Catalogue DP Mathematics Mathematics: Analysis and Approaches HL Number and algebra Proof

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Deductive proof and contradiction Practice

Build fluency in order: recall, procedure, linked reasoning, diagnosis and transfer. Check each answer for immediate feedback, then retry. Practice is low stakes.

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Practice explanations

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To prove the square of every even integer is even, how should you begin?

Let n = 2k with k an integer: This represents an arbitrary even integer.
Test n = 2,4,6: Those are only selected examples.
Assume n² is even: That assumes the conclusion.

This represents an arbitrary even integer.

For an integer n, prove that n(n+1) is even. Which approach covers all n?

Let n = 2k and stop after proving that case: This omits every odd input.
Check n = 1 and n = 2, then extend the pattern: These examples illustrate the two parities but do not prove the claim for arbitrary integers.
Split into n even and n odd; one of n,n+1 is even in each case: Consecutive integers have opposite parity, so their product contains an even factor.

Consecutive integers have opposite parity, so their product contains an even factor.

Give a positive integer counterexample to “every prime number is odd”.

Correct answer: 2

2 is prime and even.

For integers m,n, why is $5m+5n$ divisible by 5?

It equals 5(m+n), and m+n is an integer: Factoring produces the required integer multiple.
Because m and n must be positive: No positivity assumption is needed.
Because the sum is necessarily odd: Its parity can vary.

Factoring produces the required integer multiple.

Which claim can be disproved by one counterexample?

There exists an even integer: One odd example does not rule out an even integer.
Some primes are odd: An even prime does not rule out odd primes.
Every integer square is even: A single odd square refutes this universal statement.

A single odd square refutes this universal statement.

To prove √2 is irrational by contradiction, which assumption is appropriate?

$\sqrt2=p/q$ with integers p,q, then assume both are even: Evenness must be derived. It only contradicts lowest terms if a reduced representation was chosen.
$\sqrt2=p/q$ in lowest terms, with integers p,q and q nonzero: The negation of irrationality is rationality; lowest terms will matter.
$\sqrt2=p/q$ with positive real p,q: Every positive real number has such a ratio; rationality specifically requires integers.

The negation of irrationality is rationality; lowest terms will matter.

In a contradiction proof, assume √2 = p/q in lowest terms. Both p and q turn out to be even. Why is that a contradiction?

They share a factor 2 despite being in lowest terms: It contradicts the reduced-fraction assumption.
Every fraction with an even numerator is irrational: For example, 2/3 is rational; the issue is a common factor in an allegedly reduced fraction.
p and q must both be odd in every rational representation: Reduced fractions can have one even entry. They simply cannot both share factor 2.

It contradicts the reduced-fraction assumption.

Suppose r is the least positive real number. Which value contradicts that?

$r+1$: A least value may have larger values.
$-r$: This leaves the set of positive real numbers.
$r/3$: r/3 is still real, positive and smaller than r.

r/3 is still real, positive and smaller than r.

Enter an integer n that disproves “n² > n for every nonnegative integer n”, using the smallest allowed n.

Correct answer: 0

At n = 0, both sides are zero, so the strict inequality fails.

A proof assumes a largest negative integer N and uses N+1. What is wrong?

N+1 need not be negative; −1 is actually the largest negative integer: The proposed larger value can leave the specified set.
N + 1 is larger, so the argument proves the claim: Being larger is not enough; it must also remain a negative integer.
Choose N + 1/2 instead; it is still an integer: Adding 1/2 leaves the integers, so this cannot provide a member of the specified set.

The proposed larger value can leave the specified set.

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Deductive proof and contradiction Practice

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