Note:Ifappropriate,

useP(0),P(2)orother

valueinstead.

THEOREM: “Foreveryn ∈ ℕ,

PROOF:Bymathematicalinduction.

Basis:𝑃(1) assertsthat

whichistruebecause

Inductivestep:Assumeforanarbitrary𝑘 ∈ ℕ,𝑃(𝑘)istrue,i.e.,namely:

Wewillnowshowthat𝑃 𝑘 + 1 isalsotrue,i.e.:

Proofofinductivestep:

Wethushavethat𝑃(1)and∀𝑘 ∈ ℕ,𝑃 𝑘 → 𝑃 𝑘 + 1 ,sobytheprincipleofmathematicalinduction,itfollowsthat𝑃(𝑛)istrueforallnaturalnumbers𝑛.

Q.E.D.Stepsofamathematicalinductionproof:1)statethetheorem,whichisthepropositionP(n)2)showthatP(basecase)istrue.BasecaseisusuallyP(1),butsometimesP(0)orP2)orothervalueisappropriate.3)statetheinductivehypothesis(substitutekforn)4)statewhatmustbeproved(substitutek+1forn)5)statethatyouarebeginningyourproofoftheinductivestep,andproceedtomanipulatetheinductivehypothesis(whichweassumeistrue)tofindalinkbetweentheinductivehypothesisandthestatementtobeproven.Alwaysstateexplicitlywhereyouareinvokingtheinductivehypothesis.6)finishyourproofbyinvokingtheprincipleofmathematicalinductionthatallowsyoutoinferthat𝑃 𝑛 istrueforallnaturalnumbers.

∀n ∈ ℕ (𝑃(𝑛))

Stateandprove𝑃(1)

State𝑃(𝑘) (inductivehypothesis)

MathematicalInductionProofTemplate

State 𝑃(𝑘 + 1)

Prove

𝑃 ( 𝑘)⇒𝑃(𝑘+1)

𝑃(𝑛)

Alldone:wrapupproof

Statethefollowingandtrytofigureoutwhytheyaretrue.Thenseeifapatternemergesthatyoucangeneralize.Trysomemorebasecases: 𝑃(2)𝑃(3) 𝑃(4) Ifit’snotyetclearwhatmakestheinductivesteptrue(i.e.,whatisitintheinductivehypothesis𝑃 𝑘 thatcausestheconclusion𝑃(𝑘 + 1)toalsobetrue?),trysomelargerconsecutivenumbers.Asyouworktheseexamples,seeifyoucanmakeuseoftheinductivehypothesisinprovingtheconclusion(ratherthanprovingitindependently).Notethatusingexampleswithlargenumberssometimesforcesyoutotakeashortcut;thatshortcutisoftenthekeytoprovingtheinductivestep.𝑃 8 ⇒ 𝑃(9) 𝑃 25 ⇒ 𝑃(26) 𝑃 1,000,000 ⇒ 𝑃(1,000,001)

Stuckontheproofoftheinductivestep?Dosomeexamplesforinspiration!