MathematicalInductionProofTemplate
∀n ∈ ℕ (𝑃(𝑛)) THEOREM: “Foreveryn ∈ ℕ,
𝑃(𝑛)
Basis:𝑃(1) assertsthat
Note:Ifappropriate,
useP(0),P(2)orother
valueinstead.
whichistruebecause
Stateandprove𝑃(1)
PROOF:Bymathematicalinduction.
Inductivestep:Assumeforanarbitrary𝑘 ∈ ℕ,𝑃(𝑘)istrue,i.e.,namely:
State𝑃(𝑘)
(inductive
hypothesis)
Prove 𝑃 (𝑘) ⇒ 𝑃(𝑘 + 1) State
𝑃(𝑘 + 1) Alldone:wrap
upproof
Wewillnowshowthat𝑃
𝑘 + 1 isalsotrue,i.e.:
Proofofinductivestep:
Wethushavethat𝑃(1)and∀𝑘 ∈ ℕ, 𝑃 𝑘 → 𝑃 𝑘 + 1 ,sobytheprincipleof
mathematicalinduction,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.Always
stateexplicitlywhereyouareinvokingtheinductivehypothesis.
6)finishyourproofbyinvokingtheprincipleofmathematicalinductionthatallowsyoutoinferthat𝑃 𝑛 istruefor
allnaturalnumbers.
Stuckontheproofoftheinductivestep?Dosomeexamplesforinspiration!
Statethefollowingandtrytofigureoutwhytheyaretrue.Thenseeifapatternemerges
thatyoucangeneralize.
Trysomemorebasecases:
𝑃(2)
𝑃(3) 𝑃(4)
Ifit’snotyetclearwhatmakestheinductivesteptrue(i.e.,whatisitintheinductive
hypothesis𝑃 𝑘 thatcausestheconclusion𝑃(𝑘 + 1)toalsobetrue?),trysomelarger
consecutivenumbers.Asyouworktheseexamples,seeifyoucanmakeuseofthe
inductivehypothesisinprovingtheconclusion(ratherthanprovingitindependently).
Notethatusingexampleswithlargenumberssometimesforcesyoutotakeashortcut;that
shortcutisoftenthekeytoprovingtheinductivestep.
𝑃 8 ⇒ 𝑃(9) 𝑃 25 ⇒ 𝑃(26)
𝑃 1,000,000 ⇒ 𝑃(1,000,001)