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. A only checks examples (not a proof), C uses non-consecutive integers, and D both mislabels consecutive integers and makes a false claim.

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. A only tests examples, B uses even numbers, and C contains an algebra error.

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. (N need not itself be prime, so A is wrong.)

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.