A calculation answers “what?” A proof answers “why must it always be so?” The difference matters whenever a pattern continues beyond the cases we can check, a parameter can take infinitely many values, or a conclusion depends on a hidden condition. Ten thousand successful numerical tests do not prove a universal statement; one valid counterexample refutes it.
Proof is also not a separate activity reserved for abstract mathematics. Every careful equation solution contains a miniature proof: the domain is stated, each transformation is licensed, every candidate is checked, and completeness is justified. A proof simply makes the logical architecture explicit enough that another reader can verify it line by line.
This chapter develops a method-selection system rather than a collection of ritual phrases. Direct proof follows definitions forward. Cases partition the universe. Contraposition replaces an implication by a logically equivalent one. Contradiction assumes the denial of the target and derives an impossibility. Biconditional proof establishes both directions. Existence constructs or certifies an object; uniqueness proves that two alleged objects must coincide. Induction handles statements indexed by the natural numbers. Counterexamples disprove universal claims.
A demanding examination rarely announces which method to use. You must read the logical form, inspect the definitions, try a productive representation, and change course when the chosen route stalls. The habits in this chapter are therefore both mathematical and strategic.
After completing the chapter, you should be able to:
You should be able to translate quantified statements and negate implications from Sets, intervals, statements, and quantifiers. You should know exact number-system definitions from The real number system and its extensions and be able to label reversible and implication-only steps from Algebraic laws and valid transformation.
A typical theorem has the form
[ (xU),[P(x)Q(x)]. ]
The universe \(U\) determines what objects are allowed. The hypothesis \(P(x)\) is what may be assumed after choosing an arbitrary admissible \(x\). The conclusion \(Q(x)\) is what must be derived. A proof cannot assume the conclusion, choose a convenient special \(x\), or silently enlarge the hypothesis.
A reliable opening is:
Let \(x\in U\) be arbitrary and suppose \(P(x)\).
Then unpack definitions. If \(n\) is odd, write \(n=2k+1\) for some integer \(k\). If \(r\) is rational, write \(r=p/q\) with integers \(p,q\) and \(q\ne0\). If \(a\mid b\), introduce an integer witness \(k\) with \(b=ak\). These representations are not cosmetic. They convert a verbal property into algebra that can be used.
A reliable closing explicitly reconnects the derived formula to the target:
Since \(2(k^2+k)\) is twice an integer, \(n^2-1\) is even, as required.
Consider the claim:
For every integer \(n\), if \(n\) is odd, then \(n^2\) is odd.
Its components are:
Now prove it. Let \(n\) be an arbitrary odd integer. Then \(n=2k+1\) for some integer \(k\). Therefore
[ n2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1. ]
Because \(2k^2+2k\in\mathbb Z\), the final expression is one more than twice an integer. Hence \(n^2\) is odd.
The proof works for an arbitrary odd integer. Checking \(3^2\) and \(5^2\) might suggest the theorem, but the witness \(2k^2+2k\) proves every case at once.
For “if real \(x>2\), then \(x^2>4\),” name the universe, hypothesis, and conclusion. Is the converse true?
The universe is \(\mathbb R\), the hypothesis is \(x>2\), and the conclusion is \(x^2>4\). The converse is false: \(x=-3\) satisfies \(x^2>4\) but not \(x>2\).
Use direct proof when the hypothesis can be translated into a form that naturally simplifies toward the conclusion. The standard flow is:
Direct proof is especially effective for divisibility, parity, rationality, set inclusion, algebraic identities, and inequalities obtained from known nonnegative expressions.
Prove: if integers \(a,b,c,d\) satisfy \(a\mid b\) and \(c\mid d\), then \(ac\mid bd\).
By \(a\mid b\), there is \(m\in\mathbb Z\) such that \(b=am\). By \(c\mid d\), there is \(n\in\mathbb Z\) such that \(d=cn\). Multiply:
[ bd=(am)(cn)=ac(mn). ]
Because integers are closed under multiplication, \(mn\in\mathbb Z\). Thus \(bd\) is an integer multiple of \(ac\), which is precisely \(ac\mid bd\).
Notice that the proof names two witnesses and then constructs the required witness \(mn\). Writing only “obvious by multiplication” would hide the defining step.
Prove that for every real \(x>0\),
[ x+1x, ]
with equality exactly when \(x=1\).
For real \(x\), \((x-1)^2\ge0\). Expanding gives
[ x^2-2x+1. ]
Since \(x>0\), division by \(x\) preserves order:
[ x-2+1x, ]
and hence \(x+1/x\ge2\).
Equality in the final inequality occurs exactly when every reversible step began with equality, namely when \((x-1)^2=0\). Thus \(x=1\). Substitution confirms \(1+1=2\). The proof establishes both the bound and its equality case.
Prove directly that the sum of two rational numbers is rational.
Write each rational with an integer numerator and a nonzero integer denominator. The proof must explain why the new denominator is nonzero.
Let \(r,s\in\mathbb Q\). Then \(r=a/b\) and \(s=c/d\) for some integers \(a,b,c,d\) with \(b\ne0\) and \(d\ne0\). Therefore
[ r+s=ab+cd=. ]
The numerator \(ad+bc\) is an integer. The denominator \(bd\) is an integer and is nonzero because a product of nonzero real numbers is nonzero. Hence \(r+s\) is a ratio of integers with nonzero denominator, so \(r+s\in\mathbb Q\).
Some definitions change form across a boundary. Absolute value depends on sign; a parity argument may depend on whether an integer is even or odd; a parameter formula may depend on whether a coefficient is zero, positive, or negative. A proof by cases is complete only if the cases cover the entire universe. Disjoint cases are usually easier to audit because no input appears twice.
The structure is:
Prove that \(|x|\ge x\) for every real \(x\), with equality exactly when \(x\ge0\).
Every real number satisfies exactly one of \(x\ge0\) or \(x<0\).
The cases are exhaustive, so \(|x|\ge x\) for every real \(x\). Equality occurs in exactly the first case, \(x\ge0\).
Prove that \(n(n+1)\) is even for every integer \(n\).
Every integer is even or odd.
In either case one of two consecutive integers supplies a factor of \(2\). Therefore their product is even.
To prove a statement for every real parameter \(a\) when division by \(a\) may be needed, which partition is logically sufficient?
A. \(a>0\) and \(a<0\) B. \(a\ne0\) and \(a=0\) C. \(a\ge0\) only D. several tested numerical values
B is sufficient for whether division is defined. If inequality direction also matters, refine \(a\ne0\) into \(a>0\) and \(a<0\), giving three cases. Choice A omits zero.
The implication \(P\Rightarrow Q\) is logically equivalent to its contrapositive
[ QP. ]
This is not the converse \(Q\Rightarrow P\) and not the inverse \(\neg P\Rightarrow\neg Q\). Contraposition is useful when negating the conclusion produces a concrete algebraic representation, especially in parity, divisibility, injectivity, and irrationality claims.
A clean opening says:
We prove the contrapositive. Suppose \(\neg Q\).
The closing explicitly invokes equivalence:
Therefore \(\neg Q\Rightarrow\neg P\), so the original implication follows by contraposition.
The direct hypothesis \(n^2=2k\) does not immediately display a factor of \(2\) in \(n\). Prove the contrapositive: if \(n\) is odd, then \(n^2\) is odd.
Let \(n=2k+1\) for some \(k\in\mathbb Z\). Then
[ n2=4k2+4k+1=2(2k^2+2k)+1, ]
which is odd. Thus an odd integer cannot have an even square. By contraposition, if \(n^2\) is even, then \(n\) is even.
This proof relies on the integer parity dichotomy. The universe must be \(\mathbb Z\); “even real number” is not the same standard property.
Prove: if \(x,y\in\mathbb R\) and \(x+y\) is irrational, then at least one of \(x,y\) is irrational.
The conclusion is a disjunction. Its negation says both \(x\) and \(y\) are rational. We prove the contrapositive. Suppose \(x,y\in\mathbb Q\). Rational numbers are closed under addition, as shown in the guided direct proof, so \(x+y\in\mathbb Q\). Therefore
[ (xQyQ)x+yQ. ]
This is the contrapositive of the original statement, which therefore follows. The theorem does not say exactly one summand is irrational; both may be.
Prove by contraposition: if \(3\nmid n\), then \(3\nmid n^2\) for every integer \(n\).
The stated claim is already the contrapositive of “if \(3\mid n^2\), then \(3\mid n\).” Use the cases \(n=3q+1\) and \(n=3q+2\).
If \(3\nmid n\), the division algorithm gives \(n=3q+1\) or \(n=3q+2\). In the first case,
[ n2=9q2+6q+1=3(3q^2+2q)+1. ]
In the second,
[ n2=9q2+12q+4=3(3q^2+4q+1)+1. ]
In both cases the square has remainder \(1\) upon division by \(3\), so \(3\nmid n^2\). Equivalently, \(3\mid n^2\Rightarrow3\mid n\).
To prove a statement \(S\) by contradiction, assume \(\neg S\) and derive a statement known to be false. The contradiction may be \(0=1\), an integer being both even and odd, a reduced fraction having a common factor, or a violation of a strict inequality.
Contradiction is especially effective when the target denies existence (“there is no…”), asserts irrationality, or combines conditions that cannot coexist. It must not merely produce a surprising statement. The final conflict must be named, and it must arise from the negated target together with accepted hypotheses.
Assume for contradiction that \(\sqrt6\) is rational. Write
[ =pq ]
with integers \(p,q\), \(q>0\), and \(\gcd(p,q)=1\). Squaring gives
[ p2=6q2. ]
Thus \(p^2\) is even, so Example 6 implies \(p\) is even. Write \(p=2r\). Then
[ 4r2=6q22r2=3q2. ]
The right side shows \(2\mid3q^2\). Since \(2\nmid3\), parity forces \(q^2\) even, so \(q\) is even. Hence both \(p\) and \(q\) are divisible by \(2\), contradicting \(\gcd(p,q)=1\). Therefore \(\sqrt6\) is irrational.
An alternative prime-factor argument tracks both \(2\) and \(3\). The essential conflict is that a lowest-terms numerator and denominator acquire a common factor.
To prove “there is no greatest integer,” what should a contradiction proof assume, and what object should it construct?
Assume there is a greatest integer \(N\). Then \(N+1\) is an integer and \(N+1>N\), contradicting the alleged greatest property.
A biconditional \(P\Longleftrightarrow Q\) contains two claims:
[ PQ QP. ]
Prove the directions separately unless every displayed algebraic step is clearly reversible under stated conditions. One easy direction never licenses the other. In a chain of equivalences, domain restrictions must remain visible at every step.
Prove that an integer \(n\) is odd if and only if \(n^2\) is odd.
Forward direction. Suppose \(n\) is odd. Example 1 proves that \(n^2\) is odd.
Reverse direction. Suppose \(n^2\) is odd. If \(n\) were even, then \(n=2k\) and \(n^2=2(2k^2)\) would be even, impossible. Hence \(n\) is odd.
Both implications hold, so
[ nn^2. ]
The reverse direction can also be described as the contrapositive of “even \(n\) implies even \(n^2\).” Labeling directions prevents accidental repetition of the same implication.
For real \(x\), prove
[ x^2+44x(x-2)^2, ]
and then prove that equality in \(x^2+4\ge4x\) holds if and only if \(x=2\).
Subtracting \(4x\) from both sides and completing the square are reversible:
[ x^2+44x x^2-4x+4 (x-2)^2. ]
The last inequality holds for every real \(x\). Equality satisfies \((x-2)^2=0\Longleftrightarrow x-2=0\Longleftrightarrow x=2\). Thus the bound is universal and its equality case is unique.
The statement “there exists exactly one \(x\) such that \(P(x)\)” contains two obligations:
A formula may prove existence by construction, but it does not automatically prove uniqueness. Conversely, proving “if solutions exist, they are equal” does not prove that any solution exists.
Let real \(a,b\) not both be zero. Prove that there exist real \(x,y\) satisfying
[ ax+by=1. ]
Because \(a,b\) are not both zero, \(a^2+b^2>0\). Define
[ x=, y=. ]
These are real and defined. Substitution gives
[ ax+by==1. ]
Thus a solution exists. The proof does not claim uniqueness—and in fact the solution is generally not unique. If \((x,y)\) is one solution, then \((x+bt,y-at)\) is another for every real \(t\).
Prove that the equation
[ x+1x=2 ]
has exactly one positive real solution.
Existence. \(x=1\) is positive and satisfies \(1+1=2\).
Uniqueness. Let \(x>0\) be any solution. Multiplication by nonzero \(x\) gives
[ x^2+1=2x (x-1)^2=0 x=1. ]
Therefore every positive solution equals the constructed solution. Hence the unique positive real solution is \(1\).
The same algebra actually shows uniqueness over the entire real domain \(x\ne0\). Positivity arose naturally from the inequality context but is stronger than necessary for the equation alone.
A student proves: “If \(x\) and \(y\) both solve \(3t-7=0\), then \(3x-7=3y-7\), so \(x=y\).” What has been proved, and what remains?
The argument proves at most uniqueness: two solutions must coincide. Existence remains. Substituting \(t=7/3\) proves that a solution actually exists.
A universal claim
[ (xU),[P(x)Q(x)] ]
is false exactly when some \(x\in U\) makes \(P(x)\) true and \(Q(x)\) false. A value outside the universe, or one that violates the hypothesis, proves nothing.
Good counterexamples are small, exact, and diagnostic. Boundary values such as \(0\), \(1\), \(-1\), equal parameters, repeated roots, or excluded denominators often reveal the missing hypothesis. After refuting a claim, try to repair it by adding the weakest natural condition that blocks the failure.
Claim: “The product of two irrational real numbers is irrational.”
Choose \(x=y=\sqrt2\). Both factors are irrational, but
[ xy=()^2=2Q. ]
Thus the claim is false. The example works because it satisfies the hypothesis and denies the conclusion.
There is no simple repair saying “distinct” factors: \(\sqrt2\) and \(2\sqrt2\) are distinct irrational numbers whose product is \(4\). A valid narrower theorem is: if \(r\in\mathbb Q\setminus\{0\}\) and \(u\) is irrational, then \(ru\) is irrational. For if \(ru\) were rational, division by nonzero rational \(r\) would make \(u\) rational, a contradiction.
Refute the statement “if \(x^2>9\), then \(x>3\)” over \(\mathbb R\), and repair it as an equivalence.
Take \(x=-4\). Then \(x^2=16>9\), so the hypothesis is true, but \(-4>3\) is false. The correct solution of \(x^2>9\) compares distance from zero:
[ x^2>9|x|>3 x<-3x>3. ]
This is an equivalence over the reals.
Induction proves a family of statements \(P(n)\) indexed by integers from a starting value \(n_0\) onward. Ordinary induction has three obligations:
Then \(P(n)\) holds for every integer \(n\ge n_0\). The logic resembles an infinite chain: the base establishes the first link, and the step proves that every established link forces the next.
The inductive hypothesis is not the result being assumed for every \(n\). It is a temporary assumption at one arbitrary index \(k\), used to establish the next index. Always state where it is used.
Prove for every \(n\in\mathbb N\) that
[ 1+3+5++(2n-1)=n^2. ]
Base case \(n=1\). The left side is \(1\) and the right side is \(1^2=1\).
Inductive hypothesis. Assume for an arbitrary \(k\ge1\) that
[ 1+3++(2k-1)=k^2. ]
Inductive step. The next odd number is \(2(k+1)-1=2k+1\). Therefore, using the inductive hypothesis,
[ \[\begin{aligned} 1+3+\cdots+(2k-1)+(2k+1) &=k^2+2k+1\\ &=(k+1)^2. \end{aligned}\]]
Thus \(P(k)\Rightarrow P(k+1)\). By induction, the identity holds for every positive integer \(n\).
Define \(u_0=2\), \(u_1=3\), and
[ u_{n+2}=3u_{n+1}-u_n(n). ]
Prove that every \(u_n\) is an integer.
A step for \(u_{k+2}\) needs the two preceding terms, so use the strengthened statement
[ P(k):u_kZu_{k+1}Z. ]
Base case. \(u_0=2\) and \(u_1=3\) are integers, so \(P(0)\) holds.
Step. Assume \(P(k)\): both \(u_k\) and \(u_{k+1}\) are integers. Closure of the integers under multiplication and subtraction gives
[ u_{k+2}=3u_{k+1}-u_kZ. ]
We already know \(u_{k+1}\in\mathbb Z\), so both entries required for \(P(k+1)\) hold. Hence induction proves \(u_n\in\mathbb Z\) for every \(n\ge0\).
Strengthening the claim is not wasteful. It supplies exactly the information the recurrence consumes.
Prove that \(6\mid(n^3-n)\) for every integer \(n\ge0\).
Compute the difference between the expressions at \(k+1\) and \(k\).
Let \(P(n)\) state \(6\mid(n^3-n)\). For \(n=0\), the expression is \(0\), divisible by \(6\). Assume \(6\mid(k^3-k)\). Then
[ \[\begin{aligned} [(k+1)^3-(k+1)]-(k^3-k) &=3k^2+3k\\ &=3k(k+1). \end{aligned}\]]
One of consecutive integers \(k,k+1\) is even, so \(k(k+1)=2m\) for some integer \(m\). Hence the difference equals \(6m\). It is divisible by \(6\), and the inductive hypothesis says \(k^3-k\) is divisible by \(6\); their sum is divisible by \(6\). Therefore \(P(k+1)\) holds, and induction completes the proof.
Examples can reveal a conjecture or refute it, but finitely many successful cases do not prove a universal theorem. Replace the list with an arbitrary object represented through the relevant definition, or with induction when the index structure warrants it.
To prove \(P\Rightarrow Q\), beginning “suppose \(Q\)” is circular unless the proof has explicitly switched to the converse or to contradiction with \(\neg Q\). Track assumptions in the margin and discharge them at the end.
From \(P\Rightarrow Q\), proving \(Q\Rightarrow P\) addresses a different theorem. Write the implication in symbols before selecting a method. For contraposition, negate and reverse: \(\neg Q\Rightarrow\neg P\).
The cases \(a>0\) and \(a<0\) omit \(a=0\). The cases “\(n\) divisible by \(3\)” and “\(n\) odd” overlap and do not exhaust the integers. State a theorem—such as trichotomy or the division algorithm—that guarantees coverage.
Proof does not excuse invalid algebra. Division requires a nonzero divisor, square-root identities need sign conditions, and equivalence chains preserve the original domain. Cite the condition at the line where it matters.
Deriving \(n^2=4k^2+4k+1\) is not yet a proof that \(n^2\) is odd. Rewrite it as \(2(2k^2+2k)+1\) and state that the parenthesized quantity is an integer.
“If two solutions exist, they are equal” may be vacuously true when there are no solutions. Construct a solution or invoke a theorem that guarantees one.
A search over \(-1000\le n\le1000\) cannot establish a claim for every integer. Use it to discover a counterexample or test a conjecture. If it finds no counterexample, a general argument is still required.
A student tries to prove that the sum of two even integers is even: “Let \(a,b\) be even. Since \(a+b\) is even, \(a+b=2k\) for an integer \(k\). Therefore \(a+b\) is even.” Identify the failure and repair it.
The second sentence assumes the conclusion. Instead write \(a=2r\) and \(b=2s\) from the hypotheses. Then \(a+b=2(r+s)\), and \(r+s\in\mathbb Z\), proving the conclusion.
When a theorem combines divisibility by several primes, do not expand blindly. Factor, decide which coprime factors must divide the expression, and choose cases only where a residue genuinely changes the argument.
We need factors \(2\), \(3\), and \(5\). Since these are pairwise coprime, proving separate divisibility by each will prove divisibility by \(30\).
Factor:
[ n5-n=n(n4-1)=n(n-1)(n+1)(n^2+1). ]
Factor \(2\). Among the consecutive integers \(n-1,n,n+1\), at least one is even. Therefore \(2\mid(n^5-n)\).
Factor \(3\). Among three consecutive integers, exactly one is divisible by \(3\). Therefore \(3\mid(n^5-n)\).
Factor \(5\). By the division algorithm, \(n\) has residue \(0,1,2,3\), or \(4\) modulo \(5\).
Thus \(5\mid(n^5-n)\) in every case.
Because \(2,3,5\) are pairwise coprime and all divide the same integer, \(2\cdot3\cdot5=30\) divides it. Hence
[ 30(n^5-n) ]
for every integer \(n\).
The factorization makes divisibility by \(2\) and \(3\) immediate. Only the factor \(5\) requires a five-residue case split. An induction proof is possible but would hide the prime-factor structure. Method selection is part of the solution.
Prove that \(12\mid(n^4-n^2)\) for every integer \(n\).
Factor
[ n4-n2=n^2(n-1)(n+1). ]
Among \(n-1,n,n+1\), one is divisible by \(3\), so the product is divisible by \(3\). For divisibility by \(4\), if \(n\) is even then \(n^2\) is divisible by \(4\). If \(n\) is odd, both \(n-1\) and \(n+1\) are even, so their product is divisible by \(4\). Thus the expression is divisible by both \(3\) and \(4\). Since \(\gcd(3,4)=1\), it is divisible by \(12\).
Write complete proofs, not answer fragments. For each implication, identify the hypothesis and conclusion. For every divisibility or parity proof, name the integer witness. For every case proof, justify coverage. Concise answers follow, with selected full solutions for the more demanding tasks.
Suppose first that \(a,b\) have the same parity. If both are even, their squares and sum are even. If both are odd, write \(a=2r+1\), \(b=2s+1\); each square is odd, so their sum is even.
Conversely, suppose \(a^2+b^2\) is even. If \(a,b\) had different parity, one square would be even and the other odd, making the sum odd, a contradiction. Thus they have the same parity. Both directions hold.
Assume \(r^2=12\) for rational \(r=p/q\) in lowest terms with \(q>0\). Then \(p^2=12q^2\), so \(p^2\) is divisible by \(3\), hence \(p=3k\). Substitution gives \(9k^2=12q^2\), or \(3k^2=4q^2\). Therefore \(3\mid q^2\), so \(3\mid q\). This contradicts \(\gcd(p,q)=1\). Hence no rational \(r\) has square \(12\).
Prove simultaneously that \(w_{3k}\) is even and \(w_{3k+1},w_{3k+2}\) are odd. For \(k=0\), \((w_0,w_1,w_2)=(0,1,1)\) has the required parity.
Assume \((w_{3k},w_{3k+1},w_{3k+2})\) has parity even, odd, odd. Then
[ \[\begin{aligned} w_{3k+3}&=w_{3k+2}+w_{3k+1} &&\text{is even},\\ w_{3k+4}&=w_{3k+3}+w_{3k+2} &&\text{is odd},\\ w_{3k+5}&=w_{3k+4}+w_{3k+3} &&\text{is odd}. \end{aligned}\]]
Thus the strengthened statement propagates from one block of three to the next. Induction proves it for every \(k\ge0\), and in particular every \(w_{3k}\) is even.
These problems deliberately combine the set language, number systems, and transformation discipline of the preceding chapters.
Prove that for real \(x\),
[ =x+1 ]
if and only if \(x\ne1\). Explain why this is not an equality of two functions with natural real domains.
If the quotient is defined, then \(x\ne1\), and factoring permits cancellation: \((x^2-1)/(x-1)=x+1\). Conversely, for every \(x\ne1\), the same factorization and nonzero cancellation prove the equality. At \(x=1\) the left side is undefined, so no equality statement exists. The polynomial has natural domain \(\mathbb R\), while the quotient has domain \(\mathbb R\setminus\{1\}\); functions with different domains are not equal.
Prove that no two distinct real numbers are consecutive: for every \(a<b\) there exists \(c\in\mathbb R\) with \(a<c<b\). Then explain why one constructed \(c\) proves existence but not uniqueness.
Given arbitrary reals \(a<b\), choose \(c=(a+b)/2\). Since \(a<b\), adding \(a\) gives \(2a<a+b\) and division by positive \(2\) gives \(a<c\). Similarly, \(a+b<2b\), so \(c<b\). Thus \(a<c<b\). Many other choices exist—for example, \((2a+b)/3\)—so the construction proves at least one intermediate real, not a unique one.
Before using the interactive questions, select a proof method for this claim: “If integers \(a+b\) and \(a\) are even, then \(b\) is even.” A direct proof is shortest: write \(a+b=2m\) and \(a=2n\), then \(b=(a+b)-a=2(m-n)\).
The checkpoint tests method selection, logical direction, counterexample design, and induction structure. The generated notebook retains the same questions, hints, and explanations.
Complete the linked Unit 1 proof portfolio before attempting the 75-minute Unit 1 examination. These assessments combine Chapters 1.1–1.4. After correction and reassessment, continue to the next canonical unit on expression structure and canonical forms.
The fcmath.algebra proof renderer keeps statements and justifications paired. It is useful for auditing a finished proof, not for deciding whether a reason is mathematically valid.
| Step | Statement | Justification |
|---:|---|---|
| 1 | n = 2k + 1 for some integer k | definition of odd |
| 2 | n² = 4k² + 4k + 1 | expansion |
| 3 | n² = 2(2k² + 2k) + 1 | factor 2 |A counterexample to an implication must satisfy its hypothesis and fail its conclusion. Search a small, transparent domain first:
-10The first value depends on candidate order; any returned value still needs a written verification. Here the search finds a negative integer whose square is larger than \(9\).
SymPy can confirm the polynomial identity in the induction step and factor the examination expression:
(2*k + 1, n*(n - 1)*(n + 1)*(n**2 + 1))The outputs \(2k+1\) and \(n(n-1)(n+1)(n^2+1)\) support the manual work. They do not prove the induction principle, the exhaustiveness of residue cases, or the coprime-factor conclusion; those logical steps remain in the written proof.