Discrete Math Quiz: Direct Proof

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Direct ProofQuestion 1 of 20

In a direct proof of "For integers $m$ and $n$ , if $m^2 + n^2$ is divisible by $4$ , then both $m$ and $n$ have the same parity," what information is most crucial to establish early in the proof?

The possible remainders when any integer squared is divided by

4

, and their relationship to the integer's parity.

Examples showing that when

m

and

n

have different parity,

m^2 + n^2

is not divisible by

4

The fact that

4

divides

m^2 + n^2

if and only if

m^2 + n^2 \equiv 0 \pmod{4}

A proof that the sum of two squares can only be divisible by

4

under specific conditions.

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Discrete Math Quiz

Discrete Math Quiz: Direct Proof

Practice Direct Proof in Discrete Math 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 Direct Proof, giving you a quick way to practice the rules, question types, and explanations that matter most for Discrete Math.

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

In a direct proof of "For integers $m$ and $n$ , if $m^2 + n^2$ is divisible by $4$ , then both $m$ and $n$ have the same parity," what information is most crucial to establish early in the proof?

The possible remainders when any integer squared is divided by $4$ , and their relationship to the integer's parity. (correct answer)
Examples showing that when $m$ and $n$ have different parity, $m^2 + n^2$ is not divisible by $4$ .
The fact that $4$ divides $m^2 + n^2$ if and only if $m^2 + n^2 \equiv 0 \pmod{4}$ .
A proof that the sum of two squares can only be divisible by $4$ under specific conditions.

Explanation: To prove this directly, we need to understand how the parity of an integer affects the remainder when its square is divided by 4. Specifically, if n is even, n² ≡ 0 (mod 4), and if n is odd, n² ≡ 1 (mod 4). This key insight allows us to analyze all possible cases systematically. Choice B uses examples rather than general proof. Choice C restates the assumption without adding insight. Choice D is too vague and doesn't provide the specific modular arithmetic insight needed.

Question 2

Consider proving: "If $x$ and $y$ are positive real numbers and $xy = 1$ , then $x + y \geq 2$ ." A student writes: "Since $xy = 1$ and $x, y > 0$ , we have $y = \frac{1}{x}$ . Therefore $x + y = x + \frac{1}{x}$ ." What should be the next step to maintain proper direct proof structure?

Apply the AM-GM inequality directly to conclude that $\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = 1$ .
Consider the function $f(x) = x + \frac{1}{x}$ for $x > 0$ and find its minimum using calculus.
Show that $(x + \frac{1}{x}) - 2 = \frac{x^2 - 2x + 1}{x} = \frac{(x-1)^2}{x} \geq 0$ since $x > 0$ . (correct answer)
Multiply both sides by $x$ to get $x^2 + 1 \geq 2x$ , then rearrange to show $x^2 - 2x + 1 \geq 0$ .

Explanation: Choice C provides a direct algebraic approach that works entirely from the established expression x + 1/x and shows directly that it's ≥ 2 by demonstrating the difference is non-negative. This maintains the forward logical flow of direct proof. Choice A invokes AM-GM inequality which, while correct, introduces an external theorem rather than building the result directly. Choice B uses calculus, which goes beyond the algebraic scope typically expected. Choice D requires additional justification for why multiplying by x preserves the inequality direction.

Question 3

A student is constructing a direct proof of: "If $a$ and $b$ are real numbers with $a^2 + b^2 = 1$ and $a, b \geq 0$ , then $a + b \leq \sqrt{2}$ ." After establishing the hypothesis, the student writes: "We want to prove $a + b \leq \sqrt{2}$ . Squaring both sides, this is equivalent to $(a + b)^2 \leq 2$ ." What is problematic about this step in a direct proof?

The student should verify that $a + b \geq 0$ before squaring to ensure the squaring operation preserves the inequality direction.
Squaring both sides of an inequality is not a valid operation in direct proofs without additional justification.
The student is manipulating the desired conclusion rather than working forward from the given hypothesis $a^2 + b^2 = 1$ . (correct answer)
The equivalence between $a + b \leq \sqrt{2}$ and $(a + b)^2 \leq 2$ requires proving that equality cases are handled correctly.

Explanation: In a direct proof, we should work forward from our assumptions to build toward the conclusion, not manipulate the conclusion we're trying to prove. The student should start with a² + b² = 1 and a, b ≥ 0, then build algebraically toward showing a + b ≤ √2. Choice A identifies a valid concern about preserving inequality direction, but since a, b ≥ 0, we know a + b ≥ 0. Choice B is incorrect because squaring can be valid with proper justification. Choice D raises a technical point but doesn't address the main structural issue with the proof approach.

Question 4

In a direct proof of "If $a$ , $b$ , and $c$ are consecutive integers with $a < b < c$ , then $a^2 + b^2 + c^2 - ab - bc - ca = 3$ ," which representation of the consecutive integers is most conducive to direct proof?

Let $a = n$ , $b = n+1$ , $c = n+2$ for some integer $n$ , then substitute and expand.
Let $a = b-1$ and $c = b+1$ , then express everything in terms of the middle integer $b$ for symmetric algebraic manipulation. (correct answer)
Let $b = a+d$ and $c = a+2d$ where $d = 1$ , then substitute $d = 1$ at the end.
Use the general forms $a = k$ , $b = k+j$ , $c = k+2j$ with $j = 1$ to maintain generality throughout.

Explanation: Expressing a = b-1 and c = b+1 creates the most symmetric representation that simplifies algebraic manipulation. When we substitute into a² + b² + c² - ab - bc - ca, the algebra becomes more manageable due to the symmetry around b. Choice A works but is less symmetric. Choices C and D introduce unnecessary parameters (d and j) that don't simplify the problem and actually make it more complex than needed for this specific case.

Question 5

In proving "If $n$ is a positive integer and $n^3 - n$ is divisible by $6$ , then either $n \equiv 0 \pmod{6}$ or $n$ is not divisible by $3$ ," what is the most significant structural challenge for a direct proof approach?

The condition " $n$ is not divisible by $3$ " is a negative statement that cannot be proven directly.
The hypothesis involves a cubic expression, requiring factorization techniques not typically used in direct proofs.
The statement mixes divisibility by $6$ and divisibility by $3$ , creating conflicting modular arithmetic systems.
The conclusion is a disjunction ("or" statement), making it difficult to prove directly without case analysis. (correct answer)

Explanation: When analyzing proof strategies in discrete mathematics, you need to consider how the logical structure of a statement affects your approach. Direct proofs work best when you can establish a clear path from hypothesis to a single, concrete conclusion. The key insight here is recognizing that this statement's conclusion is a disjunction: "either

n \equiv 0 \pmod{6}

n

is not divisible by

3

." In a direct proof, you assume the hypothesis (

n^3 - n

is divisible by

6

) and try to prove the conclusion follows logically. But with an "or" statement, you'd need to prove that at least one of the two alternatives is true, without knowing which one applies to any given

n

. This creates an inherent structural problem—you can't directly establish which branch of the disjunction holds without breaking into cases or using indirect methods. Choice A is incorrect because negative statements can absolutely be proven directly by showing the conditions for the positive statement fail. Choice B misses the point—while

n^3 - n

does factor nicely as

n(n-1)(n+1)

, factorization is a standard technique in direct proofs. Choice C is wrong because working with different moduli simultaneously is common and manageable in number theory proofs. The correct answer is D because disjunctive conclusions fundamentally resist direct proof approaches—they naturally call for case analysis, proof by contradiction, or contrapositive methods instead. Study tip: When you see "or" statements in conclusions, immediately consider indirect proof methods rather than struggling with a direct approach that fights the logical structure.

Question 6

A student attempts to prove: "If $a$ divides $bc$ and $\gcd(a,b) = 1$ , then $a$ divides $c$ " by writing: "Assume $a | bc$ and $\gcd(a,b) = 1$ . Since $a$ and $b$ are relatively prime, $a$ cannot share any common factors with $b$ . Therefore, if $a$ divides the product $bc$ , it must divide $c$ ." What is the main logical gap in this reasoning?

The student didn't use the formal definition of divisibility to express $bc = ak$ for some integer $k$ .
The conclusion about $a$ dividing $c$ doesn't follow rigorously from the observation about common factors. (correct answer)
The student should have considered the case where $a$ , $b$ , or $c$ might be negative integers.
The student failed to prove that $\gcd(a,b) = 1$ implies $a$ and $b$ share no common factors.

Explanation: While the student's intuition is correct, they haven't provided a rigorous logical bridge from "a shares no common factors with b" to "a must divide c." A complete direct proof would need to use properties like Bézout's identity or fundamental properties of gcd to make this connection rigorous. Choice A suggests a formatting issue but doesn't identify the logical gap. Choice C is irrelevant to the logical structure. Choice D is incorrect because the definition of gcd(a,b) = 1 does imply they share no common prime factors.

Question 7

Consider the statement: "If $n$ is an integer such that $n^2 + 3n + 2$ is even, then $n$ is even." Which of the following represents the most appropriate first step in a direct proof of this statement?

Assume $n$ is an integer such that $n^2 + 3n + 2$ is even, and show that $n$ must be even by factoring the expression. (correct answer)
Assume $n$ is even and show that $n^2 + 3n + 2$ is even by substituting $n = 2k$ for some integer $k$ .
Assume $n$ is odd and derive a contradiction by showing that $n^2 + 3n + 2$ would then be odd.
Show that the contrapositive is true by assuming $n$ is odd and proving $n^2 + 3n + 2$ is odd.

Explanation: In a direct proof of an implication P → Q, we assume P is true and prove Q follows. Here P is "n² + 3n + 2 is even" and Q is "n is even", so we assume the hypothesis and prove the conclusion. Choice B describes proving the converse, while choices C and D describe proof by contradiction or contrapositive, not direct proof.

Question 8

In a direct proof that "For all integers $a$ and $b$ , if $a \equiv 1 \pmod{4}$ and $b \equiv 3 \pmod{4}$ , then $ab \equiv 3 \pmod{4}$ ," what should immediately follow the assumption that $a \equiv 1 \pmod{4}$ and $b \equiv 3 \pmod{4}$ ?

Express $a = 4k + 1$ and $b = 4j + 3$ for some integers $k$ and $j$ , then compute $ab$ . (correct answer)
Show that $a$ and $b$ are both odd integers, then use properties of odd number multiplication.
Verify the statement with specific examples like $a = 1, b = 3$ and $a = 5, b = 7$ .
Apply the theorem that congruences are preserved under multiplication to get $ab \equiv 1 \cdot 3 \pmod{4}$ .

Explanation: In a direct proof involving modular arithmetic, we should translate the congruence conditions into explicit algebraic forms that we can manipulate. Writing a = 4k + 1 and b = 4j + 3 allows us to compute ab algebraically and determine its remainder when divided by 4. Choice B makes an unnecessary detour, C uses examples rather than proof, and D assumes a theorem that may not be established.

Question 9

Consider the statement: "If $x$ and $y$ are rational numbers with $y \neq 0$ , then $\frac{x}{y}$ is rational." In a direct proof, after writing $x = \frac{p}{q}$ and $y = \frac{r}{s}$ where $p, q, r, s$ are integers with $q \neq 0$ and $s \neq 0$ , what additional condition must be established?

We must verify that $r \neq 0$ to ensure that $\frac{x}{y} = \frac{ps}{qr}$ is well-defined. (correct answer)
We must show that $\gcd(p,q) = 1$ and $\gcd(r,s) = 1$ for the fractions to be in lowest terms.
We must prove that $ps$ and $qr$ are integers to complete the definition of rational number.
We must establish that $q$ and $s$ have the same sign to ensure the division is properly defined.

Explanation: Since y ≠ 0 and y = r/s, we need r ≠ 0 to make the division x/y = (p/q)/(r/s) = ps/qr meaningful. The condition y ≠ 0 translates to r ≠ 0 in our representation, which is crucial for the proof's validity. Choice B is unnecessary because we don't need lowest terms to prove rationality. Choice C is obvious since products of integers are integers. Choice D is irrelevant to the mathematical validity of the expression.

Question 10

In proving "If $m$ and $n$ are integers such that $m^3 + n^3 + (m+n)^3$ is divisible by $9$ , then $3$ divides $mn$ ," what is the most effective way to begin the direct proof after assuming the hypothesis?

Assume $3$ does not divide $mn$ and show this leads to a contradiction with the divisibility by $9$ .
Use the identity $m^3 + n^3 = (m+n)(m^2 - mn + n^2)$ to factor the expression before expanding $(m+n)^3$ .
Consider the possible remainders when $m$ and $n$ are divided by $3$ and analyze each case systematically.
Expand $(m+n)^3$ and combine with $m^3 + n^3$ to get a simplified expression in terms of $m$ and $n$ . (correct answer)

Explanation: When approaching a direct proof in discrete mathematics, your first step after assuming the hypothesis is typically to manipulate the given conditions into a more workable form. This problem gives you that

m^3 + n^3 + (m+n)^3

is divisible by 9, so you need to transform this expression to reveal information about

mn

. The most effective approach is option D: expand

(m+n)^3

and combine terms. When you expand

(m+n)^3 = m^3 + 3m^2n + 3mn^2 + n^3

and substitute into the original expression, you get:

m^3 + n^3 + m^3 + 3m^2n + 3mn^2 + n^3 = 2m^3 + 2n^3 + 3mn(m+n)

This simplification reveals the factor

3mn

explicitly, which directly relates to what you're trying to prove. Option A describes proof by contradiction, not direct proof as specified in the question. Option B attempts factoring

m^3 + n^3

first, but this creates unnecessary complexity since you'd still need to expand

(m+n)^3

eventually. Option C suggests a case-by-case analysis based on remainders modulo 3, which is a valid proof technique but not the most direct path forward after assuming the hypothesis. The key insight is that algebraic manipulation often reveals the structure you need. When you see an expression involving cubes and need to prove something about a product, expanding and collecting terms usually exposes the relevant factors more clearly than other approaches.

Question 11

A student is proving: "If $x$ and $y$ are real numbers with $x > 0$ and $y > 0$ , then $\frac{x+y}{2} \geq \sqrt{xy}$ ." The student writes: "Assume $x > 0$ and $y > 0$ . We want to show $\frac{x+y}{2} \geq \sqrt{xy}$ . This is equivalent to showing $x + y \geq 2\sqrt{xy}$ ." What is the primary structural issue with this approach?

The student should square both sides immediately to eliminate the square root rather than multiplying by 2.
The student is working backwards from the conclusion instead of building forward from the assumptions. (correct answer)
The student should substitute specific values for $x$ and $y$ to make the proof more concrete.
The student failed to verify that both sides of the inequality are positive before manipulating them.

Explanation: In a direct proof, we start with the given assumptions and build logical steps forward to reach the conclusion. The student is starting with what they want to prove and manipulating it, which is working backwards. This approach doesn't constitute a valid direct proof structure. Choice A suggests a specific technique but doesn't address the structural issue. Choice C would make this an example-based argument, not a general proof. Choice D is incorrect because both sides are clearly positive under the given assumptions.

Question 12

A student is proving: "If $n$ is an integer and $n^2 \equiv 1 \pmod{8}$ , then $n$ is odd." The student begins: "Assume $n^2 \equiv 1 \pmod{8}$ . Then $n^2 = 8k + 1$ for some integer $k$ ." What should be the next most productive step in the direct proof?

Take square roots of both sides to get $n \equiv \pm 1 \pmod{8}$ , then conclude $n$ is odd.
Factor the right side as $n^2 = 8k + 1 = (8k) + 1$ and conclude that $n^2$ is odd.
Use the fact that $n^2 - 1 = 8k$ to factor the left side as $(n-1)(n+1) = 8k$ .
Analyze the possible values of $n \bmod 8$ by checking which values of $r$ make $r^2 \equiv 1 \pmod{8}$ . (correct answer)

Explanation: When proving statements involving modular arithmetic, you often need to systematically check all possible cases. This problem asks you to prove that if

n^2 \equiv 1 \pmod{8}

, then

n

must be odd. The most productive next step is D: checking which values of

r

make

r^2 \equiv 1 \pmod{8}

. Since any integer is congruent to 0, 1, 2, 3, 4, 5, 6, or 7 modulo 8, you can square each:

0^2 = 0 \equiv 0 \pmod{8}

1^2 = 1 \equiv 1 \pmod{8}

2^2 = 4 \equiv 4 \pmod{8}

3^2 = 9 \equiv 1 \pmod{8}

4^2 = 16 \equiv 0 \pmod{8}

5^2 = 25 \equiv 1 \pmod{8}

6^2 = 36 \equiv 4 \pmod{8}

7^2 = 49 \equiv 1 \pmod{8}

. Only when

n \equiv 1, 3, 5, 7 \pmod{8}

(all odd values) does

n^2 \equiv 1 \pmod{8}

, proving the statement directly. A is incorrect because you cannot simply take square roots in modular arithmetic—

n^2 \equiv 1 \pmod{8}

doesn't mean

n \equiv \pm 1 \pmod{8}

. As shown above,

n

could be 3, 5, or 7 modulo 8. B is unproductive because observing that

n^2

is odd doesn't directly tell you about

n

's parity without additional work. C leads to a more complex argument about factoring

(n-1)(n+1) = 8k

, which is unnecessarily complicated compared to the direct case analysis. Strategy tip: For modular arithmetic proofs, systematically checking all residue classes is often the most straightforward approach, especially when the modulus is small.

Question 13

To prove "If $n$ is an integer greater than 1 and $n$ is not prime, then $n$ has a prime divisor less than or equal to $\sqrt{n}$ ," a student writes: "Assume $n > 1$ and $n$ is composite. Then $n = ab$ where $1 < a < n$ and $1 < b < n$ ." What is the most critical next step in this direct proof?

Show that at least one of $a$ or $b$ must be prime by using the fundamental theorem of arithmetic
Establish that either $a = \sqrt{n}$ or $b = \sqrt{n}$ by using properties of integer factorization
Demonstrate that if $a \leq \sqrt{n}$ , then $a$ has a prime divisor, which is also a prime divisor of $n$
Prove that we cannot have both $a > \sqrt{n}$ and $b > \sqrt{n}$ by showing this leads to $ab > n$ (correct answer)

Explanation: When proving statements about composite numbers and their factors, the key insight is understanding the relationship between a number and its square root. This proof relies on a fundamental property: in any factorization of a composite number, at least one factor must be "small." The student has correctly set up the proof by expressing the composite number

n

n = ab

where both

a

and

b

are proper divisors. The critical next step is to establish that we cannot have both factors be large. If both

a > \sqrt{n}

and

b > \sqrt{n}

, then multiplying these inequalities gives us

ab > \sqrt{n} \cdot \sqrt{n} = n

. But this contradicts our assumption that

ab = n

. Therefore, at least one of

a

b

must be

\leq \sqrt{n}

. This makes option D correct. Option A is flawed because while the fundamental theorem guarantees prime factorizations exist, it doesn't directly help us locate a prime divisor

\leq \sqrt{n}

. Option B is incorrect because factors rarely equal

\sqrt{n}

exactly, and this isn't necessary for the proof. Option C puts the cart before the horse—we first need to show that a small factor exists before worrying about its prime divisors. Remember this pattern: proofs involving

\sqrt{n}

often use contradiction arguments about the sizes of factors. When you see composite number proofs mentioning square roots, think about what happens if all factors were "too big"—this usually leads to an impossible situation.

Question 14

A student attempts to prove: "If $x$ is a real number and $x^2 - 5x + 6 = 0$ , then $x = 2$ or $x = 3$ ." The student writes: "Assume $x^2 - 5x + 6 = 0$ . Factoring gives $(x-2)(x-3) = 0$ ." To complete this as a proper direct proof, what should the student do next?

Substitute $x = 2$ and $x = 3$ back into the original equation to verify they are solutions
Verify the factorization by expanding $(x-2)(x-3)$ and confirming it equals $x^2 - 5x + 6$
Apply the zero product property to conclude that $x - 2 = 0$ or $x - 3 = 0$ , then solve each equation (correct answer)
Use the quadratic formula to find the roots and show they match the factored form

Explanation: This question tests your understanding of direct proof structure and logical reasoning with algebraic equations. When proving a conditional statement directly, you assume the hypothesis and use valid logical steps to reach the conclusion. The student correctly assumed the hypothesis (

x^2 - 5x + 6 = 0

) and factored to get

(x-2)(x-3) = 0

. To complete the proof, you need to apply the zero product property, which states that if a product equals zero, then at least one factor must equal zero. This gives us

x - 2 = 0

x - 3 = 0

, leading directly to

x = 2

x = 3

– exactly what we wanted to prove. Choice C follows this logical progression. Choice A is incorrect because verification by substitution would be appropriate for checking solutions, but the proof's goal is to derive the conclusion from the hypothesis, not verify predetermined answers. Choice B is wrong because expanding to verify factorization is unnecessary work that doesn't advance the proof toward its conclusion – we already have the correct factored form. Choice D is incorrect because using the quadratic formula introduces an entirely different method when we already have a perfectly valid approach in progress, and direct proofs should follow a clear, linear logical path. Remember that in direct proofs, each step should logically follow from the previous one toward your target conclusion. Don't get sidetracked by verification steps or alternative methods – stay focused on the most direct logical path from hypothesis to conclusion.

Question 15

A student attempts to prove: "If $n$ is an integer and $n^3$ is odd, then $n$ is odd." The student writes: "Suppose $n$ is odd. Then $n = 2k + 1$ for some integer $k$ . So $n^3 = (2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 + 6k^2 + 3k) + 1$ , which is odd." What is wrong with this proof attempt?

The algebraic expansion of $(2k + 1)^3$ contains an error in the coefficients
The student proved the converse of the intended statement rather than the statement itself (correct answer)
The student should have considered the case when $k$ is even separately from when $k$ is odd
The final step incorrectly concludes that $2(4k^3 + 6k^2 + 3k) + 1$ is odd without justification

Explanation: The student proved "if

n

is odd, then

n^3

is odd" (the converse) instead of the given statement "if

n^3

is odd, then

n

is odd." For a direct proof, the student should assume

n^3

is odd and prove

n

is odd. Choice A is wrong because the expansion is correct. Choice C is unnecessary since the argument works for any integer

k

. Choice D is wrong because any integer of the form

2m + 1

is indeed odd by definition.

Question 16

To prove "If $n$ is divisible by 6, then $n$ is divisible by both 2 and 3," a student writes: "Assume $n$ is divisible by 6. Then $n = 6k$ for some integer $k$ . Since $6 = 2 \times 3$ , we have $n = 2 \times 3k = 2(3k)$ , so $n$ is divisible by 2." What should the student do next to complete the proof?

Show that $3k$ is an integer and conclude that $n$ is divisible by 2
Rewrite $n = 6k$ as $n = 3(2k)$ and show that $2k$ is an integer (correct answer)
Prove that if $n$ is divisible by 2, then it must also be divisible by 3
Verify the result by testing specific values of $k$ like $k = 1, 2, 3$

Explanation: The student has shown divisibility by 2. To complete the proof, they need to show divisibility by 3 using the same approach:

n = 6k = 3(2k)

, and since

k

is an integer,

2k

is also an integer, proving

n

is divisible by 3. Choice A is redundant since divisibility by 2 is already established. Choice C attempts to prove a false general statement. Choice D uses examples rather than completing the general proof.

Question 17

Consider the statement: "If $A \subseteq B$ and $B \subseteq C$ , then $A \subseteq C$ ." A student begins a direct proof by writing: "Let $x \in A$ . Since $A \subseteq B$ , we have $x \in B$ ." What should be the student's next step?

Conclude that since $x \in B$ , and $x$ was arbitrary, we have $A \subseteq B$ , which was given
Use the definition of subset to write $B \subseteq C$ as $B \cap C = B$ and apply this to element $x$
Show that $B \subseteq C$ implies there exists some element in $B$ that is also in $C$
Since $x \in B$ and $B \subseteq C$ , we can conclude $x \in C$ , completing the proof (correct answer)

Explanation: This question tests your understanding of direct proof techniques for subset relationships. When proving "if P and Q, then R" statements about sets, you need to systematically apply the definitions step by step. The student correctly started by taking an arbitrary element

x \in A

and used the given condition

A \subseteq B

to conclude

x \in B

. The logical next step is to apply the second given condition: since

x \in B

and we know

B \subseteq C

, we can conclude

x \in C

. This completes the proof because we've shown that any arbitrary element of

A

must also be in

C

, which is exactly what

A \subseteq C

means. Choice A is wrong because it's circular reasoning—restating what was already given rather than making progress toward the conclusion. The student already used

A \subseteq B

; restating it doesn't advance the proof. Choice B incorrectly uses set intersection notation. While

A \subseteq B

does imply

A \cap B = A

, this approach unnecessarily complicates the proof and doesn't directly help establish that

x \in C

. Choice C misunderstands what

B \subseteq C

means. The subset relationship means every element of

B

is in

C

, not just that some element exists in both sets. This approach would lead to an incomplete proof. Remember: In direct proofs of subset relationships, work element by element through the chain of given conditions. Take an arbitrary element from the first set, then systematically apply each subset relationship until you reach the target set.

Question 18

A student wants to prove: "If $f: \mathbb{R} \to \mathbb{R}$ is a function such that $f(x + y) = f(x) + f(y)$ for all real numbers $x$ and $y$ , then $f(0) = 0$ ." Which represents the most direct approach?

Choose specific values like $x = 1, y = -1$ and use the functional equation to determine $f(0)$
Apply the functional equation with $x = y = 0$ to obtain $f(0) = f(0) + f(0)$ , then solve for $f(0)$
Use the functional equation with $y = 0$ to get $f(x) = f(x) + f(0)$ for all $x$ , then deduce $f(0) = 0$ (correct answer)
Prove that $f$ must be linear, then use the fact that linear functions through the origin satisfy $f(0) = 0$

Explanation: Setting

y = 0

in the functional equation gives

f(x + 0) = f(x) + f(0)

, so

f(x) = f(x) + f(0)

for all

x

. This immediately implies

f(0) = 0

. Choice A works but is less general than using the functional equation directly. Choice B leads to

f(0) = 2f(0)

, which gives

f(0) = 0

but is less straightforward. Choice D proves more than necessary and doesn't represent the most direct path.

Question 19

A student wants to prove: "If $a$ and $b$ are rational numbers, then $a + b$ is rational." The student writes: "Let $a = \frac{p}{q}$ and $b = \frac{r}{s}$ where $p, q, r, s$ are integers with $q \neq 0$ and $s \neq 0$ ." What is the primary weakness in this opening?

The student should assume $a$ and $b$ are arbitrary rational numbers before expressing them in fractional form (correct answer)
The student failed to specify that $p, q, r, s$ must all be positive integers
The student should have used the same denominator for both fractions from the start
The student needs to prove that such representations exist before using them

Explanation: In a direct proof, we must clearly state that we're considering arbitrary elements satisfying the hypothesis before manipulating them. The student jumped directly to the fractional representation without establishing that

a

and

b

are arbitrary rational numbers. Choice B is incorrect because integers can be negative. Choice C is wrong because using different denominators is fine and more general. Choice D is incorrect because the definition of rational numbers guarantees such representations exist.

Question 20

Consider proving: "If $x$ and $y$ are real numbers such that $x > 0$ and $y > 0$ , then $\frac{x + y}{2} \geq \sqrt{xy}$ ." Which approach represents the most direct proof strategy?

Start with $\frac{x + y}{2} \geq \sqrt{xy}$ and manipulate it algebraically until reaching a true statement
Begin with the equivalent inequality $(\sqrt{x} - \sqrt{y})^2 \geq 0$ and work forward to the desired result (correct answer)
Assume the conclusion is false and derive a contradiction using the given conditions
Use specific examples to verify the inequality holds, then generalize the pattern

Explanation: A direct proof works forward from assumptions to conclusion. Starting with the obviously true statement

(\sqrt{x} - \sqrt{y})^2 \geq 0

and algebraically manipulating it forward to reach the desired inequality is the most direct approach. Choice A works backwards from conclusion to assumptions (not direct). Choice C describes proof by contradiction. Choice D uses inductive reasoning rather than deductive proof.