Math 3 Quiz: Constructing Mathematical Arguments

20 questions · exam conditions

0:00

Constructing Mathematical ArgumentsQuestion 1 of 20

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?

Let

f(x) = x

with domain

[0,1]

g(x) = x

with domain

[0,2]

, and

h(x) = x^2

with domain

[0,2]

. Then

f = g

[0,1]

but

h \neq g

[0,2]

Let

f(x) = \frac{x^2-1}{x-1}

with domain

\mathbb{R} \setminus \{1\}

g(x) = x+1

with domain

\mathbb{R}

, and

h(x) = x+1

with domain

\mathbb{R} \setminus \{0\}

. Here

f \neq h

due to different domains.

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

[-1,1]

and

g = h

[0,2]

, but

f \neq h

Let

f(x) = \sqrt{x}

with domain

[0,4]

g(x) = \sqrt{x}

with domain

[0,9]

, and

h(x) = \sqrt{x}

with domain

[1,9]

. Then the reasoning fails because the domains don't align properly.

← Back to quizzes

Math 3 Quiz

Math 3 Quiz: Constructing Mathematical Arguments

Practice Constructing Mathematical Arguments in Math 3 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 Constructing Mathematical Arguments, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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

Let $f(x) = x$ with domain $[0,1]$ , $g(x) = x$ with domain $[0,2]$ , and $h(x) = x^2$ with domain $[0,2]$ . Then $f = g$ on $[0,1]$ but $h \neq g$ on $[0,2]$ .
Let $f(x) = \frac{x^2-1}{x-1}$ with domain $\mathbb{R} \setminus \{1\}$ , $g(x) = x+1$ with domain $\mathbb{R}$ , and $h(x) = x+1$ with domain $\mathbb{R} \setminus \{0\}$ . Here $f \neq h$ due to different domains.
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 $f \neq h$ . (correct answer)
Let $f(x) = \sqrt{x}$ with domain $[0,4]$ , $g(x) = \sqrt{x}$ with domain $[0,9]$ , and $h(x) = \sqrt{x}$ with domain $[1,9]$ . Then the reasoning fails because the domains don't align properly.

Explanation: Choice C correctly illustrates the flaw. We have

f(x) = g(x) = |x|

[-1,1]

(domain of

f

), and

g(x) = |x|

while

h(x) = x

[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.

Question 2

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?

The student incorrectly assumes that $3y$ has the same parity as $y$ , when actually $3y$ is always odd regardless of whether $y$ is odd or even.
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 $y$ . (correct answer)
The student's parity analysis is incorrect because $7-3y$ being even requires $3y$ to be odd, which happens when $y$ is odd, but this doesn't guarantee integer solutions.
The student assumes that having infinitely many values of $y$ with the correct parity automatically means infinitely many solutions, without checking divisibility conditions.

Explanation: 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.

Question 3

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?

The student incorrectly assumes that squaring both sides of an equation preserves equivalence; this operation can introduce extraneous solutions and doesn't constitute a valid proof technique.
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. (correct answer)
The proof technique is valid but incomplete because the student should verify that both $\sqrt{a} \cdot \sqrt{b}$ and $\sqrt{ab}$ are indeed non-negative before applying the squaring step.
The argument reverses the logical direction; showing that $ab = ab$ doesn't prove the original equation, since squaring can make different expressions appear equal when they're not.

Explanation: 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.

Question 4

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?

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. (correct answer)
The proof is flawed because it assumes that $x^2 \geq 0$ without justification, and this assumption is not necessarily true for all mathematical contexts or number systems.
The proof is correct in logic but incomplete because it doesn't explicitly state that the contradiction implies the original assumption must be false, leaving the argument unfinished.
The proof is flawed because it uses circular reasoning, assuming that real numbers cannot satisfy $x^2 = -1$ in order to prove that $x^2 + 1 = 0$ has no real solutions.

Explanation: 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.

Question 5

A student proves that $\sqrt{2}$ is irrational using the following argument: "Assume $\sqrt{2} = \frac{a}{b}$ where $a$ and $b$ are integers with no common factors. Then $2 = \frac{a^2}{b^2}$ , so $2b^2 = a^2$ . This means $a^2$ is even, so $a$ is even. Let $a = 2k$ . Then $2b^2 = 4k^2$ , so $b^2 = 2k^2$ . Therefore $b$ is even. But this contradicts our assumption that $a$ and $b$ have no common factors." What makes this a strong mathematical proof?

It uses proof by contradiction with a clear assumption, derives a logical contradiction through valid algebraic steps, and properly concludes that the assumption must be false. (correct answer)
It demonstrates that both $a$ and $b$ must be even through careful case analysis, showing that rational representations of $\sqrt{2}$ always lead to reducible fractions.
It systematically eliminates all possible rational representations by showing that any fraction equal to $\sqrt{2}$ must have even numerator and denominator, creating infinite regress.
It uses the fundamental theorem of arithmetic to show that $\sqrt{2}$ cannot be expressed as a ratio of integers due to prime factorization properties.

Explanation: This is a classic proof by contradiction. The student assumes

\sqrt{2}

is rational (in lowest terms), derives valid algebraic consequences, and reaches a contradiction (both

a

and

b

are even, contradicting the assumption they have no common factors). Choice B focuses on the mechanics rather than the proof structure. Choice C mentions 'infinite regress' which isn't part of this proof. Choice D incorrectly describes the method used.

Question 6

Consider the statement: "If $n^2$ is divisible by 4, then $n$ is even." A student attempts to prove this by contrapositive: "Assume $n$ is odd. Then $n = 2k+1$ for some integer $k$ . So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4(k^2+k) + 1$ . Since this is 1 more than a multiple of 4, $n^2$ is not divisible by 4." What is the strongest aspect of this proof?

It correctly identifies that proving the contrapositive is equivalent to proving the original statement, and uses this logical equivalence to avoid the complexity of direct proof.
It provides a complete algebraic demonstration that odd numbers squared always leave remainder 1 when divided by 4, directly contradicting the hypothesis of the original statement.
It systematically applies the definition of odd numbers and uses algebraic manipulation to show that $n^2 \equiv 1 \pmod{4}$ when $n$ is odd, completing the contrapositive argument. (correct answer)
It demonstrates that $4(k^2+k) + 1$ cannot be divisible by 4 for any integer $k$ , thereby establishing that no odd number squared can be divisible by 4 through universal quantification.

Explanation: Choice C best captures the proof's strength: it properly applies the definition of odd numbers (

n = 2k+1

), performs correct algebraic manipulation, and reaches the precise conclusion needed for the contrapositive (if

n

is odd, then

n^2

is not divisible by 4). Choice A focuses on proof strategy rather than execution. Choice B is partially correct but less precise. Choice D overstates by mentioning 'universal quantification' and doesn't emphasize the contrapositive structure.

Question 7

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?

It uses the substitution method effectively by converting logarithmic to exponential form, applying exponential properties, and converting back.
It provides a constructive proof by calculating both sides independently and showing they yield identical results through manipulation.
It demonstrates the property through the fundamental relationship between logarithms and exponents, using their inverse nature to establish equality.
It employs a direct proof strategy using logarithm definitions and exponent laws, systematically building from basic principles to the conclusion. (correct answer)

Explanation: 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.

Question 8

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?

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 $x = -\frac{1}{2}$ . (correct answer)
The proof contains an error in completing the square; the correct form should be $(x + \frac{1}{2})^2 + \frac{1}{4}$ , but the conclusion about positivity would still be valid.
The proof is logically sound but should be strengthened by explicitly stating that $(x + \frac{1}{2})^2 \geq 0$ with equality only when $x = -\frac{1}{2}$ , making the minimum value clear.
The proof is incomplete because it doesn't verify the completed square form by expanding $(x + \frac{1}{2})^2 + \frac{3}{4}$ back to the original expression $x^2 + x + 1$ .

Explanation: 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.

Question 9

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?

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. (correct answer)
The argument is correct about the conclusion but incomplete because it doesn't consider that the equation might have solutions in the complex numbers where sine has an extended range.
The argument is flawed because it incorrectly states the range of sine; the actual range is $(-1,1)$ , which still excludes 2 but makes the reasoning technically incorrect.
The argument is mathematically correct but should be strengthened by providing a formal proof that the range of sine is indeed $[-1,1]$ rather than simply asserting this fact.

Explanation: 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.

Question 10

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?

The argument is logically valid because the conclusion follows necessarily from the premises, and it is sound because all premises are true and 91 is indeed prime.
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.' (correct answer)
The argument is logically valid since 'might be prime' is a weak conclusion, but it is unsound because 91 = 7 × 13 is actually composite, making the conclusion false.
The argument is both logically invalid due to faulty conditional reasoning and unsound because the factorization 91 = 7 × 13 shows the conclusion is definitively false.

Explanation: 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.'

Question 11

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?

The argument is invalid because $\sqrt{x^2}$ does not always equal $|x|$ when $x$ is negative, making the initial premise false for the given example.
The argument is valid and the conclusion is correct, but the reasoning is unnecessarily complicated since $|-3| = 3$ follows directly from the definition of absolute value.
The argument is invalid because it uses circular reasoning, attempting to prove a property of absolute value by assuming that same property in the premise.
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. (correct answer)

Explanation: 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.

Question 12

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?

The proof should explicitly verify that $m+n+1$ is indeed an integer by using the closure property of integers under addition, rather than simply asserting this fact.
The proof should include specific numerical examples to demonstrate that the algebraic manipulation works correctly for particular cases before making the general claim.
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. (correct answer)
The proof should consider the case where the odd integers might be negative separately, since the representation $2m+1$ might not apply to negative odd numbers.

Explanation: 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).

Question 13

A student argues: "The function $f(x) = \frac{1}{x-3}$ is discontinuous at $x = 3$ because $\lim_{x \to 3} f(x)$ does not exist. This limit doesn't exist because as $x$ approaches 3 from the left, $f(x) \to -\infty$ , and as $x$ approaches 3 from the right, $f(x) \to +\infty$ . Since the left and right limits are different, the limit doesn't exist." Which statement best evaluates this argument?

The argument is mathematically sound, correctly applying the definition of continuity and properly analyzing the behavior of the function near the point of discontinuity through one-sided limits. (correct answer)
The argument correctly identifies discontinuity but incorrectly states the limit behavior; both one-sided limits approach $+\infty$ since $\frac{1}{x-3}$ is always positive near $x = 3$ .
The argument is correct about the limit not existing but incomplete because it doesn't mention that $f(3)$ is undefined, which is also required for establishing discontinuity.
The argument correctly analyzes the one-sided limits but fails to recognize that infinite limits still represent a specific type of limit behavior, so the discontinuity should be classified more precisely.

Explanation: The argument is mathematically correct. As

x \to 3^-

, we have

x-3 \to 0^-

, so

\frac{1}{x-3} \to -\infty

. As

x \to 3^+

, we have

x-3 \to 0^+

, so

\frac{1}{x-3} \to +\infty

. Since the one-sided limits differ,

\lim_{x \to 3} f(x)

doesn't exist, making

f

discontinuous at

x=3

. Choice B incorrectly states both limits are

+\infty

. Choice C unnecessarily focuses on

f(3)

being undefined when the limit analysis suffices. Choice D misunderstands that infinite limits mean the limit doesn't exist in the standard sense.

Question 14

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?

The factorization $x^2-4 = (x-2)(x+2)$ is incorrect; the correct factorization should be $(x+2)^2$ or $(x-2)^2$ depending on the sign analysis.
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. (correct answer)
The algebraic manipulation is correct, but the student should verify the result by substituting several test values to confirm that $f(x) = g(x)$ for all inputs.
The argument is mathematically sound, but incomplete because it doesn't explain why the cancellation of $(x-2)$ terms is permissible under the rules of algebraic fractions.

Explanation: 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.

Question 15

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?

She needs to prove that if $p^2$ is divisible by 3, then $p$ must be divisible by 3. (correct answer)
She needs to verify that $p$ and $q$ are indeed in lowest terms before proceeding.
She needs to show that the equation $3q^2 = p^2$ follows correctly from $\sqrt{3} = \frac{p}{q}$ .
She needs to establish that 3 is prime before using divisibility arguments involving 3.

Explanation: 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

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.

Question 16

In proving that "the square of an odd integer is odd," Michael writes: "Let $n$ be odd, so $n = 2k + 1$ for some integer $k$ . Then $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1$ ." What is the most mathematically sound way for Michael to complete this proof?

Observe that $4k^2 + 4k = 4k(k + 1)$ is even, so adding 1 makes it odd.
Note that $4k^2 + 4k + 1$ is always positive, and positive odd numbers are indeed odd.
Substitute specific values of $k$ to show the formula gives odd results in each case.
Factor out 2 from the first two terms: $n^2 = 2(2k^2 + 2k) + 1$ , which has the form $2m + 1$ . (correct answer)

Explanation: When proving statements about odd and even numbers, you need to show that your result fits the precise mathematical definition. An odd number is any integer that can be written in the form

2m + 1

where

m

is an integer. Michael correctly starts by letting

n = 2k + 1

and expanding

n^2 = 4k^2 + 4k + 1

. To complete the proof rigorously, he must show this expression has the form

2m + 1

for some integer

m

. Option D provides exactly this: by factoring out 2 from the first two terms, we get

n^2 = 2(2k^2 + 2k) + 1

. Since

2k^2 + 2k

is an integer (let's call it

m

), we have

n^2 = 2m + 1

, which is the definition of an odd number. This completes the proof perfectly. Option A makes a true observation that

4k(k + 1)

is even, but saying "adding 1 makes it odd" isn't mathematically precise enough for a formal proof. Option B is completely irrelevant—whether a number is positive has nothing to do with whether it's odd. Option C suggests checking specific cases, but mathematical proofs require showing the statement works for all possible values, not just particular examples. Remember: in number theory proofs, always aim to express your final result in the exact form of the definition you're trying to prove. Don't just argue informally about properties—show the algebraic structure explicitly.

Question 17

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?

The statement is actually a biconditional, so finding one counterexample doesn't disprove it completely.
The statement is a true conditional, so no valid counterexample exists for her to have found. (correct answer)
She confused the hypothesis and conclusion, so her counterexample tests the wrong direction.
She failed to verify that her counterexample satisfies the sufficient conditions of the statement.

Explanation: 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.

Question 18

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?

Direct proof; the key step is showing that both $a$ and $b$ must have a common factor.
Proof by contradiction; the key step is deriving a contradiction to the assumption that $\sqrt{2}$ is rational. (correct answer)
Proof by contrapositive; the key step is proving the contrapositive of the original statement.
Proof by cases; the key step is considering both even and odd possibilities for $a$ and $b$ .

Explanation: 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.

Question 19

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?

$n = 6$ , because $6^2 = 36$ is divisible by 4, but 6 is not divisible by 4. (correct answer)
$n = 8$ , because $8^2 = 64$ is divisible by 4, and 8 is also divisible by 4.
$n = 5$ , because $5^2 = 25$ is not divisible by 4, and 5 is not divisible by 4.
$n = 3$ , because $3^2 = 9$ is not divisible by 4, but 3 is divisible by 3.

Explanation: 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.

Question 20

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?

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. (correct answer)
A rectangle has four right angles but not necessarily equal sides, so having equal sides alone cannot guarantee that a quadrilateral is a square without additional conditions.
The statement confuses necessary and sufficient conditions. Having four equal sides is necessary for being a square, but it is not sufficient without other requirements.
A trapezoid can have some equal sides but is clearly not a square, demonstrating that equal sides do not determine the type of quadrilateral in all cases.

Explanation: 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.