1)$$.

LHS $$= \dfrac{k(k+1)(2k+1)}{6} + (k+1)^2 = (k+1)\left[\dfrac{k(2k+1)}{6} + (k+1)\right]$$

$$= (k+1) \cdot \dfrac{k(2k+1) + 6(k+1)}{6} = (k+1) \cdot \dfrac{2k^2 + k + 6k + 6}{6}$$

$$= (k+1) \cdot \dfrac{2k^2 + 7k + 6}{6} = (k+1) \cdot \dfrac{(k+2)(2k+3)}{6}$$

$$= \dfrac{(k+1)(k+2)(2k+3)}{6} = \dfrac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$ = RHS. ✓

By PMI, the result holds for all $$n \in \mathbb{N}$$. $$\square$$

Worked Example: $$7^n - 1$$ is Divisible by 6

Prove by induction: $$7^n - 1$$ is divisible by $$6$$ for all $$n \in \mathbb{N}$$.

Let $$P(n)$$: $$6 \mid (7^n - 1)$$.

Base case ($$n = 1$$): $$7^1 - 1 = 6$$, which is divisible by 6. ✓

Inductive hypothesis: Assume $$P(k)$$: $$7^k - 1 = 6m$$ for some integer $$m$$.

Inductive step: Prove $$6 \mid (7^{k+1} - 1)$$.

$$7^{k+1} - 1 = 7 \cdot 7^k - 1 = 7(7^k - 1) + 7 - 1 = 7(7^k - 1) + 6$$

By the inductive hypothesis, $$7^k - 1 = 6m$$, so:

$$7^{k+1} - 1 = 7(6m) + 6 = 6(7m + 1)$$

Since $$7m + 1$$ is an integer, $$7^{k+1} - 1$$ is divisible by 6. ✓

By PMI, $$6 \mid (7^n - 1)$$ for all $$n \in \mathbb{N}$$. $$\square$$

Worked Example: $$n^3 - n$$ is Divisible by 3

Prove by induction: $$n^3 - n$$ is divisible by $$3$$ for all $$n \in \mathbb{N}$$.

Let $$P(n)$$: $$3 \mid (n^3 - n)$$.

Base case ($$n = 1$$): $$1^3 - 1 = 0$$, and $$3 \mid 0$$. ✓

Inductive hypothesis: Assume $$P(k)$$: $$k^3 - k = 3m$$ for some integer $$m$$.

Inductive step: Prove $$3 \mid ((k+1)^3 - (k+1))$$.

$$(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1$$

$$= (k^3 - k) + 3k^2 + 3k = (k^3 - k) + 3(k^2 + k)$$

By inductive hypothesis, $$k^3 - k = 3m$$, so:

$$(k+1)^3 - (k+1) = 3m + 3(k^2 + k) = 3(m + k^2 + k)$$

Since $$m + k^2 + k$$ is an integer, the expression is divisible by 3. ✓

By PMI, $$3 \mid (n^3 - n)$$ for all $$n \in \mathbb{N}$$. $$\square$$

Worked Example: $$2^n > n$$

Prove by induction: $$2^n > n$$ for all $$n \in \mathbb{N}$$.

Let $$P(n)$$: $$2^n > n$$.

Base case ($$n = 1$$): $$2^1 = 2 > 1$$. ✓

Inductive hypothesis: Assume $$P(k)$$: $$2^k > k$$ for some $$k \geq 1$$.

Inductive step: Prove $$2^{k+1} > k + 1$$.

$$2^{k+1} = 2 \cdot 2^k > 2 \cdot k = 2k$$ (using the inductive hypothesis)

Now we need $$2k \geq k + 1$$, i.e., $$k \geq 1$$, which is true for all $$k \in \mathbb{N}$$.

So $$2^{k+1} > 2k \geq k + 1$$, giving $$2^{k+1} > k + 1$$. ✓

By PMI, $$2^n > n$$ for all $$n \in \mathbb{N}$$. $$\square$$

Worked Example: $$n! > 2^n$$ for $$n \geq 4$$

Prove by induction: $$n! > 2^n$$ for all $$n \geq 4$$.

Let $$P(n)$$: $$n! > 2^n$$ for $$n \geq 4$$.

Base case ($$n = 4$$): $$4! = 24$$ and $$2^4 = 16$$. Since $$24 > 16$$, $$P(4)$$ is true. ✓

Inductive hypothesis: Assume $$P(k)$$: $$k! > 2^k$$ for some $$k \geq 4$$.

Inductive step: Prove $$(k+1)! > 2^{k+1}$$.

$$(k+1)! = (k+1) \cdot k! > (k+1) \cdot 2^k$$ (using the inductive hypothesis)

Since $$k \geq 4$$, we have $$k + 1 \geq 5 > 2$$, so:

$$(k+1) \cdot 2^k > 2 \cdot 2^k = 2^{k+1}$$

Therefore $$(k+1)! > 2^{k+1}$$. ✓

By PMI, $$n! > 2^n$$ for all $$n \geq 4$$. $$\square$$