IB Mathematics: Analysis and Approaches Quiz: Proof
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ProofQuestion 1 of 20

A student wants to prove that the function f(x)=x3−xf(x) = x^3 - x is an odd function. The definition of an odd function is that f(−x)=−f(x)f(-x) = -f(x) for all xx in the domain. Which line of reasoning correctly proves this?

f(−x)=(−x)3−x=−x3−x=−(x3+x)f(-x) = (-x)^3 - x = -x^3 - x = -(x^3+x), which is not equal to −f(x)-f(x).
f(−x)=(−x)3−(−x)=−x3+x=−(x3−x)=−f(x)f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3-x) = -f(x).
Since f(2)=6f(2)=6 and f(−2)=−6f(-2)=-6, we have f(−2)=−f(2)f(-2)=-f(2), so the function is odd.
−f(x)=−(x3)−(−x)=−x3+x-f(x) = -(x^3) - (-x) = -x^3+x. This is not the standard way to start the proof.
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Proof

Practice Proof in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Proof, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A student wants to prove that the function f(x)=x3−xf(x) = x^3 - x is an odd function. The definition of an odd function is that f(−x)=−f(x)f(-x) = -f(x) for all xx in the domain. Which line of reasoning correctly proves this?

  1. f(−x)=(−x)3−x=−x3−x=−(x3+x)f(-x) = (-x)^3 - x = -x^3 - x = -(x^3+x), which is not equal to −f(x)-f(x).
  2. f(−x)=(−x)3−(−x)=−x3+x=−(x3−x)=−f(x)f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3-x) = -f(x). (correct answer)
  3. Since f(2)=6f(2)=6 and f(−2)=−6f(-2)=-6, we have f(−2)=−f(2)f(-2)=-f(2), so the function is odd.
  4. −f(x)=−(x3)−(−x)=−x3+x-f(x) = -(x^3) - (-x) = -x^3+x. This is not the standard way to start the proof.
Explanation: To prove a function is odd, we must start with f(−x)f(-x) and show through algebraic manipulation that it is equivalent to −f(x)-f(x). For f(x)=x3−xf(x) = x^3 - x: Step 1: Write out f(−x)f(-x) by substituting −x-x for xx. f(−x)=(−x)3−(−x)f(-x) = (-x)^3 - (-x) Step 2: Simplify the expression. f(−x)=−x3+xf(-x) = -x^3 + x Step 3: Factor out -1 to see if the remaining expression is f(x)f(x). f(−x)=−(x3−x)f(-x) = -(x^3 - x) Step 4: Recognize that the expression in the parentheses is f(x)f(x). f(−x)=−f(x)f(-x) = -f(x). This matches the reasoning in option B. A contains an algebraic error in substitution: −x-x was not substituted for the second xx. C shows verification for a single value, which is not a general proof. D calculates −f(x)-f(x) but presents the logic awkwardly. The standard deductive flow is from f(−x)f(-x) to −f(x)-f(x) as shown in B.

Question 2

A student proves that the angles in a triangle sum to 180∘180^\circ by drawing a line through one vertex parallel to the opposite side. The proof relies on properties of angles formed by a transversal intersecting parallel lines. Which of the following is a fundamental, unstated assumption or axiom required for this proof to be valid?

  1. The Pythagorean theorem.
  2. The triangle is an isosceles triangle.
  3. Euclid's parallel postulate. (correct answer)
  4. The sum of two sides of a triangle must be greater than the third side.
Explanation: The proof described is the classic Euclidean proof of the angle sum of a triangle. This proof relies on constructing a line through a vertex parallel to the opposite side and then using the fact that alternate interior angles are equal. The existence and uniqueness of such a parallel line are not self-evident but are guaranteed by Euclid's fifth postulate, known as the parallel postulate. Without this axiom, the geometry might be non-Euclidean (e.g., spherical or hyperbolic), and the sum of angles in a triangle would not be 180∘180^\circ. A, the Pythagorean theorem, applies only to right-angled triangles. B is incorrect as the proof is general for all triangles. D is the triangle inequality for side lengths and is not directly used in this angle sum proof.

Question 3

A student presents the following "proof" that 2=12=1. Line 1: Let a=ba=b and a≠0a \neq 0. Line 2: a2=aba^2 = ab (multiply both sides by aa) Line 3: a2−b2=ab−b2a^2 - b^2 = ab - b^2 (subtract b2b^2 from both sides) Line 4: (a−b)(a+b)=b(a−b)(a-b)(a+b) = b(a-b) (factor both sides) Line 5: a+b=ba+b = b (divide both sides by a−ba-b) Line 6: b+b=bb+b = b (substitute a=ba=b) Line 7: 2b=b2b = b Line 8: 2=12 = 1 (divide both sides by bb)

In which line does the first error in reasoning occur?

  1. Line 3
  2. Line 4
  3. Line 5 (correct answer)
  4. Line 6
Explanation: The error occurs in Line 5. The step involves dividing both sides by (a−b)(a-b). However, the initial assumption in Line 1 is that a=ba=b. This means that a−b=0a-b = 0. Division by zero is an undefined operation in mathematics. Therefore, this step is invalid and leads to the false conclusion. Lines 1 through 4 contain valid algebraic manipulations. Line 6 is a valid substitution based on the initial premise. The error has already occurred in Line 5.

Question 4

A student is asked to prove the identity 1n−1n+2≡2n(n+2)\frac{1}{n} - \frac{1}{n+2} \equiv \frac{2}{n(n+2)} for n∈Z,n≠0,−2n \in \mathbb{Z}, n \neq 0, -2. Which of the following represents a correct first step in a proof starting from the left-hand side (LHS)?

  1. Combine the fractions using a common denominator: (n+2)−nn(n+2)\frac{(n+2) - n}{n(n+2)} (correct answer)
  2. Multiply both sides by n(n+2)n(n+2): n+2−n=2n+2 - n = 2
  3. Assume the identity is true and work backwards from the RHS: 2n2+2n\frac{2}{n^2+2n}
  4. Substitute a test value, for example n=1n=1: 1−13=231 - \frac{1}{3} = \frac{2}{3}
Explanation: To prove an identity by starting with the left-hand side (LHS), we must manipulate it algebraically until it is identical to the right-hand side (RHS). The LHS is 1n−1n+2\frac{1}{n} - \frac{1}{n+2}. The correct first step is to combine the two fractions by finding a common denominator, which is n(n+2)n(n+2). This gives 1(n+2)n(n+2)−1(n)n(n+2)=(n+2)−nn(n+2)\frac{1(n+2)}{n(n+2)} - \frac{1(n)}{n(n+2)} = \frac{(n+2)-n}{n(n+2)}. This simplifies to 2n(n+2)\frac{2}{n(n+2)}, which is the RHS. Thus, A is the correct first step. B is incorrect because multiplying both sides of a proposed identity assumes the identity is true. This is a method for solving equations, not proving identities. C is incorrect because a formal proof should proceed from one side to the other (e.g., LHS to RHS), not by assuming the result. D is incorrect because substituting a single value only verifies the statement for that specific value; it does not constitute a general proof.

Question 5

To prove the identity sin⁡θ1−cos⁡θ≡1+cos⁡θsin⁡θ\frac{\sin \theta}{1-\cos \theta} \equiv \frac{1+\cos \theta}{\sin \theta}, a valid first step after starting with the LHS, sin⁡θ1−cos⁡θ\frac{\sin \theta}{1-\cos \theta}, would be to:

  1. Multiply the numerator and denominator by sin⁡θ\sin \theta.
  2. Multiply the numerator and denominator by 1+cos⁡θ1+\cos \theta. (correct answer)
  3. Cross-multiply to get sin⁡2θ=(1−cos⁡θ)(1+cos⁡θ)\sin^2 \theta = (1-\cos \theta)(1+\cos \theta).
  4. Replace 11 with sin⁡2θ+cos⁡2θ\sin^2 \theta + \cos^2 \theta.
Explanation: When proving trigonometric identities involving fractions with a binomial term like (1−cos⁡θ)(1-\cos \theta) in the denominator, a common and effective strategy is to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of (1−cos⁡θ)(1-\cos \theta) is (1+cos⁡θ)(1+\cos \theta). This step is valid because it is equivalent to multiplying by 1. LHS = sin⁡θ1−cos⁡θ×1+cos⁡θ1+cos⁡θ=sin⁡θ(1+cos⁡θ)1−cos⁡2θ\frac{\sin \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta} = \frac{\sin \theta(1+\cos \theta)}{1-\cos^2 \theta}. Using the identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, we know 1−cos⁡2θ=sin⁡2θ1-\cos^2 \theta = \sin^2 \theta. The expression becomes sin⁡θ(1+cos⁡θ)sin⁡2θ=1+cos⁡θsin⁡θ\frac{\sin \theta(1+\cos \theta)}{\sin^2 \theta} = \frac{1+\cos \theta}{\sin \theta} = RHS. A leads to an expression that is difficult to simplify to the RHS. C assumes the identity is true, which is what we need to prove. D makes the expression more complex and is not a helpful step.

Question 6

A student proves that the sum of any two odd numbers is even. Let the two odd numbers be 2k+12k+1 and 2m+12m+1, where k,mk, m are integers. Step 1: Sum = (2k+1)+(2m+1)(2k+1) + (2m+1) Step 2: Sum = 2k+2m+22k + 2m + 2 Step 3: Sum = 2(k+m+1)2(k+m+1) Step 4: Since kk and mm are integers, k+m+1k+m+1 is an integer. Step 5: Therefore, the sum is a multiple of 2, so it is even.

The student then tries to prove the converse: the sum of two integers is even only if both integers are odd. Which of the following provides a counterexample to this converse statement?

  1. The integers are 3 and 5.
  2. The integers are 2 and 4. (correct answer)
  3. The integers are 2 and 3.
  4. The integers are -1 and 1.
Explanation: The converse statement is: "If the sum of two integers is even, then both integers are odd." To find a counterexample, we need to find two integers whose sum is even, but which are not both odd. Let's check the options: A: The integers are 3 and 5. The sum is 8 (even). Both integers are odd. This supports the statement, so it is not a counterexample. B: The integers are 2 and 4. The sum is 6 (even). The integers are not both odd (in fact, both are even). This fits the criteria for a counterexample: the premise (sum is even) is true, but the conclusion (both integers are odd) is false. C: The integers are 2 and 3. The sum is 5 (odd). This does not satisfy the premise of the statement, so it cannot be a counterexample. D: The integers are -1 and 1. The sum is 0 (even). Both integers are odd. This supports the statement.

Question 7

A student makes the claim: "If f′(c)=0f'(c) = 0, then the function f(x)f(x) must have a local maximum or a local minimum at x=cx=c". Which of the following functions serves as a counterexample to this claim?

  1. f(x)=x2f(x) = x^2 at c=0c=0
  2. f(x)=∣x∣f(x) = |x| at c=0c=0
  3. f(x)=x3f(x) = x^3 at c=0c=0 (correct answer)
  4. f(x)=exf(x) = e^x at c=0c=0
Explanation: To find a counterexample, we need a function where f′(c)=0f'(c) = 0 is true, but the conclusion (local max/min at x=cx=c) is false. A: For f(x)=x2f(x) = x^2, f′(x)=2xf'(x) = 2x. At c=0c=0, f′(0)=0f'(0) = 0. There is a local minimum at x=0x=0. This supports the claim. B: For f(x)=∣x∣f(x) = |x|, the derivative is undefined at c=0c=0. So the premise f′(c)=0f'(c)=0 is not met. C: For f(x)=x3f(x) = x^3, f′(x)=3x2f'(x) = 3x^2. At c=0c=0, f′(0)=0f'(0) = 0. The point x=0x=0 is a stationary point of inflection, not a local maximum or minimum. The function is increasing for x<0x<0 and for x>0x>0. This is a counterexample. D: For f(x)=exf(x) = e^x, f′(x)=exf'(x) = e^x. At c=0c=0, f′(0)=e0=1≠0f'(0) = e^0 = 1 \neq 0. The premise is not met.

Question 8

Consider the statement (x−3)2=x2−6x+9(x-3)^2 = x^2 - 6x + 9. Which of the following best describes this statement?

  1. An equation that is true only for x=3x=3.
  2. An identity that is true for all x∈Rx \in \mathbb{R}. (correct answer)
  3. An inequality that is true for all x≥3x \geq 3.
  4. An equation that has no real solutions.
Explanation: An identity is a statement of equality that is true for all possible values of the variable(s) for which the expressions are defined. By expanding the left-hand side, (x−3)2=(x−3)(x−3)=x2−3x−3x+9=x2−6x+9(x-3)^2 = (x-3)(x-3) = x^2 - 3x - 3x + 9 = x^2 - 6x + 9. Since the left-hand side is algebraically equivalent to the right-hand side, the statement is true for all real numbers xx. Therefore, it is an identity. A is incorrect; it confuses an identity with a conditional equation. While the statement is true for x=3x=3, it is not only true for x=3x=3. C is incorrect as the statement is an equality, not an inequality. D is incorrect as the statement is true for all real numbers.

Question 9

A student presents the following "proof" that 2=12=1. Line 1: Let a=ba=b and a≠0a \neq 0. Line 2: a2=aba^2 = ab (multiply both sides by aa) Line 3: a2−b2=ab−b2a^2 - b^2 = ab - b^2 (subtract b2b^2 from both sides) Line 4: (a−b)(a+b)=b(a−b)(a-b)(a+b) = b(a-b) (factor both sides) Line 5: a+b=ba+b = b (divide both sides by a−ba-b) Line 6: b+b=bb+b = b (substitute a=ba=b) Line 7: 2b=b2b = b Line 8: 2=12 = 1 (divide both sides by bb)

In which line does the first error in reasoning occur?

  1. Line 3
  2. Line 4
  3. Line 5 (correct answer)
  4. Line 6
Explanation: The error occurs in Line 5. The step involves dividing both sides by (a−b)(a-b). However, the initial assumption in Line 1 is that a=ba=b. This means that a−b=0a-b = 0. Division by zero is an undefined operation in mathematics. Therefore, this step is invalid and leads to the false conclusion. Lines 1 through 4 contain valid algebraic manipulations. Line 6 is a valid substitution based on the initial premise. The error has already occurred in Line 5.

Question 10

A student makes the claim: "If f′(c)=0f'(c) = 0, then the function f(x)f(x) must have a local maximum or a local minimum at x=cx=c". Which of the following functions serves as a counterexample to this claim?

  1. f(x)=x2f(x) = x^2 at c=0c=0
  2. f(x)=∣x∣f(x) = |x| at c=0c=0
  3. f(x)=x3f(x) = x^3 at c=0c=0 (correct answer)
  4. f(x)=exf(x) = e^x at c=0c=0
Explanation: To find a counterexample, we need a function where f′(c)=0f'(c) = 0 is true, but the conclusion (local max/min at x=cx=c) is false. A: For f(x)=x2f(x) = x^2, f′(x)=2xf'(x) = 2x. At c=0c=0, f′(0)=0f'(0) = 0. There is a local minimum at x=0x=0. This supports the claim. B: For f(x)=∣x∣f(x) = |x|, the derivative is undefined at c=0c=0. So the premise f′(c)=0f'(c)=0 is not met. C: For f(x)=x3f(x) = x^3, f′(x)=3x2f'(x) = 3x^2. At c=0c=0, f′(0)=0f'(0) = 0. The point x=0x=0 is a stationary point of inflection, not a local maximum or minimum. The function is increasing for x<0x<0 and for x>0x>0. This is a counterexample. D: For f(x)=exf(x) = e^x, f′(x)=exf'(x) = e^x. At c=0c=0, f′(0)=e0=1≠0f'(0) = e^0 = 1 \neq 0. The premise is not met.

Question 11

Let nn be an integer. A student wants to prove that the product of two consecutive even integers is always divisible by 4. Which of the following is a correct representation of the product and its subsequent manipulation?

  1. (2n)(2n+1)=4n2+2n(2n)(2n+1) = 4n^2+2n
  2. (2n)(2n+2)=4n2+4n=4(n2+n)(2n)(2n+2) = 4n^2+4n = 4(n^2+n) (correct answer)
  3. (n)(n+2)=n2+2n(n)(n+2) = n^2+2n
  4. (2n)(2n+4)=4n2+8n=4(n2+2n)(2n)(2n+4) = 4n^2+8n = 4(n^2+2n)
Explanation: To prove the statement, we must first represent two consecutive even integers algebraically. An even integer can be written as 2n2n for some integer nn. The next consecutive even integer is 2n+22n+2. Their product is (2n)(2n+2)(2n)(2n+2). Expanding this gives 4n2+4n4n^2+4n. Factoring out 4 gives 4(n2+n)4(n^2+n). Since nn is an integer, n2+nn^2+n is also an integer. Thus, the product is a multiple of 4, proving the statement. A represents the product of an even integer and an odd integer. C represents the product of two integers that differ by 2, which could be two even integers (if nn is even) or two odd integers (if nn is odd), but not necessarily two consecutive even integers in general form. D represents two even integers that are not consecutive; they differ by 4.

Question 12

Consider the following proof: Statement: For any integer nn, if n2n^2 is odd, then nn must be odd. The proof proceeds by contrapositive. Step 1: Assume nn is not odd. Therefore, nn is even. Step 2: Let n=2kn = 2k for some integer kk. Step 3: Then n2=(2k)2=4k2=2(2k2)n^2 = (2k)^2 = 4k^2 = 2(2k^2). Step 4: Since kk is an integer, 2k22k^2 is an integer. Thus, n2n^2 is a multiple of 2. Step 5: Therefore, n2n^2 is even.

What has been directly proven by this sequence of steps?

  1. If nn is even, then n2n^2 is even. (correct answer)
  2. If n2n^2 is odd, then nn is odd.
  3. If nn is odd, then n2n^2 is odd.
  4. If n2n^2 is even, then nn is even.
Explanation: The sequence of steps begins with the assumption "nn is even" (Step 1) and ends with the conclusion "n2n^2 is even" (Step 5). Therefore, the direct logical implication that has been proven by these steps is: "If nn is even, then n2n^2 is even." This statement (A) is the contrapositive of the original statement (B). By proving the contrapositive, the original statement is also proven to be true, but the question asks what the steps themselves have directly demonstrated. B is the original statement that the proof by contrapositive aims to establish, not what is directly shown by the steps. C is the converse of A. D is the converse of the original statement.

Question 13

Let nn be an integer. A student wants to prove that the product of two consecutive even integers is always divisible by 4. Which of the following is a correct representation of the product and its subsequent manipulation?

  1. (2n)(2n+1)=4n2+2n(2n)(2n+1) = 4n^2+2n
  2. (2n)(2n+2)=4n2+4n=4(n2+n)(2n)(2n+2) = 4n^2+4n = 4(n^2+n) (correct answer)
  3. (n)(n+2)=n2+2n(n)(n+2) = n^2+2n
  4. (2n)(2n+4)=4n2+8n=4(n2+2n)(2n)(2n+4) = 4n^2+8n = 4(n^2+2n)
Explanation: To prove the statement, we must first represent two consecutive even integers algebraically. An even integer can be written as 2n2n for some integer nn. The next consecutive even integer is 2n+22n+2. Their product is (2n)(2n+2)(2n)(2n+2). Expanding this gives 4n2+4n4n^2+4n. Factoring out 4 gives 4(n2+n)4(n^2+n). Since nn is an integer, n2+nn^2+n is also an integer. Thus, the product is a multiple of 4, proving the statement. A represents the product of an even integer and an odd integer. C represents the product of two integers that differ by 2, which could be two even integers (if nn is even) or two odd integers (if nn is odd), but not necessarily two consecutive even integers in general form. D represents two even integers that are not consecutive; they differ by 4.

Question 14

A student is asked to prove the identity 1n−1n+2≡2n(n+2)\frac{1}{n} - \frac{1}{n+2} \equiv \frac{2}{n(n+2)} for n∈Z,n≠0,−2n \in \mathbb{Z}, n \neq 0, -2. Which of the following represents a correct first step in a proof starting from the left-hand side (LHS)?

  1. Combine the fractions using a common denominator: (n+2)−nn(n+2)\frac{(n+2) - n}{n(n+2)} (correct answer)
  2. Multiply both sides by n(n+2)n(n+2): n+2−n=2n+2 - n = 2
  3. Assume the identity is true and work backwards from the RHS: 2n2+2n\frac{2}{n^2+2n}
  4. Substitute a test value, for example n=1n=1: 1−13=231 - \frac{1}{3} = \frac{2}{3}
Explanation: To prove an identity by starting with the left-hand side (LHS), we must manipulate it algebraically until it is identical to the right-hand side (RHS). The LHS is 1n−1n+2\frac{1}{n} - \frac{1}{n+2}. The correct first step is to combine the two fractions by finding a common denominator, which is n(n+2)n(n+2). This gives 1(n+2)n(n+2)−1(n)n(n+2)=(n+2)−nn(n+2)\frac{1(n+2)}{n(n+2)} - \frac{1(n)}{n(n+2)} = \frac{(n+2)-n}{n(n+2)}. This simplifies to 2n(n+2)\frac{2}{n(n+2)}, which is the RHS. Thus, A is the correct first step. B is incorrect because multiplying both sides of a proposed identity assumes the identity is true. This is a method for solving equations, not proving identities. C is incorrect because a formal proof should proceed from one side to the other (e.g., LHS to RHS), not by assuming the result. D is incorrect because substituting a single value only verifies the statement for that specific value; it does not constitute a general proof.

Question 15

A student wants to prove that the function f(x)=x3−xf(x) = x^3 - x is an odd function. The definition of an odd function is that f(−x)=−f(x)f(-x) = -f(x) for all xx in the domain. Which line of reasoning correctly proves this?

  1. f(−x)=(−x)3−x=−x3−x=−(x3+x)f(-x) = (-x)^3 - x = -x^3 - x = -(x^3+x), which is not equal to −f(x)-f(x).
  2. f(−x)=(−x)3−(−x)=−x3+x=−(x3−x)=−f(x)f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3-x) = -f(x). (correct answer)
  3. Since f(2)=6f(2)=6 and f(−2)=−6f(-2)=-6, we have f(−2)=−f(2)f(-2)=-f(2), so the function is odd.
  4. −f(x)=−(x3)−(−x)=−x3+x-f(x) = -(x^3) - (-x) = -x^3+x. This is not the standard way to start the proof.
Explanation: To prove a function is odd, we must start with f(−x)f(-x) and show through algebraic manipulation that it is equivalent to −f(x)-f(x). For f(x)=x3−xf(x) = x^3 - x: Step 1: Write out f(−x)f(-x) by substituting −x-x for xx. f(−x)=(−x)3−(−x)f(-x) = (-x)^3 - (-x) Step 2: Simplify the expression. f(−x)=−x3+xf(-x) = -x^3 + x Step 3: Factor out -1 to see if the remaining expression is f(x)f(x). f(−x)=−(x3−x)f(-x) = -(x^3 - x) Step 4: Recognize that the expression in the parentheses is f(x)f(x). f(−x)=−f(x)f(-x) = -f(x). This matches the reasoning in option B. A contains an algebraic error in substitution: −x-x was not substituted for the second xx. C shows verification for a single value, which is not a general proof. D calculates −f(x)-f(x) but presents the logic awkwardly. The standard deductive flow is from f(−x)f(-x) to −f(x)-f(x) as shown in B.

Question 16

A student proves that the angles in a triangle sum to 180∘180^\circ by drawing a line through one vertex parallel to the opposite side. The proof relies on properties of angles formed by a transversal intersecting parallel lines. Which of the following is a fundamental, unstated assumption or axiom required for this proof to be valid?

  1. The Pythagorean theorem.
  2. The triangle is an isosceles triangle.
  3. Euclid's parallel postulate. (correct answer)
  4. The sum of two sides of a triangle must be greater than the third side.
Explanation: The proof described is the classic Euclidean proof of the angle sum of a triangle. This proof relies on constructing a line through a vertex parallel to the opposite side and then using the fact that alternate interior angles are equal. The existence and uniqueness of such a parallel line are not self-evident but are guaranteed by Euclid's fifth postulate, known as the parallel postulate. Without this axiom, the geometry might be non-Euclidean (e.g., spherical or hyperbolic), and the sum of angles in a triangle would not be 180∘180^\circ. A, the Pythagorean theorem, applies only to right-angled triangles. B is incorrect as the proof is general for all triangles. D is the triangle inequality for side lengths and is not directly used in this angle sum proof.

Question 17

A student proves that the sum of any two odd numbers is even. Let the two odd numbers be 2k+12k+1 and 2m+12m+1, where k,mk, m are integers. Step 1: Sum = (2k+1)+(2m+1)(2k+1) + (2m+1) Step 2: Sum = 2k+2m+22k + 2m + 2 Step 3: Sum = 2(k+m+1)2(k+m+1) Step 4: Since kk and mm are integers, k+m+1k+m+1 is an integer. Step 5: Therefore, the sum is a multiple of 2, so it is even.

The student then tries to prove the converse: the sum of two integers is even only if both integers are odd. Which of the following provides a counterexample to this converse statement?

  1. The integers are 3 and 5.
  2. The integers are 2 and 4. (correct answer)
  3. The integers are 2 and 3.
  4. The integers are -1 and 1.
Explanation: The converse statement is: "If the sum of two integers is even, then both integers are odd." To find a counterexample, we need to find two integers whose sum is even, but which are not both odd. Let's check the options: A: The integers are 3 and 5. The sum is 8 (even). Both integers are odd. This supports the statement, so it is not a counterexample. B: The integers are 2 and 4. The sum is 6 (even). The integers are not both odd (in fact, both are even). This fits the criteria for a counterexample: the premise (sum is even) is true, but the conclusion (both integers are odd) is false. C: The integers are 2 and 3. The sum is 5 (odd). This does not satisfy the premise of the statement, so it cannot be a counterexample. D: The integers are -1 and 1. The sum is 0 (even). Both integers are odd. This supports the statement.

Question 18

Which of the following values of nn serves as a counterexample to the statement "For all integers n≥1n \geq 1, the expression n2−n+41n^2 - n + 41 is a prime number"?

  1. n=1n=1
  2. n=10n=10
  3. n=40n=40
  4. n=41n=41 (correct answer)
Explanation: A counterexample is a specific instance that shows a general statement is false. We need to find a value of nn for which n2−n+41n^2 - n + 41 is not a prime number. Let's test the given values: For n=1n=1: 12−1+41=411^2 - 1 + 41 = 41, which is a prime number. For n=10n=10: 102−10+41=100−10+41=13110^2 - 10 + 41 = 100 - 10 + 41 = 131, which is a prime number. For n=40n=40: 402−40+41=1600−40+41=160140^2 - 40 + 41 = 1600 - 40 + 41 = 1601, which is also a prime number. For n=41n=41: 412−41+41=412=168141^2 - 41 + 41 = 41^2 = 1681. The number 16811681 is not prime because it is divisible by 41. Therefore, n=41n=41 is a counterexample.

Question 19

To prove the identity sin⁡θ1−cos⁡θ≡1+cos⁡θsin⁡θ\frac{\sin \theta}{1-\cos \theta} \equiv \frac{1+\cos \theta}{\sin \theta}, a valid first step after starting with the LHS, sin⁡θ1−cos⁡θ\frac{\sin \theta}{1-\cos \theta}, would be to:

  1. Multiply the numerator and denominator by sin⁡θ\sin \theta.
  2. Multiply the numerator and denominator by 1+cos⁡θ1+\cos \theta. (correct answer)
  3. Cross-multiply to get sin⁡2θ=(1−cos⁡θ)(1+cos⁡θ)\sin^2 \theta = (1-\cos \theta)(1+\cos \theta).
  4. Replace 11 with sin⁡2θ+cos⁡2θ\sin^2 \theta + \cos^2 \theta.
Explanation: When proving trigonometric identities involving fractions with a binomial term like (1−cos⁡θ)(1-\cos \theta) in the denominator, a common and effective strategy is to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of (1−cos⁡θ)(1-\cos \theta) is (1+cos⁡θ)(1+\cos \theta). This step is valid because it is equivalent to multiplying by 1. LHS = sin⁡θ1−cos⁡θ×1+cos⁡θ1+cos⁡θ=sin⁡θ(1+cos⁡θ)1−cos⁡2θ\frac{\sin \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta} = \frac{\sin \theta(1+\cos \theta)}{1-\cos^2 \theta}. Using the identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, we know 1−cos⁡2θ=sin⁡2θ1-\cos^2 \theta = \sin^2 \theta. The expression becomes sin⁡θ(1+cos⁡θ)sin⁡2θ=1+cos⁡θsin⁡θ\frac{\sin \theta(1+\cos \theta)}{\sin^2 \theta} = \frac{1+\cos \theta}{\sin \theta} = RHS. A leads to an expression that is difficult to simplify to the RHS. C assumes the identity is true, which is what we need to prove. D makes the expression more complex and is not a helpful step.

Question 20

Consider the following proof: Statement: For any integer nn, if n2n^2 is odd, then nn must be odd. The proof proceeds by contrapositive. Step 1: Assume nn is not odd. Therefore, nn is even. Step 2: Let n=2kn = 2k for some integer kk. Step 3: Then n2=(2k)2=4k2=2(2k2)n^2 = (2k)^2 = 4k^2 = 2(2k^2). Step 4: Since kk is an integer, 2k22k^2 is an integer. Thus, n2n^2 is a multiple of 2. Step 5: Therefore, n2n^2 is even.

What has been directly proven by this sequence of steps?

  1. If nn is even, then n2n^2 is even. (correct answer)
  2. If n2n^2 is odd, then nn is odd.
  3. If nn is odd, then n2n^2 is odd.
  4. If n2n^2 is even, then nn is even.
Explanation: The sequence of steps begins with the assumption "nn is even" (Step 1) and ends with the conclusion "n2n^2 is even" (Step 5). Therefore, the direct logical implication that has been proven by these steps is: "If nn is even, then n2n^2 is even." This statement (A) is the contrapositive of the original statement (B). By proving the contrapositive, the original statement is also proven to be true, but the question asks what the steps themselves have directly demonstrated. B is the original statement that the proof by contrapositive aims to establish, not what is directly shown by the steps. C is the converse of A. D is the converse of the original statement.