Proof

Consider the statement: "For any two consecutive integers, their sum is odd." Which working is a correct proof by deduction?

• ATry 3 + 4 = 7 and 10 + 11 = 21; both are odd, so the statement is true.
• BLet the integers be n and n + 1. Their sum is 2n + 1, which is one more than an even number, so the sum is odd.✓
• CLet the integers be n and n + 2. Their sum is 2n + 2 = 2(n + 1), which is even.
• DLet the integers be 2n and 2n + 1. Their sum is 4n + 1, so the sum is always prime.

Answer: Consecutive integers are n and n + 1 for some integer n. Their sum is n + (n + 1) = 2n + 1, which is an even number (2n) plus 1, hence odd. This is a general deductive argument. Testing 3 + 4 and 10 + 11 only checks examples rather than proving the statement; taking the integers as n and n + 2 uses non-consecutive integers; and writing them as 2n and 2n + 1 restricts the proof to pairs beginning with an even integer, then draws the false conclusion that 4n + 1 is always prime.

A student claims: "For every positive integer n, the value n² − n + 11 is a prime number." Which single value of n disproves this claim by counter-example?

• An = 5, giving 31
• Bn = 1, giving 11
• Cn = 11, giving 121✓
• Dn = 10, giving 101

Answer: A counter-example needs one n for which n² − n + 11 is not prime. At n = 11: 11² − 11 + 11 = 121 = 11 × 11, which is composite. So n = 11 disproves the claim. For n = 5 (31), n = 1 (11) and n = 10 (101) the result is prime, so they do not disprove it.

In a proof by contradiction that √2 is irrational, we assume √2 = a/b where a and b are integers with no common factor, and reach 2b² = a². What is the key contradiction that completes the proof?

• Aa and b are both even, so they share the factor 2, contradicting "no common factor"✓
• Ba² is odd, which is impossible for an integer
• Cb must equal a, so √2 = 1
• D2b² can never be a perfect square for any integers

Answer: From 2b² = a², a² is even so a is even; write a = 2k, giving 2b² = 4k², so b² = 2k² and b is also even. Then a and b share the factor 2, contradicting the assumption that a/b is in lowest terms. Hence no such a, b exist and √2 is irrational.

Which argument correctly proves by deduction that the sum of any two odd numbers is even?

• ASince 3 + 5 = 8 and 7 + 9 = 16 are even, the sum of two odd numbers is always even.
• BLet the numbers be 2m and 2n. Their sum is 2(m + n), which is even.
• CLet the numbers be 2m + 1 and 2n + 1. Their sum is 4mn + 1, which is odd.
• DLet the numbers be 2m + 1 and 2n + 1. Their sum is 2m + 2n + 2 = 2(m + n + 1), a multiple of 2, so it is even.✓

Answer: Any odd number can be written 2m + 1 (and 2n + 1) for integers m, n. The sum is 2m + 1 + 2n + 1 = 2(m + n + 1), which is divisible by 2 and therefore even. Checking 3 + 5 and 7 + 9 only tests examples rather than proving the general case; writing the numbers as 2m and 2n uses even numbers, not odd ones; and expanding the sum as 4mn + 1 is an algebra error, since the terms are added, not multiplied.

Prove the identity for all real n by selecting the line that correctly expands the left-hand side of (n + 3)² − (n − 3)².

• A(n² + 6n + 9) − (n² − 6n + 9) = 12n✓
• B(n² + 9) − (n² − 9) = 18
• C(n² + 6n + 9) − (n² − 6n − 9) = 12n + 18
• D(n² + 3n + 9) − (n² − 3n + 9) = 6n

Answer: Expand each square: (n + 3)² = n² + 6n + 9 and (n − 3)² = n² − 6n + 9. Subtracting: (n² + 6n + 9) − (n² − 6n + 9) = 12n. The other options drop the middle term, mis-sign the −9, or use 3n instead of 6n.

Euclid's proof that there are infinitely many primes assumes there are finitely many, p₁, p₂, …, pₙ, and considers N = p₁ × p₂ × … × pₙ + 1. Which statement gives the contradiction?

• AN is always a prime number larger than every pₙ.
• BN leaves remainder 1 when divided by each listed prime, so it has a prime factor not in the list (or is itself a new prime).✓
• CN is even, so 2 must have been left out of the list.
• DN is a perfect square, which no product of primes can be.

Answer: Dividing N by any listed prime pᵢ leaves remainder 1, so no listed prime divides N. Therefore N is either prime itself or has a prime factor outside the list — either way a prime not in the supposedly complete list exists, contradicting finiteness. Claiming that N must always be prime itself overstates the argument: N need only have a prime factor that is not on the list.

A student writes: "x = 2 satisfies x² = 4. Therefore x² = 4 implies x = 2." Why is this reasoning invalid?

• AIt ignores the other solution x = −2, since x² = 4 also holds when x = −2.✓
• BIt is valid; squaring both sides always gives a unique solution.
• CBecause 2² = 4 is false, so the premise is wrong.
• DBecause x² = 4 has no real solutions.

Answer: Verifying that x = 2 works shows it is one solution, but "x² = 4 ⟹ x = 2" is a stronger claim that fails because x = −2 also satisfies x² = 4. Confirming one case does not establish the statement for all cases — a common logical error.

Which of these is a valid counterexample to the statement 'n² > n for all integers n'?

• An = 2, since 2² = 4 > 2
• Bn = 1, since 1² = 1 which is not greater than 1✓
• Cn = −3, since (−3)² = 9 > −3
• Dn = 10, since 10² = 100 > 10

Answer: A counterexample must make the statement FALSE. At n = 1, n² = 1 and 1 > 1 is false, so the claim fails. The other three all satisfy the statement, so they support it rather than disprove it. One counterexample is enough to destroy a universal claim.

In a proof by contradiction that √2 is irrational, what is assumed at the start?

• AThat a and b share a common factor
• BThat √2 is irrational
• CThat √2 can be written as a fully simplified fraction a/b✓
• DThat a² is odd

Answer: Proof by contradiction assumes the OPPOSITE of what you want to show. So you assume √2 is rational — expressible as a/b in lowest terms — and then derive a contradiction, namely that a and b must both be even and so were not in lowest terms after all.

Prove or disprove: 'the sum of two consecutive integers is always odd'. Which reasoning is correct?

• AIncorrect: 2 + 3 = 5 is odd but 4 + 5 = 9 is odd, so it only sometimes holds
• BCorrect: n + (n + 1) = 2n + 1, which is odd for every integer n✓
• CCorrect, but only for positive integers
• DIncorrect, because 0 + 1 = 1 is not odd

Answer: Take the integers as n and n + 1. Their sum is 2n + 1, which is one more than an even number and therefore odd for every integer n, positive, negative or zero. An algebraic argument covers all cases at once, which is exactly why it beats checking examples.

What is required to disprove a statement of the form 'for all n, P(n)'?

• AA proof by induction covering every value of n from the base case upwards
• BOne counterexample✓
• CShowing P(n) fails for infinitely many different values of n
• DDemonstrating that P(n) is false for every single n without exception

Answer: A universal claim is destroyed by a single case where it fails. You do not need infinitely many, and you certainly do not need it to fail everywhere — one is enough, which is why counterexamples are the cheapest disproof available.

In proof by exhaustion, what must be shown?

• AThat the claim holds in a representative sample of the cases you have chosen to check
• BThat the claim holds in every possible case✓
• CThat assuming the opposite leads directly to a logical contradiction
• DThat the claim holds for n = 1 and then for n + 1 whenever it holds for n

Answer: Exhaustion splits the problem into finitely many cases and verifies each one. A sample is not enough — that is evidence, not proof. Assuming the opposite and reaching a contradiction is proof by contradiction; showing the case n = 1 and then n + 1 is induction.