Proof - IB Mathematics Analysis and Approaches

1. A student is proving that the sum of three consecutive integers is always divisible by 3. If the integers are represented as $n-1, n, n+1$, what is the correct formulation of the proof?
   Answer: $(n-1) + n + (n+1) = 3n$. Since $n$ is an integer, $3n$ is a multiple of 3. The question specifies using the representation $n-1, n, n+1$ for the three consecutive integers. The proof should calculate their sum and show it is a multiple of 3. The sum is $(n-1) + n + (n+1)$. Adding these terms gives $n-1+n+n+1 = 3n$. Since $n$ is an integer by definition, $3n$ is an integer multiple of 3. This proves the statement and matches option A. B uses a different, though also valid, representation ($n, n+1, n+2$). However, the question specified the representation to be used. C calculates the product, not the sum. D states the conclusion that needs to be proven but does not show the steps of the proof.
2. Given the statement: "If it is raining, then the ground is wet." Which of the following can be logically deduced?
   Answer: If the ground is not wet, then it is not raining. This question tests understanding of logical implications. The original statement is of the form $P \implies Q$, where P is "it is raining" and Q is "the ground is wet". A: "If the ground is wet, then it is raining" is the converse, $Q \implies P$. This is not logically equivalent. The ground could be wet for other reasons (e.g., a sprinkler). B: "If it is not raining, then the ground is not wet" is the inverse, $\neg P \implies \neg Q$. This is also not logically equivalent. It could have just stopped raining. C: "If the ground is not wet, then it is not raining" is the contrapositive, $\neg Q \implies \neg P$. A statement and its contrapositive are always logically equivalent. If the ground is not wet, it certainly cannot be raining (based on the original statement). D: This asserts that P and Q are both true, which is a specific scenario, not a logical deduction from the conditional statement itself.
3. Let $n$ 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?
   Answer: $(2n)(2n+2) = 4n^2+4n = 4(n^2+n)$ To prove the statement, we must first represent two consecutive even integers algebraically. An even integer can be written as $2n$ for some integer $n$. The next consecutive even integer is $2n+2$. Their product is $(2n)(2n+2)$. Expanding this gives $4n^2+4n$. Factoring out 4 gives $4(n^2+n)$. Since $n$ is an integer, $n^2+n$ 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 $n$ is even) or two odd integers (if $n$ 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.
4. What has been directly proven by this sequence of steps?
   Answer: If $n$ is even, then $n^2$ is even. The sequence of steps begins with the assumption "$n$ is even" (Step 1) and ends with the conclusion "$n^2$ is even" (Step 5). Therefore, the direct logical implication that has been proven by these steps is: "If $n$ is even, then $n^2$ 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.
5. To prove the identity $\frac{\sin \theta}{1-\cos \theta} \equiv \frac{1+\cos \theta}{\sin \theta}$, a valid first step after starting with the LHS, $\frac{\sin \theta}{1-\cos \theta}$, would be to:
   Answer: Multiply the numerator and denominator by $1+\cos \theta$. When proving trigonometric identities involving fractions with a binomial term like $(1-\cos \theta)$ in the denominator, a common and effective strategy is to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $(1-\cos \theta)$ is $(1+\cos \theta)$. This step is valid because it is equivalent to multiplying by 1. LHS = $\frac{\sin \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta} = \frac{\sin \theta(1+\cos \theta)}{1-\cos^2 \theta}$. Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, we know $1-\cos^2 \theta = \sin^2 \theta$. The expression becomes $\frac{\sin \theta(1+\cos \theta)}{\sin^2 \theta} = \frac{1+\cos \theta}{\sin \theta}$ = 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.
6. To prove the identity $(a+b)^3 - (a-b)^3 \equiv 2b(3a^2+b^2)$, a student starts by expanding the left-hand side (LHS). Which of the following is the correct expansion and simplification of the LHS?
   Answer: $(a^3+3a^2b+3ab^2+b^3) - (a^3-3a^2b+3ab^2-b^3) = 6a^2b + 2b^3$ The proof begins by expanding the cubic terms on the LHS. Using the binomial expansion $(x+y)^3 = x^3+3x^2y+3xy^2+y^3$: $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$ $(a-b)^3 = a^3-3a^2b+3ab^2-b^3$ Subtracting the second from the first: LHS = $(a^3+3a^2b+3ab^2+b^3) - (a^3-3a^2b+3ab^2-b^3)$ LHS = $a^3+3a^2b+3ab^2+b^3 - a^3+3a^2b-3ab^2+b^3$ Combine like terms: $(a^3-a^3) + (3a^2b+3a^2b) + (3ab^2-3ab^2) + (b^3+b^3) = 6a^2b + 2b^3$. This matches option A. The final step would be to factor this expression: $2b(3a^2+b^2)$, which is the RHS. B incorrectly uses $(a+b)^3 = a^3+b^3$. C shows a possible intermediate step of expansion, but A shows the direct binomial expansion and its result. Also, the result in C is correct, but A represents the most direct expansion step itself. D contains sign errors in the subtraction. For example, $3a^2b - (-3a^2b) = 6a^2b$, but $a^3-a^3=0$, not $2a^3$.
7. Which of the following values of $n$ serves as a counterexample to the statement "For all integers $n \geq 1$, the expression $n^2 - n + 41$ is a prime number"?
   Answer: $n=41$ A counterexample is a specific instance that shows a general statement is false. We need to find a value of $n$ for which $n^2 - n + 41$ is not a prime number. Let's test the given values: For $n=1$: $1^2 - 1 + 41 = 41$, which is a prime number. For $n=10$: $10^2 - 10 + 41 = 100 - 10 + 41 = 131$, which is a prime number. For $n=40$: $40^2 - 40 + 41 = 1600 - 40 + 41 = 1601$, which is also a prime number. For $n=41$: $41^2 - 41 + 41 = 41^2 = 1681$. The number $1681$ is not prime because it is divisible by 41. Therefore, $n=41$ is a counterexample.
8. Given $a, b \in \mathbb{R}$, a student wants to prove that if $a>b>0$, then $a^2 > b^2$. Which of the following is a key step in a direct deductive proof?
   Answer: Since $a>b$, it follows that $a-b>0$. Factor $a^2-b^2$ as $(a-b)(a+b)$. A direct deductive proof starts from the premises and proceeds through logical steps to the conclusion. To prove $a^2 > b^2$, we can aim to show that $a^2 - b^2 > 0$. The expression $a^2 - b^2$ can be factored as $(a-b)(a+b)$. From the premise $a>b>0$, we can deduce two things: 1) $a-b > 0$ (a positive number) and 2) $a+b > 0$ (the sum of two positive numbers is positive). The product of two positive numbers is positive, so $(a-b)(a+b) > 0$, which implies $a^2 - b^2 > 0$, and therefore $a^2 > b^2$. Option A outlines this crucial line of reasoning. B describes proof by contradiction, which is a valid but different proof technique. C is verification with an example, not a general proof. D states the conclusion but does not provide the reasoning or justification for why squaring both sides is a valid step that preserves the inequality (which is because both sides are positive).
9. In which line does the first error in reasoning occur?
   Answer: Line 5 The error occurs in Line 5. The step involves dividing both sides by $(a-b)$. However, the initial assumption in Line 1 is that $a=b$. This means that $a-b = 0$. 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.
10. A student is asked to prove the identity $\frac{1}{n} - \frac{1}{n+2} \equiv \frac{2}{n(n+2)}$ for $n \in \mathbb{Z}, n \neq 0, -2$. Which of the following represents a correct first step in a proof starting from the left-hand side (LHS)?
    Answer: Combine the fractions using a common denominator: $\frac{(n+2) - n}{n(n+2)}$ 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 $\frac{1}{n} - \frac{1}{n+2}$. The correct first step is to combine the two fractions by finding a common denominator, which is $n(n+2)$. This gives $\frac{1(n+2)}{n(n+2)} - \frac{1(n)}{n(n+2)} = \frac{(n+2)-n}{n(n+2)}$. This simplifies to $\frac{2}{n(n+2)}$, 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.
11. Consider the statement $(x-3)^2 = x^2 - 6x + 9$. Which of the following best describes this statement?
    Answer: An identity that is true for all $x \in \mathbb{R}$. 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, $(x-3)^2 = (x-3)(x-3) = x^2 - 3x - 3x + 9 = x^2 - 6x + 9$. Since the left-hand side is algebraically equivalent to the right-hand side, the statement is true for all real numbers $x$. Therefore, it is an identity. A is incorrect; it confuses an identity with a conditional equation. While the statement is true for $x=3$, it is not only true for $x=3$. C is incorrect as the statement is an equality, not an inequality. D is incorrect as the statement is true for all real numbers.
12. Which of the following statements is an identity?
    Answer: $(x+1)^2 - (x-1)^2 = 4x$ An identity is true for all values of the variable for which both sides are defined. Let's check each option: A: This is not an identity. The correct identity is $\sin(2x) = 2\sin(x)\cos(x)$. For example, when $x = \frac{\pi}{4}$, $\sin(\frac{\pi}{2}) = 1$ but $2\sin(\frac{\pi}{4}) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$. B: This is incorrect. The correct property is $e^{x+y} = e^x \cdot e^y$. C: This is a valid logarithm property, but requires $x > 0$ and $y > 0$, making it more restrictive than option D. D: Expanding the left side: $(x^2+2x+1) - (x^2-2x+1) = x^2+2x+1-x^2+2x-1 = 4x$. This holds for all real numbers $x$, making it a true identity.
13. In a proof of the identity $\log_a(x^n) \equiv n \log_a(x)$, where $a>0, a \neq 1, x>0$, a key deductive step involves:
    Answer: Using the definition of a logarithm: let $y = \log_a(x)$, which implies $x = a^y$. A fundamental proof of the power rule for logarithms stems directly from the definition of a logarithm. The standard proof is as follows: 1. Let $y = \log_a(x)$. 2. By the definition of a logarithm, this is equivalent to $x = a^y$. 3. Raise both sides to the power of $n$: $x^n = (a^y)^n = a^{ny}$. 4. Convert this exponential form back to a logarithmic form: $\log_a(x^n) = ny$. 5. Substitute the original expression for $y$: $\log_a(x^n) = n \log_a(x)$. Option A describes the first two crucial steps of this proof. B is incorrect because it involves circular reasoning. C uses the property to be proven (or a version of it for natural logs) within the 'proof', which is not a fundamental demonstration. D states an incorrect logarithm property.
14. 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?
    Answer: The integers are 2 and 4. 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.
15. A student wants to prove that for any real numbers $x$ and $y$, $|x+y| \leq |x| + |y|$ (the triangle inequality). The proof often begins by squaring both sides. Why is this a valid first step?
    Answer: Both sides of the inequality are non-negative, and for non-negative numbers $a$ and $b$, $a \leq b$ is equivalent to $a^2 \leq b^2$. The validity of squaring both sides of an inequality depends on the sign of the terms. The function $f(t) = t^2$ is strictly increasing only for non-negative values. Since the absolute value of any number is non-negative, both $|x+y|$ and $|x|+|y|$ are greater than or equal to zero. Therefore, squaring both sides will preserve the direction of the inequality. Option B provides the correct and complete justification for this step. A is false; squaring can reverse the inequality if negative numbers are involved (e.g., $-3 < 2$, but $9 > 4$). C describes a step used after squaring, but it does not justify the validity of the squaring step itself. D confuses the process of solving an equation with proving an inequality. While squaring can introduce extraneous roots when solving equations, the concern in a proof is whether the logical implication is preserved.
16. A student makes the claim: "If $f'(c) = 0$, then the function $f(x)$ must have a local maximum or a local minimum at $x=c$". Which of the following functions serves as a counterexample to this claim?
    Answer: $f(x) = x^3$ at $c=0$ To find a counterexample, we need a function where $f'(c) = 0$ is true, but the conclusion (local max/min at $x=c$) is false. A: For $f(x) = x^2$, $f'(x) = 2x$. At $c=0$, $f'(0) = 0$. There is a local minimum at $x=0$. This supports the claim. B: For $f(x) = |x|$, the derivative is undefined at $c=0$. So the premise $f'(c)=0$ is not met. C: For $f(x) = x^3$, $f'(x) = 3x^2$. At $c=0$, $f'(0) = 0$. The point $x=0$ is a stationary point of inflection, not a local maximum or minimum. The function is increasing for $x<0$ and for $x>0$. This is a counterexample. D: For $f(x) = e^x$, $f'(x) = e^x$. At $c=0$, $f'(0) = e^0 = 1 \neq 0$. The premise is not met.
17. A student proves that the angles in a triangle sum to $180^\circ$ 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?
    Answer: Euclid's parallel postulate. 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 $180^\circ$. 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.
18. A student wants to prove that the function $f(x) = x^3 - x$ is an odd function. The definition of an odd function is that $f(-x) = -f(x)$ for all $x$ in the domain. Which line of reasoning correctly proves this?
    Answer: $f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3-x) = -f(x)$. To prove a function is odd, we must start with $f(-x)$ and show through algebraic manipulation that it is equivalent to $-f(x)$. For $f(x) = x^3 - x$: Step 1: Write out $f(-x)$ by substituting $-x$ for $x$. $f(-x) = (-x)^3 - (-x)$ Step 2: Simplify the expression. $f(-x) = -x^3 + x$ Step 3: Factor out -1 to see if the remaining expression is $f(x)$. $f(-x) = -(x^3 - x)$ Step 4: Recognize that the expression in the parentheses is $f(x)$. $f(-x) = -f(x)$. This matches the reasoning in option B. A contains an algebraic error in substitution: $-x$ was not substituted for the second $x$. C shows verification for a single value, which is not a general proof. D calculates $-f(x)$ but presents the logic awkwardly. The standard deductive flow is from $f(-x)$ to $-f(x)$ as shown in B.

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