Unit 1 Proof Examination

75-minute closed-resource paper with solutions and marking scheme

Instructions

This original examination assesses cumulative work from Chapters 1.1–1.4. Allow 75 minutes and attempt every question. The paper has 60 marks. Use no notes, calculator, computer algebra system, or external resource during the attempt.

Write complete arguments. A correct final statement without supporting reasoning may earn little credit. You may cite results proved in the four Unit 1 chapters, but state the result and verify that its hypotheses apply.

Assessment status

This paper and marking scheme are review candidates. They provide local learning evidence, not a credential or a prediction of performance in another examination.

Suggested timing

Question	Focus	Marks	Minutes
1	quantifiers and cancellation	10	12
2	contrapositive divisibility proof	12	15
3	counterexample and theorem repair	10	12
4	equivalence and domain audit	14	17
5	synthesis divisibility proof	14	19
Total		60	75

Do not read beyond the end-of-paper marker until the timed attempt is complete.

Question 1 — language and cancellation [10]

Let \(a,b,c\in\mathbb R\).

(a) [3] Write with quantifiers and logical connectives the multiplicative cancellation theorem: if \(ac=bc\) and \(c\ne0\), then \(a=b\).

(b) [3] Negate your complete statement from part (a), pushing the negation through every quantifier and connective.

(c) [4] Remove the condition \(c\ne0\). Give an exact counterexample to the resulting claim and explain why it satisfies the hypothesis but not the conclusion.

Question 2 — contraposition [12]

For every integer \(n\), prove by contraposition that

[ 4n^2n. ]

You may use the definition of odd integer, but you must derive the relevant remainder of an odd square rather than citing it without explanation.

Question 3 — counterexample and repair [10]

Consider the claim:

If real \(x+y\) is rational, then both \(x\) and \(y\) are rational.

(a) [4] Disprove the claim with an exact counterexample.

(b) [2] Explain why choosing \(x=1\), \(y=2\) would not be a counterexample.

(c) [4] State and prove a valid implication about the sum of two rational numbers that is closely related to the false claim.

Question 4 — earliest-invalid-step audit [14]

A student solves over \(\mathbb R\):

[ \[\begin{aligned} \frac{x^2-4}{x-2}&=x+2,\\ x^2-4&=(x+2)(x-2),\\ x^2-4&=x^2-4,\\ 0&=0. \end{aligned}\]

]

The student concludes that every real number is a solution.

(a) [3] State the original domain.

(b) [4] Audit each displayed transformation and identify the earliest missing condition or invalid inference.

(c) [3] Give the exact solution set.

(d) [4] Decide whether the functions \(f(x)=(x^2-4)/(x-2)\) and \(g(x)=x+2\), each with its natural real domain, are equal. Justify your answer.

Question 5 — synthesis proof [14]

Prove that for every integer \(n\),

[ 24n(n+1)(n+2)(n+3). ]

Your proof must establish the required factors of \(3\) and \(8\), explain why the relevant cases are exhaustive, and justify combining the factors into \(24\).

End of timed paper

Stop the timer. Record your confidence from 0–3 beside each question before opening the solutions:

0: no viable method;
1: partial method with major gaps;
2: likely correct but not fully audited;
3: independently verifiable.

Then use a different color to mark and correct the paper.

Solutions and marking scheme

Question 1 [10]

Solution and marks

(a) [3] One correct form is

[ (a,b,cR) [(ac=bcc)a=b]. ]

Award 1 mark for the universal real quantifiers, 1 for the complete conjunctive hypothesis, and 1 for the implication to \(a=b\).

(b) [3] The negation is

[ (a,b,cR) [ac=bccab]. ]

Award 1 for changing universal to existential, 1 for retaining both hypothesis conditions, and 1 for negating the conclusion to \(a\ne b\).

(c) [4] For example, choose \(a=1\), \(b=2\), \(c=0\). Then \(ac=0=bc\), so the reduced hypothesis is true, while \(a=b\) is false. Award 1 per value, 1 for checking equal products, and 1 for explicitly denying the conclusion.

Question 2 [12]

Solution and marks

We prove the contrapositive: if \(n\) is odd, then \(4\nmid n^2\). Let \(n=2k+1\) for some integer \(k\). Then

[ n^2=(2k+1)2=4k^2+4k+1=4(k2+k)+1. ]

Thus \(n^2\) has remainder \(1\) upon division by \(4\), so \(4\nmid n^2\). This proves the contrapositive and therefore the original implication.

Marks: 2 for stating the correct contrapositive; 2 for the odd representation with integer witness; 3 for correct expansion; 2 for the remainder-\(1\) form; 1 for interpreting nondivisibility; and 2 for closing by contraposition.

Question 3 [10]

Solution and marks

(a) [4] Let \(x=\sqrt2\) and \(y=-\sqrt2\). Both are irrational, but \(x+y=0\) is rational. Award 2 for exact values, 1 for verifying the rational sum, and 1 for verifying that the claimed conclusion fails.

(b) [2] The values \(1\) and \(2\) make both the hypothesis and conclusion true. A supporting example cannot refute a universal implication.

(c) [4] A valid related theorem is: if \(x,y\) are rational, then \(x+y\) is rational. Write \(x=a/b\) and \(y=c/d\) with integer numerators and nonzero integer dominators. Then

[ x+y=. ]

The numerator is an integer and the denominator is a nonzero integer, so the sum is rational. Award 1 for a correct theorem, 1 for representations, 1 for common denominator algebra, and 1 for closure/nonzero justification.

Question 4 [14]

Solution and marks

(a) [3] The denominator requires \(x\ne2\), so the original domain is \(\mathbb R\setminus\{2\}\).

(b) [4] Multiplication by \(x-2\) is reversible on the original domain because there \(x-2\ne0\). Expansion is valid, and \(0=0\) confirms that every admissible input works. The earliest failure is not an algebra line but the final inference from “every admissible \(x\)” to “every real \(x\),” which drops the standing domain restriction. Award full credit to an answer that calls the first multiplication incomplete because the restriction was not stated, provided it explains that the step becomes valid under \(x\ne2\).

(c) [3] The solution set is \(\mathbb R\setminus\{2\}\).

(d) [4] They are not equal as functions with natural domains. The rational formula is undefined at \(2\), while \(g(2)=4\). They agree at every point of the smaller domain, but equality of functions also requires equal domains.

Question 5 [14]

Solution and marks

Among four consecutive integers, one is divisible by \(3\), because their residues contain at least one complete run of three consecutive classes modulo \(3\). Thus \(3\) divides the product.

For the factor \(8\), split by parity of \(n\).

If \(n\) is even, then \(n\) and \(n+2\) are the two even factors. One of them is divisible by \(4\), while the other is divisible by \(2\); their product is therefore divisible by \(8\).
If \(n\) is odd, then \(n+1\) and \(n+3\) are even. One is divisible by \(4\) and the other by \(2\), so again their product is divisible by \(8\).

The two parity cases exhaust the integers. Hence the four-factor product is divisible by both \(3\) and \(8\). Since \(\gcd(3,8)=1\), it is divisible by \(24\).

Marks: 3 for the factor-\(3\) argument; 2 for an exhaustive parity split; 5 for the factor-\(8\) argument including a factor \(4\) and another factor \(2\); 2 for combining coprime divisors; and 2 for a clear universal conclusion.

Correction and reassessment record

For every lost mark, record:

Question	Error category	Earliest unsupported line	Correct principle	Revised line	New check
	interpretation / definition / logic / algebra / completeness

Suggested local readiness threshold: at least 42/60, no score below half on Questions 4 or 5, and a complete correction of every logical or domain error. This threshold guides study only. If the conditions are not met, revise the portfolio artifact matching the weak proof form and retake a fresh variant written by an instructor or peer rather than memorizing this marking scheme.