Reason with direct and indirect proof methods

Connect logical methods across domains

Use domains,quantifiers,negations,necessary/sufficient conditions and equivalence to construct complete arguments in algebra,complex numbers,vectors,matrices,graphs and elementary calculus

Record the logical contract

State the domain and hypotheses before the conclusion. For P⇒Q,P is sufficient for Q and Q is necessary for P. To prove P⇔Q,prove both directions or show every step is reversible on the stated domain. Keep a proof ledger:allowed assumptions,new deductions,and any nonzero or sign condition needed for an operation

To refute a proposed equivalence P⇔Q, what is enough?

Conjecture, test, refine, prove

A pattern can suggest a statement but cannot certify an infinite universal claim. For example,AB=0⇒A=0 or B=0 is true for real scalars but false for square matrices. Before transferring a theorem to another domain,check which properties its proof uses. A failed proof may reveal a missing hypothesis rather than a false final statement

Construct a quantified statement, ungraded

For real x and positive a,formulate |x|<a as an equivalence involving a double inequality. Give its exact negation. Prove both directions,and explain what changes at a=0 and at a<0. Do not silently assume that squaring preserves a strict inequality

Model: an equivalence with a sign condition

For a>0,|x|<a iff−a<x<a. If x≥0,|x|=x and the lower bound follows from−a<0≤x; if x<0,|x|=−x and the upper bound follows from x<0<a. Conversely,either sign of x gives |x|<a from the corresponding bound. The negation is x≤−a or x≥a. At a=0 and a<0,both sides of the strict equivalence have empty solution sets,but the positive-radius interpretation and a squaring argument need separate care. For instance,x²<a² has solutions when a<0 although |x|<a has none

For a=3,what is the lower boundary excluded by |x|<a?

For x=0,a=−2,which statement is true?

Direct proof with a complete equality case

For real a,b,c,the identity2(a²+b²+c²−ab−bc−ca)=(a−b)²+(b−c)²+(c−a)² proves a²+b²+c²≥ab+bc+ca. Equality requires every displayed real square to be zero,so a=b=c. Merely finding one equality example does not prove the full equality condition

Evaluate a²+b²+c²−ab−bc−ca at a=1,b=2,c=3.

Prove an inequality by cases, ungraded

Prove |x+y|≤|x|+|y| for all real x,y. Give the complete equality condition and demonstrate strict inequality with one pair of opposite nonzero signs. State why your cases cover every pair

Model: separate equal and opposite signs

If xy≥0,the numbers have the same sign or one is zero,and |x+y|=|x|+|y|. If xy<0,write their magnitudes as u,v>0. Then |x+y|=|u−v|=u+v−2min(u,v)<u+v. The cases xy≥0 and xy<0 are exhaustive. Therefore equality holds exactly when xy≥0. For x=2,y=−5,the two sides are3 and7.

Give |2+(−5)|

Use a contrapositive when it exposes structure

For an integer n,if n² is not divisible by3,then n is not divisible by3. Prove the contrapositive:if3 divides n,write n=3k for integer k,so n²=9k² is divisible by3. This needs only one direction. The different claim3|n²⇒3|n requires a separate prime or residue argument

Which statement is the contrapositive of P⇒Q?

Prove a complex equivalence, ungraded

For z≠0,prove |z|=1 iff1/z=conjugate(z). State where the nonzero condition is used. Then refute the claim that Re(z)>0 implies z is a positive real number,using an explicit complex counterexample

Model: multiply only on the permitted domain

The identity z·conjugate(z)=|z|² holds for every complex z. If |z|=1,then z·conjugate(z)=1 and division by nonzero z gives conjugate(z)=1/z. Conversely,1/z=conjugate(z) implies1=|z|²;since modulus is nonnegative,|z|=1. The expression1/z is undefined at0. For the second claim,z=2+i has positive real part but a nonzero imaginary part,so it is not real

For z=(3+4i)/5,give z·conjugate(z)

Does checking z=(3+4i)/5 prove that |z|=1 if and only if 1/z=conjugate(z) for every nonzero z?

Prove a vector equivalence, ungraded

For real plane vectors u,v,prove |u+v|=|u−v| iff u·v=0. Include zero vectors. Explain why perpendicular nonzero vectors do not justify the false conclusion that one vector must be zero

Model: compare nonnegative squared lengths

The squared lengths are |u|²+2u·v+|v|² and |u|²−2u·v+|v|². Their difference is4u·v,so the squares are equal iff u·v=0. Both lengths are nonnegative,so equality of their squares is equivalent to equality of lengths. Zero vectors satisfy the identity. Nonzero u=(1,0),v=(0,1) have dot product0,providing a counterexample to “zero dot product means a zero vector”

For u=(2,1),v=(1,−2),give u·v.

Repair a matrix claim, ungraded

Refute AB=0⇒A=0 or B=0 for real2×2 matrices. Then prove that AB=0 and invertible A imply B=0. Explain which scalar cancellation property failed in the original claim

Model: supply a counterexample and a sufficient hypothesis

Take A=diag(1,0) and B=diag(0,1). Both are nonzero and AB=0. A is singular,so cancellation by A is unavailable. If A is invertible,pre-multiply AB=0 by A⁻¹:associativity gives(A⁻¹A)B=IB=B=0. Multiplication order matters;no commutativity assumption is needed. Invertibility is sufficient for this cancellation argument,not necessary for every individual pair to happen to have B=0.

Give the(1,1)entry of diag(1,0)diag(0,1)

Use contradiction with a clearly identified conflict

In a finite undirected graph with E edges,the degree sum is2E. Suppose the number of odd-degree vertices were odd. The sum of their odd degrees would be odd,while the remaining even degrees contribute an even total. The total degree sum would be odd,contradicting2E being even. Thus the number of odd-degree vertices is even. A loop contributes2 to degree,so it does not invalidate the parity argument

In a finite undirected graph with E edges, suppose the total degree sum is odd. Which established fact gives a contradiction?

A calculus conjecture needs the correct direction

For the familiar polynomial f(x)=x³,the function is strictly increasing on R,but f′(0)=0. Therefore “strictly increasing implies derivative strictly positive at every point” is false. Strict increase can be proved without that false test:for a<b,b³−a³=(b−a)(a²+ab+b²)>0,because the second factor equals(a+b/2)²+3b²/4 and cannot vanish when a<b.

Give f′(0) for f(x)=x³

Audit an argument across domains, ungraded

Write a complete response that identifies one valid equivalence,one false converse and one invalid cancellation from this lesson. State each domain,provide a proof or counterexample,and repair every false claim with a precise hypothesis or direction. Include the x³ example and distinguish a derivative fact from a general monotonicity theorem

Model synthesis and future use

The unit-modulus complex equivalence is proved in both directions on C excluding0. The converse “positive real part implies positive real number” fails at2+i. Matrix cancellation fails for singular A but is justified for invertible A. The x³ example refutes a claimed necessary derivative condition;it does not by itself prove a general sufficient derivative theorem. In later domains,keep this same discipline:quantify,record assumptions,justify each implication,test boundaries,and distinguish complete proof from examples. Induction across structures is the separate next section

Written assessment boundary

Complete arguments and conjecture development are ungraded. State the domain and hypotheses,justify every implication,and give a complete equality condition or counterexample where required