Praxis Math Quiz: Identify Arithmetic Counterexamples
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Identify Arithmetic CounterexamplesQuestion 1 of 18

A mathematical statement claims: "For any two distinct prime numbers, p and q, their sum (p + q) is always an even number." Which of the following pairs of values for p and q provides a counterexample to this statement?

p = 3, q = 5
p = 2, q = 7
p = 7, q = 11
p = 1, q = 5
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Praxis Math Quiz

Praxis Math Quiz: Identify Arithmetic Counterexamples

Practice Identify Arithmetic Counterexamples in Praxis Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Identify Arithmetic Counterexamples, giving you a quick way to practice the rules, question types, and explanations that matter most for Praxis Math.

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Question 1

A mathematical statement claims: "For any two distinct prime numbers, p and q, their sum (p + q) is always an even number." Which of the following pairs of values for p and q provides a counterexample to this statement?

  1. p = 3, q = 5
  2. p = 2, q = 7 (correct answer)
  3. p = 7, q = 11
  4. p = 1, q = 5
Explanation: A counterexample must satisfy the conditions (p and q are distinct prime numbers) but violate the conclusion (their sum is even). The number 2 is the only even prime number. All other prime numbers are odd. The sum of two odd numbers is even. The statement fails only when one of the primes is 2. For p = 2 and q = 7, the sum is 2 + 7 = 9, which is an odd number. This disproves the statement. Choice A (3+5=8) and C (7+11=18) both result in even sums, supporting the statement. Choice D is incorrect because 1 is not a prime number, so it does not meet the conditions of the statement.

Question 2

A student proposes the following rule: "If an integer is divisible by 4 and is also divisible by 6, then it must be divisible by 24." Which of the following integers serves as a counterexample to this rule?

  1. 16
  2. 30
  3. 36 (correct answer)
  4. 48
Explanation: A counterexample must meet the conditions (divisible by 4 and 6) but fail the conclusion (not divisible by 24). The integer 36 is divisible by 4 (36/4=9) and by 6 (36/6=6), so it meets the conditions. However, 36 is not divisible by 24 (36/24 = 1.5). Therefore, 36 is a counterexample. Choice A, 16, is not divisible by 6. Choice B, 30, is not divisible by 4. Choice D, 48, is divisible by 4, 6, and 24, so it supports the rule.

Question 3

A student asserts, "For any two integers a and b, if the absolute value of a is greater than the absolute value of b, then a must be greater than b." Which pair of integers is a counterexample to this assertion?

  1. a = 5, b = 3
  2. a = -3, b = -5
  3. a = 5, b = -3
  4. a = -5, b = 3 (correct answer)
Explanation: A counterexample must satisfy the condition |a| > |b| but violate the conclusion a > b. For a = -5 and b = 3, the condition is |-5| > |3|, which simplifies to 5 > 3. This is true. The conclusion is -5 > 3, which is false. Therefore, this pair is a counterexample. Choices A and C support the assertion. In choice B, the condition |-3| > |-5| (or 3 > 5) is not met, so it cannot be a counterexample.

Question 4

A student conjectures: "The square root of any non-negative integer is always a rational number." Which of the following integers, when its square root is taken, serves as a counterexample to this conjecture?

  1. 0
  2. 1
  3. 8 (correct answer)
  4. 9
Explanation: The conjecture claims that for any integer x ≥ 0, √x is rational. A rational number can be expressed as a fraction of two integers. A counterexample would be a non-negative integer whose square root is irrational. Let's check the options: √0 = 0 (rational), √1 = 1 (rational), √9 = 3 (rational). These all support the conjecture. However, √8 = √(4×2) = 2√2. Since √2 is an irrational number, 2√2 is also irrational. Therefore, 8 is a counterexample. The conjecture is only true for perfect squares.

Question 5

A common statement is that "multiplying a number by 10 is equivalent to adding a zero to the end of the number." This works for integers. Which of the following numbers serves as a counterexample to show this rule does not always apply to all numbers?

  1. 450
  2. 4.5 (correct answer)
  3. -45
  4. 0.8
Explanation: A counterexample will be a number where multiplying by 10 does not result in simply appending a zero. For choice B, the number is 4.5. Multiplying by 10 gives 4.5 × 10 = 45. Adding a zero to the end of 4.5 gives 4.50, which equals 4.5, not 45. Thus, 4.5 is a counterexample. For integers like 450 and -45, the rule works correctly (4500 and -450). For 0.8, multiplying by 10 gives 8, while adding a zero gives 0.80 = 0.8.

Question 6

A statement about divisibility claims: "If the sum of the digits of an integer is divisible by 6, then the integer itself is divisible by 6." Which of the following integers is a counterexample to this claim?

  1. 18
  2. 24
  3. 36
  4. 33 (correct answer)
Explanation: A counterexample must satisfy the condition (sum of digits is divisible by 6) but violate the conclusion (the integer is not divisible by 6). For choice D: The integer 33 has digit sum 3 + 3 = 6, which is divisible by 6. However, 33 ÷ 6 = 5.5, so 33 is not divisible by 6. Thus, 33 is a counterexample. The other choices (18, 24, 36) are all divisible by 6 and support the statement rather than contradict it.

Question 7

A student generalizes from the facts 2³ = 8 and 3² = 9 to claim: "For any two distinct positive integers a and b, aᵇ is never equal to bᵃ." Which of the following pairs of values for a and b is a counterexample to this claim?

  1. a = 2, b = 3
  2. a = 1, b = 5
  3. a = 2, b = 4 (correct answer)
  4. a = 3, b = 1
Explanation: A counterexample must be a pair of distinct positive integers a and b for which aᵇ = bᵃ. Let's test choice C: a = 2, b = 4. The value of aᵇ is 2⁴ = 16. The value of bᵃ is 4² = 16. Since 16 = 16, this pair is a counterexample. For choice A, 2³ = 8 and 3² = 9. For choice B, 1⁵ = 1 and 5¹ = 5. For choice D, 3¹ = 3 and 1³ = 1. None of these show equality.

Question 8

Consider the statement: "For any two rational numbers a and b, if a < b, then a² < b²." Which pair of values for a and b is a counterexample to this statement?

  1. a = 2, b = 3
  2. a = -2, b = 3
  3. a = 0.2, b = 0.5
  4. a = -4, b = 3 (correct answer)
Explanation: A counterexample must satisfy the premise (a < b) but violate the conclusion (a² < b²). For a = -4 and b = 3, the premise -4 < 3 is true. However, a² = (-4)² = 16 and b² = 3² = 9. Since 16 is not less than 9, the conclusion a² < b² is false. Thus, this pair is a counterexample. For choice A, 2 < 3 and 4 < 9 (supports). For choice B, -2 < 3 and 4 < 9 (supports). For choice C, 0.2 < 0.5 and 0.04 < 0.25 (supports).

Question 9

A property of numbers states: "For any non-zero rational number x, the value of x² is always greater than x." Which of the following values for x is a counterexample to this property?

  1. x = 2
  2. x = -3
  3. x = 0.5 (correct answer)
  4. x = -0.5
Explanation: A counterexample must be a non-zero rational number (the condition) for which x² is not greater than x (violates the conclusion). For x = 0.5, x² = (0.5)² = 0.25. Since 0.25 is not greater than 0.5, this is a counterexample. For x = 2, x² = 4, and 4 > 2 (supports). For x = -3, x² = 9, and 9 > -3 (supports). For x = -0.5, x² = 0.25, and 0.25 > -0.5 (supports). The property fails for rational numbers between 0 and 1.

Question 10

Consider the statement: "The sum of any three consecutive integers is always an odd number." Which set of integers is a counterexample to this statement?

  1. {2, 3, 4}
  2. {3, 4, 5} (correct answer)
  3. {2, 4, 6}
  4. {-2, -1, 0}
Explanation: A counterexample must be a set of three consecutive integers whose sum is not odd (i.e., is even). For the set {3, 4, 5}, the integers are consecutive and their sum is 3 + 4 + 5 = 12. Since 12 is an even number, this set is a counterexample. For choice A, the sum is 2 + 3 + 4 = 9 (odd), which supports the statement. Choice C is not a set of consecutive integers. For choice D, the sum is -2 + (-1) + 0 = -3 (odd), which supports the statement.

Question 11

A mathematics principle states: "For any positive integer k, the expression k² + k + 41 generates a prime number." Which of the following values of k demonstrates that this principle is false?

  1. k = 1
  2. k = 10
  3. k = 39
  4. k = 40 (correct answer)
Explanation: We need to find a value of k for which k² + k + 41 is a composite (not prime) number. For k = 40, the expression is 40² + 40 + 41 = 1600 + 40 + 41 = 1681. We can also factor this: 40² + 40 + 41 = 40(40 + 1) + 41 = 40(41) + 41 = 41(40 + 1) = 41 × 41. Since 1681 has factors other than 1 and itself (namely 41), it is not a prime number. Therefore, k = 40 is a counterexample. For A, 1²+1+41=43 (prime). For B, 10²+10+41=151 (prime). For C, 39²+39+41 = 1521+39+41 = 1601 (which is prime).

Question 12

A student makes the common mistake of assuming that for any two positive numbers a and b, √(a + b) = √a + √b. Which pair of values provides a counterexample to this incorrect assumption?

  1. a = 25, b = 0
  2. a = 9, b = 16 (correct answer)
  3. a = 0, b = 49
  4. a = 1, b = 0
Explanation: A counterexample will show that √(a + b) is not equal to √a + √b. For a = 9, b = 16: √(a + b) = √(9 + 16) = √25 = 5. However, √a + √b = √9 + √16 = 3 + 4 = 7. Since 5 ≠ 7, this pair is a counterexample. In choices A, C, and D, one of the values is 0. In these special cases, the incorrect formula happens to work: for a=25, b=0, √25 = √25 + √0, which is 5=5. These are plausible distractors because they represent boundary conditions where the false statement appears true.

Question 13

Consider the statement: "If the product of two integers is positive, then both integers must be positive." Which pair of integers is a counterexample to this statement?

  1. 3 and 5
  2. -3 and -5 (correct answer)
  3. -3 and 5
  4. 3 and -5
Explanation: A counterexample must satisfy the condition (product is positive) but violate the conclusion (both integers are positive). For -3 and -5, the product is (-3) × (-5) = 15, which is positive. This satisfies the condition. However, both integers are negative, not positive, which violates the conclusion. Therefore, this is a counterexample. Choice A supports the statement. Choices C and D have negative products, so they do not satisfy the condition.

Question 14

Consider the statement: "If a number is a multiple of 9, it must also be a multiple of 6." Which of the following numbers is a counterexample to this statement?

  1. 12
  2. 18
  3. 27 (correct answer)
  4. 36
Explanation: A counterexample must satisfy the condition (be a multiple of 9) but violate the conclusion (not be a multiple of 6). The number 27 is a multiple of 9 (27 = 9 × 3). However, 27 is not a multiple of 6 (27 ÷ 6 = 4.5). Therefore, 27 is a counterexample. Choice A, 12, is a multiple of 6 but not of 9, so it doesn't meet the condition. Choices B (18) and D (36) are multiples of both 9 and 6, so they support the statement.

Question 15

A property of exponents is sometimes misremembered as (a - b)² = a² - b². Which pair of non-zero values for a and b provides a counterexample to this false statement?

  1. a = 3, b = 3
  2. a = 3, b = 0
  3. a = 4, b = 2 (correct answer)
  4. a = -2, b = -2
Explanation: We need to find a pair of non-zero values where (a - b)² is not equal to a² - b². Let's test C: a = 4, b = 2. The left side is (4 - 2)² = 2² = 4. The right side is 4² - 2² = 16 - 4 = 12. Since 4 ≠ 12, this is a counterexample. For choices A and D, a=b, so a-b=0 and (a-b)²=0. Also a²-b²=0. So these support the false statement. For choice B, b=0, which violates the 'non-zero' condition. The correct expansion is (a-b)² = a² - 2ab + b².

Question 16

Consider the claim: "When a positive number is divided by another positive number, the quotient is always smaller than the original number." Which calculation provides a counterexample to this claim?

  1. 10 ÷ 2 = 5
  2. 10 ÷ 0.5 = 20 (correct answer)
  3. 10 ÷ 20 = 0.5
  4. 10 ÷ 1 = 10
Explanation: The claim is that for a > 0 and b > 0, a ÷ b < a. A counterexample would be a case where a ÷ b ≥ a. In choice B, the original number is 10 and the quotient is 20. Since 20 is not smaller than 10, this is a counterexample. This occurs because dividing by a positive number between 0 and 1 results in a quotient larger than the original number. Choices A and C support the claim. Choice D results in a quotient equal to the original number, which also contradicts the 'always smaller' part of the claim, but B is a stronger counterexample showing the quotient can be larger.

Question 17

A geometric statement claims: "If a quadrilateral has four equal sides, then it must be a square." Which of the following shapes is a counterexample to this claim?

  1. A rectangle with sides 4 and 6
  2. A kite with adjacent sides of 3 and 5
  3. A rhombus with angles of 80° and 100° (correct answer)
  4. A parallelogram with sides 5 and 5
Explanation: A counterexample must meet the condition (four equal sides) but violate the conclusion (not a square). A square is a quadrilateral with four equal sides AND four right (90°) angles. A rhombus has four equal sides. If its angles are not 90°, it is not a square. A rhombus with angles of 80° and 100° has four equal sides but is not a square. Thus, it is a counterexample. Choice A (rectangle) and B (kite) do not have four equal sides. Choice D describes a rhombus, but it doesn't specify non-right angles, while C explicitly does, making it the clearest counterexample.

Question 18

A statement about inequalities reads: "For any three non-zero rational numbers a, b, and c, if a > b, then ac > bc." Which set of values provides a counterexample to this statement?

  1. a = 5, b = 2, c = 3
  2. a = 5, b = 2, c = -3 (correct answer)
  3. a = 2, b = 5, c = 3
  4. a = -2, b = -5, c = 3
Explanation: A counterexample must satisfy the condition (a > b) but violate the conclusion (ac > bc). For choice B, a = 5, b = 2, c = -3. The condition is 5 > 2, which is true. For the conclusion, we calculate ac = 5 × (-3) = -15 and bc = 2 × (-3) = -6. The conclusion ac > bc becomes -15 > -6, which is false. Thus, this set of values is a counterexample. The rule fails when multiplying an inequality by a negative number. Choices A and D support the statement. Choice C does not satisfy the condition a > b.