Unit 1 Proof Portfolio

Mathematical language, numbers, transformations, and proof

Purpose

This untimed portfolio asks you to produce six complete arguments using the proof forms developed in Unit 1. It is not a speed test. The goal is to create work that another reader can verify without reconstructing missing definitions, cases, or domain conditions.

Complete the portfolio after studying:

  1. Sets, intervals, statements, and quantifiers;
  2. The real number system and its extensions;
  3. Algebraic laws and valid transformation;
  4. Proof methods for algebra.

All tasks are original to this course. You may use notes and the linked chapters. Label any computational experiment as exploration or verification; the submitted proof must remain mathematical and self-contained.

WarningAssessment status

This portfolio is a review candidate, not a credential or prediction of performance elsewhere. Its rubric and model analyses still require independent mathematical and editorial review.

Submission protocol

For each artifact:

  1. copy the theorem exactly and name its universe;
  2. identify the proof method you intend to use;
  3. list definitions or earlier results you expect to need;
  4. write a complete first attempt without reading the model analysis;
  5. audit every assumption, implication, case boundary, and cancellation;
  6. optionally perform a separate computational check;
  7. use the rubric to annotate one strength and one required revision; and
  8. submit a corrected proof alongside the first attempt.

A portfolio demonstrates learning through revision. Do not erase the first attempt: the correction record is part of the evidence.

Artifact 1 — direct proof from definitions

Let \(a,b,c\in\mathbb Z\). Prove directly that if \(a\mid b\) and \(a\mid c\), then \(a\mid(3b-2c)\).

Your proof must introduce witnesses for both divisibility hypotheses and name the integer that witnesses the conclusion.

What do \(a\mid b\) and \(a\mid c\) mean separately? After substitution, can the result be written as \(a\) times an integer? Where is integer closure used?

Let \(a,b,c\in\mathbb Z\) and suppose \(a\mid b\) and \(a\mid c\). There exist \(m,n\in\mathbb Z\) such that \(b=am\) and \(c=an\). Hence

[ 3b-2c=3am-2an=a(3m-2n). ]

Because integers are closed under multiplication and subtraction, \(3m-2n\in\mathbb Z\). It is the required witness, so \(a\mid(3b-2c)\).

A complete proof does not divide by \(a\) and therefore does not need \(a\ne0\). Under the standard definition, \(0\mid b\) forces \(b=0\); the witness argument still works.

Artifact 2 — exhaustive cases

Prove for every real \(x\) that

[ |x-1|+|x+1|. ]

Use an exhaustive case argument, identify the intervals that determine the absolute-value formulas, and classify every equality case.

The expressions inside the absolute values change sign at \(-1\) and \(1\). Use three intervals. Do not use a graph as the proof.

The critical values partition \(\mathbb R\) into \(x<-1\), \(-1\le x\le1\), and \(x>1\).

  • If \(x<-1\), then both \(x-1\) and \(x+1\) are negative, so the sum is \((1-x)+(-x-1)=-2x>2\).
  • If \(-1\le x\le1\), then \(x-1\le0\) and \(x+1\ge0\), so the sum is \((1-x)+(x+1)=2\).
  • If \(x>1\), both expressions are positive, so the sum is \((x-1)+(x+1)=2x>2\).

The cases are exhaustive. Therefore the inequality holds for all real \(x\), and equality holds exactly on the interval \([-1,1]\).

Artifact 3 — contraposition

For \(n\in\mathbb Z\), prove by contraposition:

[ 5n^2n. ]

Use the division algorithm rather than citing a prime-divisor theorem without proof.

The contrapositive begins \(5\nmid n\). Which four nonzero residues modulo \(5\) are possible, and what are their squares modulo \(5\)?

We prove the contrapositive. If \(5\nmid n\), the residue of \(n\) modulo \(5\) is \(1,2,3\), or \(4\). The corresponding square residues are

[ 12,2,2,2. ]

None is \(0\), so \(5\nmid n^2\). This proves \(5\nmid n\Rightarrow5\nmid n^2\), the contrapositive of the stated theorem. Hence \(5\mid n^2\Rightarrow5\mid n\).

Artifact 4 — contradiction

Prove that \(\sqrt5\) is irrational. You may use the result from Artifact 3, but you must state where it enters and why a lowest-terms representation exists.

Assume \(\sqrt5=p/q\) with \(q>0\) and \(\gcd(p,q)=1\). After squaring, apply the prime-square result twice.

Assume for contradiction that \(\sqrt5\) is rational. Every rational number has a lowest-terms representation, so write \(\sqrt5=p/q\) with integers \(p,q\), \(q>0\), and \(\gcd(p,q)=1\). Squaring gives \(p^2=5q^2\). Thus \(5\mid p^2\), and Artifact 3 gives \(5\mid p\). Write \(p=5r\). Substitution yields \(25r^2=5q^2\), hence \(q^2=5r^2\). Therefore \(5\mid q^2\), so Artifact 3 gives \(5\mid q\). This contradicts \(\gcd(p,q)=1\). Thus \(\sqrt5\) is irrational.

Artifact 5 — biconditional, existence, and uniqueness

Let \(a,b\in\mathbb R\). Prove the biconditional:

[ ax=b a. ]

The forward direction must handle both possibilities for \(b\) when \(a=0\). The reverse direction must prove existence and uniqueness separately.

For the forward direction, contraposition is efficient. If \(a=0\), compare \(b=0\) and \(b\ne0\). For the reverse direction, construct \(x=b/a\) and use multiplicative cancellation.

Forward direction. We prove the contrapositive. Suppose \(a=0\). If \(b=0\), the equation \(0x=0\) is true for every real \(x\), so the solution is not unique. If \(b\ne0\), the equation \(0x=b\) has no solution. Thus \(a=0\) prevents exactly one solution, proving that exactly one solution implies \(a\ne0\).

Reverse direction. Suppose \(a\ne0\). Existence follows because \(x=b/a\) is real and satisfies \(a(b/a)=b\). For uniqueness, if \(x_1,x_2\) both solve the equation, then \(ax_1=b=ax_2\). Since \(a\ne0\), multiplicative cancellation gives \(x_1=x_2\). Hence the solution exists and is unique.

Artifact 6 — induction and a recurrence

Define \(u_0=0\), \(u_1=1\), and

[ u_{n+2}=3u_{n+1}-2u_n(n). ]

Prove by induction that \(u_n=2^n-1\) for every integer \(n\ge0\). Explain why the inductive statement or base information must control two consecutive terms.

Check the claimed formula at both \(n=0\) and \(n=1\). Assume formulas for \(u_k\) and \(u_{k+1}\), then compute \(u_{k+2}\).

Use the strengthened statement

[ P(k):u_k=2^k-1u_{k+1}=2^{k+1}-1. ]

For \(k=0\), \(u_0=0=2^0-1\) and \(u_1=1=2^1-1\). Suppose \(P(k)\) holds. Then

[ \[\begin{aligned} u_{k+2} &=3u_{k+1}-2u_k\\ &=3(2^{k+1}-1)-2(2^k-1)\\ &=6\cdot2^k-3-2\cdot2^k+2\\ &=4\cdot2^k-1\\ &=2^{k+2}-1. \end{aligned}\]

]

Together with the assumed formula for \(u_{k+1}\), this proves \(P(k+1)\). By induction, \(u_n=2^n-1\) for every \(n\ge0\). Two consecutive formulas are needed because the recurrence consumes two preceding terms.

Counterexample and computation audit

After the six proofs, complete this short audit.

A search reports no counterexample to

[ n4+n2+1 ]

for integers \(1\le n\le100\). Explain why this is not a proof. Find an exact counterexample or factorization, and state how computation may legitimately support the revised argument.

The search covers only finitely many integers, so it cannot prove a universal claim. In fact, at \(n=2\) the expression is \(16+4+1=21\), which is composite. More structurally,

[ n4+n2+1=(n2+n+1)(n2-n+1). ]

For \(n\ge2\), both factors exceed \(1\), so the expression is composite, not prime. Software may suggest or verify the expansion of this factorization. The written proof must still show why both factors exceed \(1\) on the stated domain.

Analytic rubric

Score each of the six required artifacts from 0 to 4 in each dimension, for a maximum of 24 points per artifact and 144 points total.

Dimension 4 — independent 3 — secure 2 — developing 1 — fragmentary 0 — absent
Interpretation Universe, quantifiers, hypotheses, and target are precise One minor notation lapse Meaning mostly right but a condition is implicit Major misreading needs repair No relevant interpretation
Method selection Method is efficient and its use is explained Valid method with modest inefficiency Method can work but structure is unclear Method addresses a different claim No coherent method
Definitions and assumptions Every definition and assumption is introduced where used One noncritical omission Several implied steps remain recoverable A decisive condition is missing Circular or undefined start
Reasoning and algebra Every inference is valid and ordered Correct with a minor local slip Core idea works after repair Major gap or invalid transformation No viable chain
Completeness and exceptions All cases/directions/equality conditions are closed Complete except one minor closing sentence Main claim proven but a branch or obligation is thin Only one substantial branch Target not established
Communication and verification Readable proof plus meaningful audit/check Clear proof with limited audit Understandable after reconstruction Difficult to follow No assessable response

Portfolio readiness interpretation

  • 122–144: strong independent evidence; proceed to the timed examination after correcting every marked issue.
  • 101–121: generally secure; revise every dimension below 3 before timing.
  • 78–100: developing; repeat the affected proof forms with new examples.
  • Below 78: return to the relevant chapter sections and complete guided proofs before reassessment.

A total score cannot hide a structural zero. Do not proceed if any artifact scores 0 in reasoning or completeness.

Peer-review protocol

A peer reviewer should not rewrite the proof. Instead, annotate:

  1. the exact sentence where each assumption is introduced;
  2. the line where each definition is closed;
  3. whether every case is exhaustive and every biconditional direction appears;
  4. whether any division, root, or implication needs a missing condition;
  5. one question the current proof leaves a careful reader; and
  6. one concrete revision that would make the argument independently verifiable.

After revision, the author answers each annotation with either a change or a mathematical justification for retaining the original line.

Computational verification appendix

Use computation only after completing the proofs. The following examples check finite evidence or algebraic identities; neither replaces the quantified argument.

from fcmath.algebra import find_implication_counterexample

# Search for a failure of Artifact 3 on a deliberately finite range.
result = find_implication_counterexample(
    range(-100, 101),
    hypothesis=lambda n: n**2 % 5 == 0,
    conclusion=lambda n: n % 5 == 0,
)
result

A result of None means only that this finite search found no failure. The residue proof establishes the universal theorem.

import sympy as sp

n = sp.symbols("n", integer=True)
sp.factor(n**4 + n**2 + 1)

\(\displaystyle \left(n^{2} - n + 1\right) \left(n^{2} + n + 1\right)\)

The factorization is a useful discovery. The proof still supplies the domain argument showing compositeness for every integer \(n\ge2\).

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