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Question Number 156951 by mr W last updated on 17/Oct/21
solve for n∈N (n−1)!+1=n^2
solveforn∈N(n−1)!+1=n2
Commented by mr W last updated on 17/Oct/21
thanks sir! if n=1, then (n−2)!−(n+1) is not defined. therefore n≠1.
thankssir!ifn=1,then(n−2)!−(n+1)isnotdefined.thereforen≠1.
Commented by Rasheed.Sindhi last updated on 17/Oct/21
(n−1)!+1=n^2 (n−1)!−(n^2 −1)=0 (n−1){(n−2)!−(n+1)}=0 n=1 ∣ (n−2)!−(n+1)=0 But n=1 is false because it doesn′t fulfill the original.Why is it false? (n−2)!=n+1 ;n≥2 2≤n≤4: (n−2)!<n+1 n=5: (5−2)!=5+1✓ n>5: (n−2)!>n+1 n=5
(n−1)!+1=n2(n−1)!−(n2−1)=0(n−1){(n−2)!−(n+1)}=0n=1∣(n−2)!−(n+1)=0Butn=1isfalsebecauseitdoesn′tfulfilltheoriginal.Whyisitfalse?(n−2)!=n+1;n⩾22⩽n⩽4:(n−2)!<n+1n=5:(5−2)!=5+1✓n>5:(n−2)!>n+1n=5
Commented by Rasheed.Sindhi last updated on 18/Oct/21
mr W Sir^(Thanks) ! You′ve always helped me. ...I thought it only matter of the definition: 0!=1 If it were defined 0!=0,the solution x=1 were not false, although the other factor be undefined for it. Afterall ′0!=1′ is a definition which is made for some necessities.What′s your openion sir?Your openion matters to me.
mrWSirThanks!You′vealwayshelpedme.…Ithoughtitonlymatterofthedefinition:0!=1Ifitweredefined0!=0,thesolutionx=1werenotfalse,althoughtheotherfactorbeundefinedforit.Afterall′0!=1′isadefinitionwhichismadeforsomenecessities.What′syouropenionsir?Youropenionmatterstome.
Commented by mr W last updated on 18/Oct/21
i think if 0! is defined, then 0! must be equal to 1. we can not further define 0!=0, since otherwise we would get 1!=1×0!=1×0=0, 2!=2×1!=0 etc.
ithinkif0!isdefined,then0!mustbeequalto1.wecannotfurtherdefine0!=0,sinceotherwisewewouldget1!=1×0!=1×0=0,2!=2×1!=0etc.
Commented by Rasheed.Sindhi last updated on 18/Oct/21
Yes Sir,you′re right!
YesSir,you′reright!
Commented by mr W last updated on 18/Oct/21
0! is defined! so it is ok with (n−1)!−(n^2 −1)=0 and you can put n=1 into it. but if you factorise it to (n−1){(n−2)!−(n+1)}=0 (n−2)! must be defined, that means n−2≥0.
0!isdefined!soitisokwith(n−1)!−(n2−1)=0andyoucanputn=1intoit.butifyoufactoriseitto(n−1){(n−2)!−(n+1)}=0(n−2)!mustbedefined,thatmeansn−2⩾0.
Commented by Rasheed.Sindhi last updated on 18/Oct/21
Thanks very much! But I think n≥2 is only for other values of n. Sir I only talk about supposition. If 0!=0 ,then... Then the solution may contain 1 or not?
Thanksverymuch!ButIthinkn⩾2isonlyforothervaluesofn.SirIonlytalkaboutsupposition.If0!=0,then…Thenthesolutionmaycontain1ornot?
Answered by mindispower last updated on 17/Oct/21
⇔n^2 −1−(n−1)!=0,n≥1,n=1not solution ∀n,N−{0,1}=E ⇔(n−1)(n+1−(n−2)!)=0 (n+1)=(n−2)! n≥5 (n−2)!≥n+1...? n=5 3!≥6 true suppose ∀n≥5 (n−2)!≥n+1,⇒(n−1)!≥n+2 (n−1)!=(n−2)!.(n−1)≥(n+1)(n−1)=n^2 −1 n.n−1≥5n−1=4n+n−1≥20−1+n=n+19>2n+2 ⇒(n−1)!>n+2 ⇒∀n≥5 (n−2)!≥(n+1),∀n>6 (n−2)!>n+1 just try n∈{2,3,4,5) 2⇒1+1=2^2 ,n=3⇒3=9,n=4⇒7=16 n=5⇒25=5^2 ⇒n=5
⇔n2−1−(n−1)!=0,n⩾1,n=1notsolution∀n,N−{0,1}=E⇔(n−1)(n+1−(n−2)!)=0(n+1)=(n−2)!n⩾5(n−2)!⩾n+1…?n=53!⩾6truesuppose∀n⩾5(n−2)!⩾n+1,⇒(n−1)!⩾n+2(n−1)!=(n−2)!.(n−1)⩾(n+1)(n−1)=n2−1n.n−1⩾5n−1=4n+n−1⩾20−1+n=n+19>2n+2⇒(n−1)!>n+2⇒∀n⩾5(n−2)!⩾(n+1),∀n>6(n−2)!>n+1justtryn∈{2,3,4,5)2⇒1+1=22,n=3⇒3=9,n=4⇒7=16n=5⇒25=52⇒n=5
Commented by mr W last updated on 17/Oct/21
thanks sir!
thankssir!
Commented by mindispower last updated on 18/Oct/21
withe pleasur sir
withepleasursir

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