Proof Techniques I: Direct, Contrapositive, and Contradiction

🎯 Learning Objectives

• Construct direct proofs by logical deduction from premises
• Use proof by contrapositive to prove conditional statements
• Apply proof by contradiction when direct approaches fail
• Recognise which technique is appropriate for different statements
• Write complete, well-structured proofs

3.1 Intuition: The Art of Mathematical Argument

🔑 Key Insight: Different proof techniques are different strategies for building an irrefutable case. The right technique depends on the structure of what you're trying to prove.

3.2 Direct Proof

Structure

1. Assume $P$ is true.
2. Use logical reasoning to deduce $Q$.
3. Conclude $P \to Q$.

Example 1: Sum of Even Numbers

• By definition, $m = 2a$ and $n = 2b$ for some integers $a, b$.
• Then $m + n = 2a + 2b = 2(a + b)$.
• Since $a + b$ is an integer, $m + n$ is even (divisible by 2). $\square$

Example 2: Product of Odd Numbers

• By definition, $m = 2a + 1$ and $n = 2b + 1$ for some integers $a, b$.
• $mn = (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1$.
• Since $2ab + a + b$ is an integer, $mn$ is odd. $\square$

Example 3: Divisibility

• By definition, there exist integers $k_1, k_2$ such that $b = ak_1$ and $c = bk_2$.
• Then $c = (ak_1)k_2 = a(k_1k_2)$.
• Since $k_1k_2$ is an integer, $a \mid c$. $\square$

3.3 Proof by Contrapositive

When to Use

Example 1: Squares and Parity

Example 2: Inequality

Example 3: Function Injectivity

3.4 Proof by Contradiction

Structure

1. Assume $\neg S$ (the negation of what you want to prove).
2. Reason logically until you reach a contradiction ($P \land \neg P$).
3. Conclude $S$ must be true.

When to Use

Example 1: Irrationality of $\sqrt{2}$

• Then $\sqrt{2} = \frac{a}{b}$ where $a, b$ are integers in lowest terms (coprime).
• Squaring: $2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$.
• So $a^2$ is even, hence $a$ is even (by contrapositive example above). Write $a = 2k$.
• Then $(2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies b^2 = 2k^2$.
• So $b^2$ is even, hence $b$ is even.
• But then $a$ and $b$ are both even, contradicting that $\frac{a}{b}$ was in lowest terms. Thus our assumption was false — $\sqrt{2}$ is irrational. $\square$

Example 2: Infinitely Many Primes

• Consider $N = p_1 \cdot p_2 \cdot \dots \cdot p_n + 1$.
• $N$ is greater than any $p_i$, so $N$ is not prime (by our assumption that the list is complete).
• Therefore $N$ must have a prime divisor $p$.
• $p$ must be one of $p_1, \dots, p_n$ (since we assumed these are all primes).
• But $p$ divides both $p_1 \cdots p_n$ and $N$, so $p$ divides $N - p_1 \cdots p_n = 1$.
• Contradiction: a prime cannot divide 1. Thus our assumption was false — there are infinitely many primes. $\square$

Example 3: Pigeonhole Principle

• Parity can be even or odd. By pigeonhole principle, among 3 numbers, at least two have the same parity (both even or both odd).
• The sum of two evens is even; the sum of two odds is even.
• Contradiction to our assumption. $\square$

3.5 Choosing the Right Technique

📊 Technique Comparison

✅ Practice Questions

Solution

We'll use contrapositive: If $n$ is odd, then $3n+2$ is odd. Assume $n$ is odd, so $n = 2k+1$. $3n+2 = 3(2k+1) + 2 = 6k + 3 + 2 = 6k + 5 = 2(3k+2) + 1$. Thus $3n+2$ is odd. By contrapositive, if $3n+2$ is even then $n$ is even. Q2: Prove that if $a$ and $b$ are rational numbers, then $a + b$ is rational. Solution

Direct proof: Let $a = \frac{p}{q}$ and $b = \frac{r}{s}$ where $p,q,r,s$ are integers, $q \neq 0$, $s \neq 0$. $a + b = \frac{p}{q} + \frac{r}{s} = \frac{ps + qr}{qs}$. Since $ps + qr$ and $qs$ are integers and $qs \neq 0$, $a+b$ is rational. Q3: Prove that $\sqrt[3]{2}$ is irrational. Solution

Assume $\sqrt[3]{2} = \frac{a}{b}$ in lowest terms. Then $2 = \frac{a^3}{b^3}$, so $a^3 = 2b^3$. Thus $a^3$ is even, so $a$ is even. Write $a = 2k$. Then $8k^3 = 2b^3 \implies 4k^3 = b^3$, so $b^3$ is even, so $b$ is even. Contradiction — $a$ and $b$ both even, but they were in lowest terms. Q4: Prove that between any two distinct rational numbers there is another rational number. Solution

Let $r < s$ be rational numbers. Their average $m = \frac{r+s}{2}$ is rational (sum of rationals divided by 2). And $r < m < s$ because $m - r = \frac{s-r}{2} > 0$ and $s - m = \frac{s-r}{2} > 0$. Q5: Prove that for any integer $n$, $n^2$ is divisible by 3 iff $n$ is divisible by 3. Solution

($\leftarrow$) If $n = 3k$, then $n^2 = 9k^2 = 3(3k^2)$, divisible by 3. ($\rightarrow$) We prove the contrapositive: if $n$ is not divisible by 3, then $n^2$ is not divisible by 3. If $n$ is not divisible by 3, then $n = 3k+1$ or $n = 3k+2$. Case 1: $n = 3k+1$, then $n^2 = 9k^2 + 6k + 1 = 3(3k^2+2k) + 1$, remainder 1 mod 3. Case 2: $n = 3k+2$, then $n^2 = 9k^2 + 12k + 4 = 3(3k^2+4k+1) + 1$, remainder 1 mod 3. In both cases, $n^2 \equiv 1 \pmod{3}$, so not divisible by 3. Join Discord PreviousQuantifier ApplicationsNextLogical Fallacies