Constructing Mathematical Arguments - Math 3

A student claims: "The function $f(x) = \frac{x^2-4}{x-2}$ is equivalent to $g(x) = x+2$ because when we factor and cancel, we get $\frac{(x-2)(x+2)}{x-2} = x+2$ ." What is the most significant mathematical issue with this argument?
Answer: The cancellation step is invalid because it assumes $x \neq 2$ , but the argument fails to address that the functions have different domains and are therefore not equivalent. The key issue is that $f(x)$ is undefined at $x=2$ (domain is all real numbers except 2) while $g(x)=x+2$ is defined everywhere (domain is all real numbers). The functions are not equivalent because they have different domains. The algebraic manipulation is correct for $x \neq 2$ , but function equivalence requires identical domains and outputs. Choice A is wrong because the factorization is correct. Choice C misses the domain issue. Choice D incorrectly suggests the argument is sound.
Tom argues: "Since $f(x) = x^2$ is increasing on $(0, \infty)$ and $3 > 2$ , we can conclude that $f(3) > f(2)$ ." Which mathematical principle most directly supports Tom's reasoning?
Answer: The definition of an increasing function, which preserves the order of inputs in the outputs. An increasing function is defined as one where $x_1 < x_2$ implies $f(x_1) < f(x_2)$ . Since $2 < 3$ and both are in $(0, \infty)$ where $f$ is increasing, we get $f(2) < f(3)$ . Choice A is irrelevant to the inequality reasoning. Choice B doesn't apply since we're not chaining inequalities. Choice D is about continuity, not the order-preserving property being used.
Elena wants to prove: "If $x^2 - 5x + 6 = 0$ , then $x = 2$ or $x = 3$ ." She begins by factoring: $x^2 - 5x + 6 = (x-2)(x-3) = 0$ . What should be her next step to complete a valid proof?
Answer: Apply the zero product property: if $(x-2)(x-3) = 0$ , then $x-2 = 0$ or $x-3 = 0$ . To complete the logical chain, Elena must explicitly invoke the zero product property, which states that if a product equals zero, then at least one factor must be zero. This gives $x-2 = 0$ or $x-3 = 0$ , leading to $x = 2$ or $x = 3$ . Choice A skips this crucial logical step. Choice C is verification, not proof of the conditional. Choice D is unnecessary since the factorization is correct.
Sarah claims: "For any integer $n$ , if $n^2$ is divisible by 4, then $n$ is divisible by 4." To construct a counterexample, which value of $n$ would be most effective, and why?
Answer: $n = 6$ , because $6^2 = 36$ is divisible by 4, but 6 is not divisible by 4. A counterexample must satisfy the hypothesis ( $n^2$ divisible by 4) but fail the conclusion ( $n$ divisible by 4). Since $6^2 = 36 = 4 \times 9$ , the hypothesis is satisfied, but 6 is not divisible by 4, so the conclusion fails. Choice B satisfies both hypothesis and conclusion, supporting rather than refuting the claim. Choices C and D don't satisfy the hypothesis ( $25$ and $9$ are not divisible by 4), so they're irrelevant to the conditional statement.
A student claims: "Since $\sqrt{x^2} = |x|$ for all real numbers $x$ , and $\sqrt{(-3)^2} = \sqrt{9} = 3$ , this proves that $|-3| = 3$ ." Which statement best evaluates the logical structure of this argument?
Answer: The argument is valid in structure and the conclusion is correct, demonstrating proper application of the relationship between square roots and absolute values through a specific example. The argument is logically valid: it correctly applies the general principle $\sqrt{x^2} = |x|$ to the specific case $x = -3$ , performs the calculation $\sqrt{(-3)^2} = \sqrt{9} = 3$ , and concludes $|-3| = 3$ . Choice A is incorrect because $\sqrt{x^2} = |x|$ is always true. Choice B misses that the argument demonstrates understanding of the relationship. Choice C incorrectly identifies circular reasoning when the argument actually applies a general principle to a specific case.
To prove that $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ for non-negative real numbers $a$ and $b$ , a student writes: "Since both sides are non-negative, I can square both sides to get equivalent statements. $\left(\sqrt{a} \cdot \sqrt{b}\right)^2 = \left(\sqrt{ab}\right)^2$ becomes $a \cdot b = ab$ , which is clearly true." What is the main weakness in this argument?
Answer: The argument is circular because it assumes the property $\left(\sqrt{a} \cdot \sqrt{b}\right)^2 = a \cdot b$ without justification, which essentially assumes what needs to be proven. The main flaw is circularity. The student uses the fact that $\left(\sqrt{a} \cdot \sqrt{b}\right)^2 = a \cdot b$ , but this property is essentially equivalent to what needs to be proven. A proper proof would need to establish this step using definitions or previously proven properties of square roots. Choice A overstates the problem with squaring (it's valid here since both sides are non-negative). Choice C misses that non-negativity is given. Choice D misunderstands the logical structure.
A student attempts to prove that $f(x) = x^2 + 1$ has no real zeros by arguing: "Suppose $f(x) = 0$ for some real number $x$ . Then $x^2 + 1 = 0$ , so $x^2 = -1$ . Since squares of real numbers are non-negative, $x^2 \geq 0$ . But $-1 < 0$ , which contradicts $x^2 = -1$ . Therefore, no such $x$ exists." Which best describes this proof?
Answer: The proof is correct and uses proof by contradiction effectively, establishing that the assumption leads to a mathematical impossibility and therefore the original claim must be true. This is a correct proof by contradiction. The student properly assumes the negation of what they want to prove, derives a logical contradiction ( $x^2 = -1$ where $x^2 \geq 0$ ), and concludes the assumption must be false. Choice B is incorrect because $x^2 \geq 0$ is a fundamental property of real numbers. Choice C is wrong because the contradiction clearly implies the assumption is false. Choice D misidentifies circular reasoning; the proof uses the property of real numbers to reach a contradiction, not the conclusion itself.
To prove that the equation $2x + 3y = 7$ has infinitely many integer solutions, a student argues: "Since we can solve for $x = \frac{7-3y}{2}$ , we just need $7-3y$ to be even. Since $7$ is odd and $3y$ has the same parity as $y$ , we need $y$ to be odd. Every odd integer gives a valid solution." What is the primary flaw in this reasoning?
Answer: The student correctly identifies that $y$ must be odd but fails to verify that the resulting $x$ values are actually integers for all odd values of $y$ . The student's parity analysis is actually correct: for $x = \frac{7-3y}{2}$ to be an integer, $7-3y$ must be even. Since 7 is odd, $3y$ must be odd, which occurs when $y$ is odd. However, the student fails to verify that $x$ is actually an integer for all odd $y$ . For example, when $y=1$ , $x = \frac{7-3}{2} = 2$ (integer), but the student should verify this pattern holds. Choice A is wrong because $3y$ is odd exactly when $y$ is odd. Choices C and D misidentify where the gap in reasoning occurs.
Lisa proves that $\sqrt{3}$ is irrational by assuming $\sqrt{3} = \frac{p}{q}$ in lowest terms, then showing both $p$ and $q$ must be divisible by 3. However, her proof contains an error in this step: "Since $3q^2 = p^2$ , we know $p^2$ is divisible by 3, so $p$ is divisible by 3." What additional justification does Lisa need?
Answer: She needs to prove that if $p^2$ is divisible by 3, then $p$ must be divisible by 3. Lisa's step assumes that divisibility of $p^2$ by 3 implies divisibility of $p$ by 3, but this requires proof. For a prime $p$ , if $p|ab$ , then $p|a$ or $p|b$ . Since $3|p^2 = p \cdot p$ and 3 is prime, we get $3|p$ . Choice B is wrong because lowest terms was already assumed. Choice C is incorrect because squaring both sides of $\sqrt{3} = \frac{p}{q}$ clearly gives $3 = \frac{p^2}{q^2}$ . Choice D is unnecessary since the primality of 3 is well-established.
Consider the statement: "If a quadrilateral has four equal sides, then it is a square." Which of the following represents the most complete mathematical argument for why this statement is false?
Answer: A rhombus has four equal sides but is not necessarily a square, since a square requires both equal sides and right angles. A rhombus with angles of 60° and 120° serves as a counterexample. Choice A provides the most complete argument by: (1) identifying a specific counterexample (rhombus), (2) explaining why the counterexample satisfies the hypothesis (four equal sides), (3) explaining why it fails the conclusion (lacks right angles), and (4) providing a concrete example. Choice B discusses rectangles which don't have four equal sides, making it irrelevant. Choice C correctly identifies the logical error but doesn't provide a counterexample. Choice D mentions trapezoids which don't typically have four equal sides.
To prove that $\log_2(xy) = \log_2(x) + \log_2(y)$ for positive real numbers $x$ and $y$ , a student writes: "Let $a = \log_2(x)$ and $b = \log_2(y)$ . Then $x = 2^a$ and $y = 2^b$ . So $xy = 2^a \cdot 2^b = 2^{a+b}$ . Taking $\log_2$ of both sides: $\log_2(xy) = a + b = \log_2(x) + \log_2(y)$ ." What makes this proof effective?
Answer: It employs a direct proof strategy using logarithm definitions and exponent laws, systematically building from basic principles to the conclusion. This proof effectively uses a direct approach, starting with the definition of logarithm ( $\log_2(x) = a$ means $x = 2^a$ ), applying the law of exponents ( $2^a \cdot 2^b = 2^{a+b}$ ), and using the inverse relationship between logarithm and exponential functions. Choice A describes the method but doesn't capture why it's effective. Choice B incorrectly describes this as calculating both sides independently. Choice C is partially correct but less complete than D.
In a proof that the sum of two even integers is even, which of the following represents the most rigorous approach to the argument?
Answer: Let $a = 2m$ and $b = 2n$ where $m, n$ are integers. Then $a + b = 2m + 2n = 2(m + n)$ , which is even. Choice C provides a rigorous algebraic proof using the definition of even numbers and showing the sum has the required form. Choice A only proves the result for one specific case, not the general statement. Choice B relies on decimal representation rules rather than the algebraic definition of even numbers. Choice D invokes the 'closure property' without proving that this property actually applies to this situation.
To prove that the sum of any two odd integers is even, a student writes: "Let the odd integers be $2m+1$ and $2n+1$ where $m$ and $n$ are integers. Their sum is $(2m+1) + (2n+1) = 2m + 2n + 2 = 2(m+n+1)$ . Since $m+n+1$ is an integer, the sum is even." Which aspect of this proof could be strengthened?
Answer: The proof should begin with a clear statement of what needs to be proven and explicitly connect the final form $2(m+n+1)$ to the definition of even numbers. While the algebraic work is correct, the proof would be stronger with explicit bookends: stating what needs to be proven and clearly invoking the definition that a number is even if it can be written as $2k$ for some integer $k$ . Choice A is unnecessary since integer closure is fundamental. Choice B suggests examples aren't needed in a general proof. Choice D is incorrect because $2m+1$ represents all odd integers when $m$ ranges over all integers (including negative).
Consider the argument: "All prime numbers greater than 2 are odd. The number 91 is odd. Therefore, 91 might be prime." Which statement best analyzes the logical validity and soundness of this argument?
Answer: The argument is logically invalid because it commits the fallacy of affirming the consequent, incorrectly reasoning from 'if prime then odd' to 'if odd then possibly prime.' The argument commits the logical fallacy of affirming the consequent. The first premise establishes 'if prime (and >2) then odd,' but the argument incorrectly tries to conclude something about primality from oddness. Even the weak conclusion 'might be prime' doesn't follow logically from the premises. Choice A is wrong because the argument is invalid and 91 is not prime. Choice C incorrectly claims the argument is valid. Choice D is partially correct about invalidity but overstates the problem with the conclusion 'might be prime.'
A student claims that if $f(x) = g(x)$ for all $x$ in the domain of $f$ , and if $h(x) = g(x)$ for all $x$ in the domain of $g$ , then $f(x) = h(x)$ for all $x$ in both domains. Which scenario best illustrates why this reasoning is flawed?
Answer: Let $f(x) = |x|$ with domain $[-1,1]$ , $g(x) = |x|$ with domain $[-2,2]$ , and $h(x) = x$ with domain $[0,2]$ . Then $f = g$ on $[-1,1]$ and $g = h$ on $[0,2]$ , but $f \neq h$ . Choice C correctly illustrates the flaw. We have $f(x) = g(x) = |x|$ on $[-1,1]$ (domain of $f$ ), and $g(x) = |x|$ while $h(x) = x$ on $[0,2]$ (domain of $g$ ), so $g = h$ only on $[0,2]$ . But $f(-0.5) = 0.5$ while $h(-0.5) = -0.5$ , so $f \neq h$ . The flaw is assuming transitivity without considering domain restrictions. Choice A has $h \neq g$ . Choice B has correct domains making $f = h$ where both are defined. Choice D doesn't show the logical flaw clearly.
Consider the statement: "If a quadrilateral has four right angles, then it is a rectangle." Maria claims this statement is false because she found a counterexample. Which of the following best describes the flaw in Maria's reasoning?
Answer: The statement is a true conditional, so no valid counterexample exists for her to have found. The given statement is a true conditional: any quadrilateral with four right angles must be a rectangle by definition. Since the statement is true, no valid counterexample can exist. Maria's error is in believing she found a counterexample to a true statement. Choice A is wrong because the statement is conditional, not biconditional. Choice C is incorrect because the direction of reasoning isn't the issue. Choice D misses the point that no counterexample should exist at all.
Alex proves that $\sqrt{2}$ is irrational using the following argument: "Assume $\sqrt{2} = \frac{a}{b}$ where $a$ and $b$ are integers with $\gcd(a,b) = 1$ . Then $2b^2 = a^2$ , so $a^2$ is even, which means $a$ is even. Let $a = 2k$ , then $2b^2 = 4k^2$ , so $b^2 = 2k^2$ . Therefore $b$ is also even, contradicting $\gcd(a,b) = 1$ ." What type of proof technique is Alex using, and what is the key logical step?
Answer: Proof by contradiction; the key step is deriving a contradiction to the assumption that $\sqrt{2}$ is rational. Alex uses proof by contradiction by assuming the opposite of what he wants to prove (that $\sqrt{2}$ is rational) and deriving a logical contradiction (that $\gcd(a,b) = 1$ and $\gcd(a,b) \neq 1$ ). Choice A is wrong because this isn't a direct proof—it assumes the opposite. Choice C is incorrect because no contrapositive is being proved. Choice D is wrong because the proof doesn't systematically consider different cases.
A student argues: "The equation $\sin(x) = 2$ has no solutions because the range of sine is $[-1,1]$ , and since $2 > 1$ , the value 2 is outside this range. Therefore, there is no real number $x$ such that $\sin(x) = 2$ ." What best describes the quality of this mathematical argument?
Answer: The argument is logically sound and mathematically correct, properly applying the definition of range to conclude that the equation has no solutions in the real numbers. The argument is logically valid and mathematically correct. It properly applies the definition of the range of a function to determine when an equation has solutions. The reasoning is: if $y$ is not in the range of $f$ , then $f(x) = y$ has no solutions. Since $2 \notin [-1,1]$ , the equation $\sin(x) = 2$ has no real solutions. Choice B introduces complex numbers unnecessarily. Choice C incorrectly states the range of sine. Choice D suggests unnecessary additional proof of a standard fact.
Consider this attempted proof: "To show that $x^2 + x + 1 > 0$ for all real $x$ , I'll complete the square: $x^2 + x + 1 = (x + \frac{1}{2})^2 + \frac{3}{4}$ . Since squares are non-negative and $\frac{3}{4} > 0$ , we have $x^2 + x + 1 \geq \frac{3}{4} > 0$ ." Which statement best evaluates this proof?
Answer: The proof is correct and complete, successfully demonstrating that the expression is always positive by showing it has a minimum value of $\frac{3}{4}$ achieved when $x = -\frac{1}{2}$ . The proof is mathematically correct and complete. The completing the square is accurate: $x^2 + x + 1 = (x + \frac{1}{2})^2 + \frac{3}{4}$ . Since $(x + \frac{1}{2})^2 \geq 0$ , the expression is at least $\frac{3}{4} > 0$ . Choice B is incorrect because the completion is right. Choice C suggests the proof needs more detail, but it's sufficiently rigorous. Choice D suggests unnecessary verification since the algebraic work can be checked directly.
Consider the statement: "For all real numbers $x$ , if $x > 0$ , then $\frac{1}{x} > 0$ ." Which approach would constitute a complete proof of this statement?
Answer: Show that the contrapositive "if $\frac{1}{x} \leq 0$ , then $x \leq 0$ " is true for all relevant values. Proving the contrapositive is a valid proof technique. If $\frac{1}{x} \leq 0$ , then $x$ must be negative (since $\frac{1}{x}$ is undefined at $x = 0$ and negative when $x < 0$ ), so $x \leq 0$ . This proves the contrapositive, hence the original statement. Choice B only provides examples, not a general proof. Choice C assumes what it's trying to prove (that we can multiply by $\frac{1}{x}$ ). Choice D appeals to a 'quotient rule' without justification.

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