Mathematical Reasoning • Topic 3 of 3

Validating Statements

Once we can read and combine statements, the central question of the topic is: how do we know a statement is true? Validating a statement means proving it, and the form of the statement decides the method. There are four standard techniques in the syllabus.

1. The direct method. To prove "if $p$ then $q$", assume $p$ is true and, through valid steps, deduce that $q$ is true. This is the most common method and matches the natural flow of an argument. To prove a "for all" statement you argue with an arbitrary (general) element so the conclusion holds for every case, not just an example.

$$\text{Direct: assume } p \text{ true } \longrightarrow \text{ deduce } q \text{ true } \longrightarrow p \Rightarrow q \text{ holds}$$

2. The method of contrapositive. Because $p \Rightarrow q$ is logically equivalent to $\sim q \Rightarrow \sim p$, you may instead assume $\sim q$ and deduce $\sim p$. This is invaluable when the negation $\sim q$ is easier to work with than $q$ itself — for example, statements about "odd/even" or "irrational" are often cleaner from the contrapositive end.

$$p \Rightarrow q \ \equiv\ \sim q \Rightarrow \sim p$$

3. The method of contradiction (reductio ad absurdum). To prove a statement $p$, assume the opposite — that $\sim p$ is true — and show this assumption forces a logical contradiction (something known to be false, such as $a = b$ and $a \ne b$ together). Since the assumption leads to an absurdity, it must be wrong, so $p$ is true. The classic example is the proof that $\sqrt{2}$ is irrational: assuming it is rational ($\sqrt{2} = p/q$ in lowest terms) leads to $p$ and $q$ both being even, contradicting "lowest terms".

$$\text{Assume } \sim p \longrightarrow \text{derive a contradiction} \longrightarrow \therefore p \text{ is true}$$

4. Disproof by counter-example. The previous three methods establish truth. To disprove a "for all" statement, you do not need a general argument — a single counter-example is enough, because one exception destroys a universal claim. Note the asymmetry: a counter-example can disprove but can never prove a "for all" statement, since checking finitely many cases never covers all of them.

GoalMethodYou assume…
Prove $p \Rightarrow q$Direct$p$, then deduce $q$
Prove $p \Rightarrow q$Contrapositive$\sim q$, then deduce $\sim p$
Prove $p$Contradiction$\sim p$, reach absurdity
Disprove $\forall$ claimCounter-examplenothing — exhibit one exception

For a biconditional "$p$ if and only if $q$", validity requires proving both directions separately: $p \Rightarrow q$ and $q \Rightarrow p$. Proving only one direction is a common and serious gap.

Deeper Insight — match the method to the shape of the claim: The skill being trained here is not memorising four recipes but diagnosing which one fits. Ask what kind of statement you face. A conditional invites the direct method first; if the conclusion is awkward but its negation is clean, switch to the contrapositive — they are the same proof seen from opposite ends. An existence or "no such object" claim, or anything where assuming the opposite gives you something concrete to manipulate, suits contradiction. And the moment you suspect a universal claim is false, stop trying to prove it and hunt for one counter-example. Two warnings carry most of the marks: a counter-example never proves a universal statement, and an "if and only if" demands both directions. Internalise the pairing between claim-shape and method, and validation stops being guesswork.

1
Worked Example
By the direct method, prove: "If $n$ is an odd integer, then $n^2$ is odd."
Solution
  1. Assume $p$: $n$ is odd, so $n = 2k + 1$ for some integer $k$.
  2. Then $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$.
  3. This is $2m + 1$ with $m = 2k^2 + 2k$, an integer — hence odd.

Answer: $n^2$ is odd, so the implication holds by the direct method.

2
Worked Example
Disprove by counter-example: "For every prime $p$, $p$ is odd."
Solution
  1. The claim is a "for all" statement, so one exception suffices.
  2. $p = 2$ is prime but even.

Answer: Counter-example $p = 2$ shows the statement is false.

3
Worked Example
Using the contrapositive, prove: "If $n^2$ is even, then $n$ is even."
Solution
  1. Contrapositive: "If $n$ is odd, then $n^2$ is odd."
  2. Assume $n$ odd: $n = 2k + 1$, so $n^2 = 2(2k^2 + 2k) + 1$ is odd.
  3. The contrapositive holds, so the original does too.

Answer: Proved by the method of contrapositive.

4
Worked Example
By contradiction, prove that $\sqrt{2}$ is irrational.
Solution
  1. Assume $\sim p$: $\sqrt{2}$ is rational, so $\sqrt{2} = \dfrac{a}{b}$ with $a,b$ integers, $b \ne 0$, in lowest terms.
  2. Then $2b^2 = a^2$, so $a^2$ is even, hence $a$ is even; write $a = 2c$.
  3. Then $2b^2 = 4c^2 \Rightarrow b^2 = 2c^2$, so $b$ is even too.
  4. But then $a$ and $b$ share the factor $2$, contradicting "lowest terms".

Answer: The assumption is absurd, so $\sqrt{2}$ is irrational.

5
Worked Example
Disprove: "For all real numbers $x$, $x^2 > x$."
Solution
  1. A "for all" claim falls to a single counter-example.
  2. Try $x = \tfrac{1}{2}$: $x^2 = \tfrac{1}{4}$, which is not greater than $\tfrac{1}{2}$.

Answer: Counter-example $x = \tfrac{1}{2}$ shows the statement is false.

6
Worked Example
By the direct method, prove: "The sum of two even integers is even."
Solution
  1. Let the integers be $a = 2m$ and $b = 2n$ for integers $m,n$.
  2. Then $a + b = 2m + 2n = 2(m + n)$.
  3. This is $2 \times$ an integer.

Answer: $a + b$ is even, proving the statement directly.

7
Worked Example
Prove by contradiction: "There is no greatest natural number."
Solution
  1. Assume the opposite: there is a greatest natural number $N$.
  2. Consider $N + 1$, which is also a natural number and $N + 1 > N$.
  3. This contradicts $N$ being the greatest.

Answer: The contradiction proves there is no greatest natural number.

8
Worked Example
Check the validity of: "If $x$ is a real number such that $x^3 + 4x = 0$, then $x = 0$." Use the direct method.
Solution
  1. Assume $x^3 + 4x = 0$.
  2. Factorise: $x(x^2 + 4) = 0$.
  3. For real $x$, $x^2 + 4 \ge 4 > 0$, so the only possibility is $x = 0$.

Answer: The statement is valid: $x = 0$ is forced.

9
Worked Example
Disprove: "For all integers $n$, $n^2 - n + 41$ is prime."
Solution
  1. A universal claim needs only one counter-example.
  2. Try $n = 41$: $41^2 - 41 + 41 = 41^2 = 1681 = 41 \times 41$, which is not prime.

Answer: Counter-example $n = 41$ gives $1681 = 41^2$, so the statement is false.

10
Worked Example
Prove the biconditional: "An integer $n$ is even if and only if $n^2$ is even."
Solution
  1. Forward ($n$ even $\Rightarrow n^2$ even): $n = 2k$ gives $n^2 = 4k^2 = 2(2k^2)$, even.
  2. Backward ($n^2$ even $\Rightarrow n$ even): use the contrapositive — $n$ odd $\Rightarrow n^2$ odd (shown earlier).
  3. Both directions hold.

Answer: Both $\Rightarrow$ are established, so the biconditional is valid.

Key Points

  • Direct method: to prove $p \Rightarrow q$, assume $p$ and deduce $q$.
  • Contrapositive: prove $\sim q \Rightarrow \sim p$ instead — equivalent to $p \Rightarrow q$, and often easier.
  • Contradiction (reductio ad absurdum): assume $\sim p$ and derive an absurdity, forcing $p$ true.
  • A single counter-example disproves a "for all" statement; it can never prove one.
  • Universal claims need a general argument; one example is not a proof.
  • A biconditional "$p$ iff $q$" requires proving both $p \Rightarrow q$ and $q \Rightarrow p$.
  • The classic contradiction proof shows $\sqrt{2}$ is irrational; the classic counter-example to "all primes are odd" is $p = 2$.
Tap an option to check your answer0 / 4
Q1.To prove "if $p$ then $q$" by the direct method you:
Explanation: The direct method assumes the hypothesis and deduces the conclusion.
Q2.The method of contradiction proves $p$ by assuming:
Explanation: Assume the negation, derive a contradiction, conclude the statement is true.
Q3.A counter-example can be used to:
Explanation: One exception disproves a universal claim; it cannot prove one.
Q4.Proving the contrapositive $\sim q \Rightarrow \sim p$ establishes:
Explanation: The contrapositive is logically equivalent to the original conditional.