:: Irrationality of e
:: by Freek Wiedijk
::
:: Received July 2, 1999
:: Copyright (c) 1999-2018 Association of Mizar Users
:: (Stowarzyszenie Uzytkownikow Mizara, Bialystok, Poland).
:: This code can be distributed under the GNU General Public Licence
:: version 3.0 or later, or the Creative Commons Attribution-ShareAlike
:: License version 3.0 or later, subject to the binding interpretation
:: detailed in file COPYING.interpretation.
:: See COPYING.GPL and COPYING.CC-BY-SA for the full text of these
:: licenses, or see http://www.gnu.org/licenses/gpl.html and
:: http://creativecommons.org/licenses/by-sa/3.0/.
environ
vocabularies NUMBERS, SUBSET_1, INT_1, SEQ_1, FINSEQ_1, RAT_1, ORDINAL4,
POWER, INT_2, SQUARE_1, XXREAL_0, CARD_1, ARYTM_3, RELAT_1, NAT_1,
FUNCT_1, ARYTM_1, REALSET1, SEQ_2, ORDINAL2, COMPLEX1, VALUED_1,
SERIES_1, CARD_3, XBOOLE_0, RFINSEQ, PARTFUN1, FINSEQ_2, NEWTON, SIN_COS,
IRRAT_1, REAL_1, ASYMPT_1;
notations XBOOLE_0, SUBSET_1, ORDINAL1, NUMBERS, XCMPLX_0, XXREAL_0, XREAL_0,
REAL_1, INT_1, COMPLEX1, SQUARE_1, NAT_1, FUNCT_1, PARTFUN1, FUNCOP_1,
VALUED_0, VALUED_1, SEQ_1, COMSEQ_2, SEQ_2, POWER, FINSEQ_1, RVSUM_1,
SERIES_1, FINSEQ_2, RFINSEQ, SIN_COS, RAT_1, INT_2, PEPIN, NEWTON;
constructors ARYTM_0, REAL_1, NAT_1, NAT_D, SEQ_2, PARTFUN1, RVSUM_1,
LIMFUNC1, COMSEQ_3, RFINSEQ, BINARITH, SIN_COS, PEPIN, SERIES_1, POWER,
RELSET_1, COMSEQ_2, SEQ_1, NUMBERS, FUNCOP_1;
registrations ORDINAL1, RELSET_1, NUMBERS, XXREAL_0, XREAL_0, NAT_1, INT_1,
MEMBERED, NEWTON, XBOOLE_0, VALUED_0, VALUED_1, FUNCT_2, POWER, FINSEQ_1,
SQUARE_1, RVSUM_1, SEQ_1, SEQ_2;
requirements REAL, NUMERALS, BOOLE, SUBSET, ARITHM;
definitions SEQ_2;
equalities SQUARE_1, XCMPLX_0;
expansions SEQ_2;
theorems SEQ_1, POWER, NEWTON, NAT_1, SEQ_2, SEQ_4, ABSVALUE, RVSUM_1,
FINSEQ_2, SERIES_1, RFINSEQ, FINSEQ_3, FINSEQ_5, SIN_COS, SQUARE_1,
INT_1, RAT_1, INT_2, FUNCT_2, XCMPLX_0, XCMPLX_1, XREAL_1, COMPLEX1,
XXREAL_0, NAT_D, PARTFUN1, CARD_1, VALUED_0, ORDINAL1, XREAL_0, TARSKI;
schemes SEQ_1, NAT_1;
begin :: Square roots of primes are irrational
reserve k, m, n, p, K, N for Nat;
reserve i for Integer;
reserve x, y, eps for Real;
reserve seq, seq1, seq2 for Real_Sequence;
reserve sq for FinSequence of REAL;
notation
let x be Real;
antonym x is irrational for x is rational;
end;
notation
let x, y be Real;
synonym x ^ y for x to_power y;
end;
::$N The Irrationality of the Square Root of 2
theorem Th1:
p is prime implies sqrt p is irrational
proof
assume
A1: p is prime;
then
A2: p>1 by INT_2:def 4;
assume sqrt p is rational;
then consider i, n such that
A3: n<>0 and
A4: sqrt p=i/n and
A5: for i1 being Integer, n1 being Nat st n1<>0 & sqrt p=i1/
n1 holds n<=n1 by RAT_1:9;
A6: i=sqrt p*n by A3,A4,XCMPLX_1:87;
sqrt p>=0 by SQUARE_1:def 2;
then reconsider m = i as Element of NAT by A6,INT_1:3;
A7: m^2 = (sqrt p)^2*n^2 by A6
.= p*n^2 by SQUARE_1:def 2;
then p divides m^2 by NAT_D:def 3;
then p divides m by A1,NEWTON:80;
then consider m1 being Nat such that
A8: m=p*m1 by NAT_D:def 3;
n^2 = p*(p*m1^2)/p by A2,A7,A8,XCMPLX_1:89
.= p*m1^2 by A2,XCMPLX_1:89;
then p divides n^2 by NAT_D:def 3;
then p divides n by A1,NEWTON:80;
then consider n1 being Nat such that
A9: n=p*n1 by NAT_D:def 3;
reconsider n1 as Element of NAT by ORDINAL1:def 12;
A10: m1/n1 = sqrt p by A2,A4,A8,A9,XCMPLX_1:91;
A11: n1<>0 by A3,A9;
then p*n1>1*n1 by A2,XREAL_1:98;
hence contradiction by A5,A9,A11,A10;
end;
theorem
ex x, y st x is irrational & y is irrational & x ^ y is rational
proof
set w = sqrt 2;
A1: (w ^ w) ^ w = w ^ (w^2) by POWER:33,SQUARE_1:19
.= w ^ 2 by SQUARE_1:def 2
.= w^2 by POWER:46
.= 2 by SQUARE_1:def 2;
reconsider dwa = 2 as Real;
per cases;
suppose
A2: w ^ w is rational;
take w, w;
thus thesis by A2,Th1,INT_2:28;
end;
suppose
A3: w ^ w is irrational;
take w ^ w, w;
thus thesis by A1,A3,Th1,INT_2:28;
end;
end;
begin :: A proof that e = e
scheme
LambdaRealSeq{F(set)->Real}: (ex seq st for n holds seq.n=F(n)) & for
seq1, seq2 st (for n holds seq1.n=F(n)) & (for n holds seq2.n=F(n)) holds seq1=
seq2 proof
thus ex seq st for n holds seq.n=F(n) from SEQ_1:sch 1;
let seq1, seq2;
assume
A1: for n holds seq1.n=F(n);
assume
A2: for n holds seq2.n=F(n);
now
let n be Element of NAT;
thus seq1.n = F(n) by A1
.= seq2.n by A2;
end;
hence thesis by FUNCT_2:63;
end;
definition
let k be Nat;
func aseq(k) -> Real_Sequence means
:Def1:
for n holds it.n=(n-k)/n;
correctness
proof
deffunc F(Nat)=($1-k)/$1;
thus (ex seq being Real_Sequence st for n being Nat holds seq.n
=F(n)) & for seq1, seq2 being Real_Sequence st
(for n being Nat
holds seq1.n=F(n)) & (for n being Nat holds seq2.n=F(n)) holds seq1=
seq2 from LambdaRealSeq;
end;
func bseq(k) -> Real_Sequence means
:Def2:
for n holds it.n=(n choose k)*(n ^ (-k));
correctness
proof
deffunc F(Nat)=($1 choose k)*($1 ^ (-k));
thus (ex seq being Real_Sequence st for n being Nat holds seq.n
=F(n)) & for seq1, seq2 being Real_Sequence st
(for n being Nat
holds seq1.n=F(n)) & (for n being Nat holds seq2.n=F(n)) holds seq1=
seq2 from LambdaRealSeq;
end;
end;
definition
let n be Nat;
func cseq(n) -> Real_Sequence means
:Def3:
for k holds it.k=(n choose k)*(n ^ (-k));
correctness
proof
deffunc F(Nat)=(n choose $1)*(n ^ (-$1));
thus (ex seq being Real_Sequence st for n being Nat holds seq.n
=F(n)) & for seq1, seq2 being Real_Sequence st
(for n being Nat
holds seq1.n=F(n)) & (for n being Nat holds seq2.n=F(n)) holds seq1=
seq2 from LambdaRealSeq;
end;
end;
theorem Th3:
cseq(n).k=bseq(k).n
proof
thus cseq(n).k = (n choose k)*(n ^ (-k)) by Def3
.= bseq(k).n by Def2;
end;
definition
func dseq -> Real_Sequence means
:Def4:
for n holds it.n=(1+(1/n)) ^ n;
correctness
proof
deffunc F(Nat)=(1+(1/$1)) ^ $1;
thus (ex seq being Real_Sequence st for n being Nat holds seq.n
=F(n)) & for seq1, seq2 being Real_Sequence st
(for n being Nat
holds seq1.n=F(n)) & (for n being Nat holds seq2.n=F(n)) holds seq1=
seq2 from LambdaRealSeq;
end;
end;
definition
func eseq -> Real_Sequence means
:Def5:
for k holds it.k=1/(k!);
correctness
proof
deffunc F(Nat)=1/($1!);
thus (ex seq being Real_Sequence st for n being Nat holds seq.n
=F(n)) & for seq1, seq2 being Real_Sequence st
(for n being Nat
holds seq1.n=F(n)) & (for n being Nat holds seq2.n=F(n)) holds seq1=
seq2 from LambdaRealSeq;
end;
end;
:: lim(\n:(n choose k)*(n ^ (-k))) = 1/(k!)
theorem Th4:
n>0 implies (n ^ (-(k+1)))=(n ^ (-k))/n
proof
assume
A1: n>0;
thus (n ^ (-(k+1))) = (n ^ ((-k)-1)) .= (n ^ (-k))/(n ^ 1) by A1,POWER:29
.= (n ^ (-k))/n;
end;
theorem Th5:
(n choose (k+1))=((n-k)/(k+1))*(n choose k)
proof
per cases;
suppose
A1: k+1<=n;
then reconsider l = n-(k+1) as Element of NAT by INT_1:5;
l+1 = n-k;
then reconsider l1 = n-k as Element of NAT;
k <= k+1 by NAT_1:11;
then
A2: k <= n by A1,XXREAL_0:2;
thus (n choose (k+1)) = n!/((k+1)!*(l!)) by A1,NEWTON:def 3
.= n!/((k!*(k+1))*(l!)) by NEWTON:15
.= n!/((k!*(k+1))*((l!)*(l+1)/(l+1))) by XCMPLX_1:89
.= n!/((k!*(k+1))*((l+1)!/(l+1))) by NEWTON:15
.= (l1/(k+1))*(n!/((k!)*(l1!))) by XCMPLX_1:233
.= ((n-k)/(k+1))*(n choose k) by A2,NEWTON:def 3;
end;
suppose
A3: k+1>n & k<=n;
then k>=n by NAT_1:13;
then k=n by A3,XXREAL_0:1;
hence thesis by A3,NEWTON:def 3;
end;
suppose
A4: k+1>n & k>n;
hence (n choose (k+1)) = ((n-k)/(k+1))*0 by NEWTON:def 3
.= ((n-k)/(k+1))*(n choose k) by A4,NEWTON:def 3;
end;
end;
theorem Th6:
n>0 implies bseq(k+1).n=(1/(k+1))*(bseq(k).n)*(aseq(k).n)
proof
assume
A1: n>0;
thus bseq(k+1).n = (n choose (k+1))*(n ^ (-(k+1))) by Def2
.= ((n-k)/(k+1))*(n choose k)*(n ^ (-(k+1))) by Th5
.= ((n-k)/(k+1))*(n choose k)*((n ^ (-k))/n) by A1,Th4
.= (n-k)*(k+1)"*(n choose k)*((n ^ (-k))/n)
.= (n-k)*(k+1)"*(n choose k)*((n ^ (-k))*n")
.= (k+1)"*((n choose k)*(n ^ (-k)))*((n-k)*n")
.= (1/(k+1))*((n choose k)*(n ^ (-k)))*((n-k)*n")
.= (1/(k+1))*((n choose k)*(n ^ (-k)))*((n-k)/n)
.= (1/(k+1))*(bseq(k).n)*((n-k)/n) by Def2
.= (1/(k+1))*(bseq(k).n)*(aseq(k).n) by Def1;
end;
theorem Th7:
n>0 implies aseq(k).n=1-(k/n)
proof
assume
A1: n>0;
thus aseq(k).n = (n-k)/n by Def1
.= (n/n)-(k/n)
.= 1-(k/n) by A1,XCMPLX_1:60;
end;
theorem Th8:
aseq(k) is convergent & lim(aseq(k))=1
proof
A1: for eps be Real st 00;
consider N such that
A3: N>k/eps by SEQ_4:3;
take N;
let n;
assume
A4: n>=N;
then n>(k/eps) by A3,XXREAL_0:2;
then (k/eps)*eps0 by A2,A3,A4;
then |.aseq(k).n-1.| = |.(1-(k/n))-1.| by Th7
.= |.-(k/n).|
.= |.k/n.| by COMPLEX1:52
.= k/n by ABSVALUE:def 1;
hence thesis by A6,A5,XREAL_1:83;
end;
thus aseq(k) is convergent
by A1;
hence thesis by A1,SEQ_2:def 7;
end;
theorem Th9:
for seq st for n being Nat holds seq.n=x holds seq is convergent & lim(seq)=x
proof
let seq;
assume
A1: for n being Nat holds seq.n=x;
x in REAL by XREAL_0:def 1;
then
A2: seq is constant by A1,VALUED_0:def 18;
hence seq is convergent;
thus lim(seq) = seq.0 by A2,SEQ_4:26
.= x by A1;
end;
theorem Th10:
for n holds bseq(0).n=1
proof
let n;
thus bseq(0).n = (n choose 0)*(n ^ (-0)) by Def2
.= 1*(n ^ (-0)) by NEWTON:19
.= 1 by POWER:24;
end;
theorem Th11:
(1/(k+1))*(1/(k!))=1/((k+1)!)
proof
thus (1/(k+1))*(1/(k!)) = 1/((k+1)*(k!)) by XCMPLX_1:102
.= 1/((k+1)!) by NEWTON:15;
end;
theorem Th12:
bseq(k) is convergent & lim(bseq(k))=1/(k!) & lim(bseq(k))=eseq. k
proof
defpred P[Nat] means bseq($1) is convergent & lim(bseq($1))=1/($1!);
A1: now
let k;
assume
A2: P[k];
thus P[k+1]
proof
deffunc F(Nat)=(1/(k+1))*(bseq(k).$1);
consider seq such that
A3: for n holds seq.n = F(n) from SEQ_1:sch 1;
deffunc G(Nat)=seq.$1*(aseq(k).$1);
consider seq1 such that
A4: for n holds seq1.n=G(n) from SEQ_1:sch 1;
A5: for n st n>=1 holds bseq(k+1).n=seq1.n
proof
let n;
assume n>=1;
hence bseq(k+1).n = (1/(k+1))*(bseq(k).n)*(aseq(k).n) by Th6
.= (seq.n)*(aseq(k).n) by A3
.= seq1.n by A4;
end;
A6: seq = (1/(k+1))(#)(bseq(k)) by A3,SEQ_1:9;
then
A7: seq is convergent by A2;
A8: lim(seq) = (1/(k+1))*(1/(k!)) by A2,A6,SEQ_2:8
.= 1/((k+1)!) by Th11;
A9: aseq(k) is convergent by Th8;
A10: seq1=seq(#)(aseq(k)) by A4,SEQ_1:8;
then
A11: seq1 is convergent by A7,A9;
hence bseq(k+1) is convergent by A5,SEQ_4:18;
lim(seq1) = lim(seq)*lim(aseq(k)) by A7,A10,A9,SEQ_2:15
.= 1/((k+1)!) by A8,Th8;
hence thesis by A11,A5,SEQ_4:19;
end;
end;
A12: P[0]
proof
reconsider jj = 1 as Element of REAL by XREAL_0:def 1;
set bseq0 = seq_const 1;
A13: for n being Nat holds bseq0.n=1 by SEQ_1:57;
A14: for n st n>=1 holds bseq(0).n=bseq0.n
proof
let n;
assume n>=1;
thus bseq(0).n = 1 by Th10
.= bseq0.n by SEQ_1:57;
end;
hence bseq(0) is convergent by SEQ_4:18;
lim(bseq0)=1 by A13,Th9;
hence thesis by A14,NEWTON:12,SEQ_4:19;
end;
for k holds P[k] from NAT_1:sch 2(A12,A1);
hence bseq(k) is convergent & lim(bseq(k))=1/(k!);
hence thesis by Def5;
end;
:: 0 <= (n choose k)*(n ^ (-k))) <= 1/(k!)
theorem Th13:
k0 by XREAL_1:50;
hence aseq(k).n>0 by A2,A1,XREAL_1:139;
1-(k/n)<=1-0 by XREAL_1:6;
hence thesis by A2,Th7;
end;
theorem Th14:
n>0 implies 0<=bseq(k).n & bseq(k).n<=1/(k!) & bseq(k).n<=eseq.k
& 0<=cseq(n).k & cseq(n).k<=1/(k!) & cseq(n).k<=eseq.k
proof
defpred P[Nat] means bseq($1).n<=1/($1!);
assume
A1: n>0;
then
A2: n ^ (-k)>0 by POWER:34;
A3: now
let k;
assume
A4: P[k];
thus P[k+1]
proof
per cases;
suppose
A5: k=0 by A5,Th13;
then
A7: (1/(k+1))*(bseq(k).n)*(aseq(k).n)<=(1/((k+1)!))*(aseq(k).n ) by A6,
XREAL_1:64;
aseq(k).n<=1 by A5,Th13;
then
A8: 1/((k+1)!)*(aseq(k).n)<=1/((k+1)!) by XREAL_1:153;
bseq(k+1).n=(1/(k+1))*(bseq(k).n)*(aseq(k).n) by A1,Th6;
hence thesis by A7,A8,XXREAL_0:2;
end;
suppose
k>=n;
then
A9: k+1>n by NAT_1:13;
bseq(k+1).n = (n choose (k+1))*(n ^ (-(k+1))) by Def2
.= 0*(n ^ (-(k+1))) by A9,NEWTON:def 3
.= 0;
hence thesis;
end;
end;
end;
A10: bseq(k).n=(n choose k)*(n ^ (-k)) by Def2;
hence 0<=bseq(k).n by A2;
A11: P[0] by Th10,NEWTON:12;
for k holds P[k] from NAT_1:sch 2(A11,A3);
hence
A12: bseq(k).n<=1/(k!);
hence bseq(k).n<=eseq.k by Def5;
hence thesis by A10,A2,A12,Th3;
end;
:: n>0 implies (1+(1/n)) ^ n = Sum (\k:(n choose k)*(n ^ (-k)))
theorem Th15:
for seq st seq^\1 is summable holds seq is summable & Sum(seq)=(
seq.0)+Sum(seq^\1)
proof
let seq;
assume
A1: seq^\1 is summable;
hence seq is summable by SERIES_1:13;
thus Sum(seq) = Partial_Sums(seq).0+Sum(seq^\(1+0)) by A1,SERIES_1:13,15
.= (seq.0)+Sum(seq^\1) by SERIES_1:def 1;
end;
theorem Th16:
for D being non empty set, sq being FinSequence of D st 1<=k & k
=1 by A1,A2,XXREAL_0:2;
then len(sq/^1)=len sq-1 by RFINSEQ:def 1;
then k in dom(sq/^1) by A1,A4,FINSEQ_3:25;
hence thesis by A5,RFINSEQ:def 1;
end;
theorem Th17:
for sq st len(sq)>0 holds Sum(sq)=(sq.1)+Sum(sq/^1)
proof
let sq;
assume
A1: len sq>0;
then 0+1<=len sq by NAT_1:13;
then
A2: 1 in dom sq by FINSEQ_3:25;
thus Sum(sq) = Sum(<*sq/.1*>^(sq/^1)) by A1,CARD_1:27,FINSEQ_5:29
.= Sum(<*sq.1*>^(sq/^1)) by A2,PARTFUN1:def 6
.= (sq.1)+Sum(sq/^1) by RVSUM_1:76;
end;
theorem Th18:
for n holds for seq, sq st len(sq)=n & (for k st k=n holds seq.k=0) holds seq is summable & Sum(seq)=Sum
(sq)
proof
defpred P[Nat] means
for seq, sq st len(sq)=$1 & (for k st k<$1
holds seq.k=sq.(k+1)) & (for k st k>=$1 holds seq.k=0) holds seq is summable &
Sum(seq)=Sum(sq);
now
let n;
assume
A1: for seq, sq st len(sq)=n & (for k st k=n holds seq.k=0) holds seq is summable & Sum(seq)=Sum(sq);
let seq, sq;
assume that
A2: len(sq)=n+1 and
A3: for k st k=n+1 holds seq.k=0;
A5: now
let k;
A6: k+1>=0+1 by XREAL_1:6;
assume k=n;
then
A9: k+1>=n+1 by XREAL_1:6;
thus (seq^\1).k = seq.(k+1) by NAT_1:def 3
.= 0 by A4,A9;
end;
n+1>=0+1 by XREAL_1:6;
then
A10: len(sq/^1) = len(sq)-1 by A2,RFINSEQ:def 1
.= n by A2;
then
A11: Sum(seq^\1)=Sum(sq/^1) by A1,A5,A8;
A12: seq^\1 is summable by A1,A10,A5,A8;
hence seq is summable by Th15;
thus Sum(seq) = (seq.0)+Sum(seq^\1) by A12,Th15
.= (sq.(0+1))+Sum(seq^\1) by A3
.= Sum(sq) by A2,A11,Th17;
end;
then
A13: P[k] implies P[k+1];
now
let seq, sq;
assume that
A14: len(sq)=0 and
for k st k<0 holds seq.k=sq.(k+1) and
A15: for k st k>=0 holds seq.k=0;
sq is Element of 0-tuples_on REAL by A14,FINSEQ_2:92;
then
A16: Sum(sq)=0 by RVSUM_1:79;
defpred P[Nat] means Partial_Sums(seq).$1=0;
A17: now
let k be Nat;
A18: Partial_Sums(seq).(k+1) = (Partial_Sums(seq).k)+(seq.(k+1)) by
SERIES_1:def 1;
assume P[k];
hence P[k+1] by A15,A18;
end;
seq.0=0 by A15;
then
A19: P[0] by SERIES_1:def 1;
A20: for k being Nat holds P[k] from NAT_1:sch 2(A19,A17);
then Partial_Sums(seq) is convergent by Th9;
hence seq is summable by SERIES_1:def 2;
lim(Partial_Sums(seq))=0 by A20,Th9;
hence Sum(seq) = Sum(sq) by A16,SERIES_1:def 3;
end;
then
A21: P[0];
thus P[n] from NAT_1:sch 2(A21,A13);
end;
theorem Th19:
k<=n implies ((x,y) In_Power n).(k+1)=(n choose k)*(x ^ (n-k))*( y ^ k)
proof
reconsider i1 = (k+1)-1 as Element of NAT by ORDINAL1:def 12;
A1: k+1>=0+1 & len((x,y) In_Power n)=n+1 by NEWTON:def 4,XREAL_1:6;
assume
A2: k<=n;
then reconsider l = n-i1 as Element of NAT by INT_1:5;
k+1<=n+1 by A2,XREAL_1:6;
then k+1 in dom((x,y) In_Power n) by A1,FINSEQ_3:25;
hence ((x,y) In_Power n).(k+1) = (n choose i1)*(x ^ l)*(y|^i1) by
NEWTON:def 4
.= (n choose k)*(x ^ (n-k))*(y ^ k);
end;
theorem Th20:
n>0 & k<=n implies cseq(n).k=((1,1/n) In_Power n).(k+1)
proof
assume that
A1: n>0 and
A2: k<=n;
thus ((1,1/n) In_Power n).(k+1) = (n choose k)*(1 ^ (n-k))*((1/n) ^ k) by A2
,Th19
.= (n choose k)*1*((1/n) ^ k) by POWER:26
.= (n choose k)*(n ^ (-k)) by A1,POWER:32
.= cseq(n).k by Def3;
end;
theorem Th21:
n>0 implies cseq(n) is summable & Sum(cseq(n))=(1+(1/n)) ^ n &
Sum(cseq(n))=dseq.n
proof
A1: now
let k;
assume k>=n+1;
then
A2: k>n by NAT_1:13;
thus cseq(n).k = (n choose k)*(n ^ (-k)) by Def3
.= 0*(n ^ (-k)) by A2,NEWTON:def 3
.= 0;
end;
assume
A3: n>0;
A4: now
let k;
assume k0 holds ex
N st for n st n>=N holds seq.n>x-eps
proof
assume
A1: seq is convergent & lim(seq)=x;
let eps;
assume eps>0;
then consider N such that
A2: for n st n>=N holds |.seq.n-x.|=N;
then |.seq.n-x.|-eps by SEQ_2:1;
then (seq.n-x)+x>-eps+x by XREAL_1:6;
hence thesis;
end;
theorem Th26:
(for eps st eps>0 holds ex N st for n st n>=N holds seq.n>x-eps)
& (ex N st for n st n>=N holds seq.n<=x) implies seq is convergent & lim(seq)=x
proof
assume that
A1: for eps st eps>0 holds ex N st for n st n>=N holds seq.n>x-eps and
A2: ex N st for n st n>=N holds seq.n<=x;
A3: for eps be Real st eps>0 ex N st for n st n>=N holds |.seq.n-x.|
0;
then
A5: x+eps>x+0 by XREAL_1:6;
consider N2 being Nat such that
A6: for n st n>=N2 holds seq.n<=x by A2;
consider N1 being Nat such that
A7: for n st n>=N1 holds seq.n>x-eps by A1,A4;
reconsider N = max(N1,N2) as Nat by TARSKI:1;
take N;
let n;
assume
A8: n>=N;
then n>=N1 by XXREAL_0:30;
then seq.n>x-eps by A7;
then
A9: seq.n-x>(x-eps)-x by XREAL_1:9;
seq.n<=x by A6,A8,XXREAL_0:30;
then seq.n0 holds ex K st
Partial_Sums(seq).K>Sum(seq)-eps
proof
assume seq is summable;
then
A1: Partial_Sums(seq) is convergent by SERIES_1:def 2;
let eps;
assume eps>0;
then consider K such that
A2: for k st k>=K holds Partial_Sums(seq).k>lim(Partial_Sums(seq))-eps
by A1,Th25;
take K;
Sum(seq)=lim(Partial_Sums(seq)) by SERIES_1:def 3;
hence thesis by A2;
end;
theorem Th28:
n>=1 implies dseq.n<=Sum(eseq)
proof
assume
A1: n>=1;
then for k holds 0<=cseq(n).k & cseq(n).k<=eseq.k by Th14;
then Sum(cseq(n))<=Sum(eseq) by Th23,SERIES_1:20;
hence thesis by A1,Th21;
end;
theorem Th29:
seq is summable & (for k holds seq.k>=0) implies Sum(seq)>=
Partial_Sums(seq).K
proof
assume that
A1: seq is summable and
A2: for k holds seq.k>=0;
A3: now
let k;
(seq^\(K+1)).k = seq.(K+1+k) by NAT_1:def 3;
hence (seq^\(K+1)).k>=0 by A2;
end;
seq^\(K+1) is summable by A1,SERIES_1:12;
then Sum(seq^\(K+1))>=0 by A3,SERIES_1:18;
then Partial_Sums(seq).K+Sum(seq^\(K+1))>=Partial_Sums(seq).K+0 by XREAL_1:6;
hence thesis by A1,SERIES_1:15;
end;
theorem Th30:
dseq is convergent & lim(dseq)=Sum(eseq)
proof
for eps st eps>0 holds ex N st for n st n>=N holds dseq.n>Sum(eseq)-eps
proof
let eps;
assume
A1: eps>0;
then consider K such that
A2: Partial_Sums(eseq).K>Sum(eseq)-eps/2 by Th23,Th27,XREAL_1:139;
A3: Partial_Sums(eseq).K-eps/2>Sum(eseq)-eps/2-eps/2 by A2,XREAL_1:9;
deffunc F(Nat)=Partial_Sums(cseq($1)).K;
consider dseqK being Real_Sequence such that
A4: for n holds dseqK.n=F(n) from SEQ_1:sch 1;
dseqK is convergent & lim(dseqK)=Partial_Sums(eseq).K by A4,Th24;
then consider N such that
A5: for n st n>=N holds dseqK.n>Partial_Sums(eseq).K-eps/2 by A1,Th25,
XREAL_1:139;
take N1 = N+1;
let n;
assume
A6: n>=N1;
then ( for k holds cseq(n).k>=0)& cseq(n) is summable by Th14,Th21;
then Sum(cseq(n))>=Partial_Sums(cseq n).K by Th29;
then dseq.n>=Partial_Sums(cseq n).K by A6,Th21;
then
A7: dseq.n>=dseqK.n by A4;
N+1>=N+0 by XREAL_1:6;
then n>=N by A6,XXREAL_0:2;
then dseqK.n>Partial_Sums(eseq).K-eps/2 by A5;
then dseq.n>Partial_Sums(eseq).K-eps/2 by A7,XXREAL_0:2;
hence thesis by A3,XXREAL_0:2;
end;
hence thesis by Th26,Th28;
end;
:: number_e = exp(1)
definition
redefine func number_e equals
Sum eseq;
compatibility by Th22,Th30;
end;
definition
redefine func number_e equals
exp_R(1);
compatibility by Th23;
end;
begin :: number_e is irrational
theorem Th31:
x is rational implies ex n st n>=2 & n!*x is integer
proof
assume x is rational;
then consider i, n such that
A1: n<>0 and
A2: x=i/n by RAT_1:8;
per cases;
suppose
A3: n<1+1;
A4: n>=0+1 by A1,NAT_1:13;
n<=1 by A3,NAT_1:13;
then n=1 by A4,XXREAL_0:1;
then reconsider x1 = x as Integer by A2;
take n = 2;
n!*x1 is integer;
hence thesis;
end;
suppose
A5: n>=2;
take n;
thus n>=2 by A5;
reconsider m = n-1 as Element of NAT by A5,INT_1:5,XXREAL_0:2;
n!*x = (m+1)*(m!)*x by NEWTON:15
.= (n*x)*(m!)
.= i*(m!) by A1,A2,XCMPLX_1:87;
hence thesis;
end;
end;
theorem Th32:
n!*eseq.k = (n!)/(k!)
proof
thus n!*eseq.k = n!*(1/(k!)) by Def5
.= (n!)/(k!);
end;
theorem
(n!)/(k!)>0 by XREAL_1:139;
theorem Th34:
seq is summable & (for n holds seq.n>0) implies Sum(seq)>0
proof
assume that
A1: seq is summable and
A2: for n holds seq.n>0;
A3: Sum(seq)=(Partial_Sums(seq).0)+Sum(seq^\(0+1)) by A1,SERIES_1:15
.= seq.0+Sum(seq^\1) by SERIES_1:def 1;
A4: now
let n;
(seq^\1).n = seq.(1+n) by NAT_1:def 3;
hence (seq^\1).n>=0 by A2;
end;
seq^\1 is summable by A1,SERIES_1:12;
then Sum(seq^\1)>=0 by A4,SERIES_1:18;
then Sum(seq)>0+0 by A2,A3,XREAL_1:8;
hence thesis;
end;
theorem Th35:
n!*Sum(eseq^\(n+1))>0
proof
A1: now
let k;
(eseq^\(n+1)).k = eseq.(n+1+k) by NAT_1:def 3
.= 1/((n+1+k)!) by Def5;
hence (eseq^\(n+1)).k>0;
end;
eseq^\(n+1) is summable by Th23,SERIES_1:12;
then n!>0 & Sum(eseq^\(n+1))>0 by A1,Th34;
hence thesis by XREAL_1:129;
end;
theorem Th36:
k<=n implies (n!)/(k!) is integer
proof
defpred P[Nat] means (($1+k)!)/(k!) is integer;
assume k<=n;
then reconsider m = n-k as Element of NAT by INT_1:5;
A1: n=m+k;
now
let m;
A2: (((m+1)+k)!)/(k!) = (m+k+1)*((m+k)!)*(k!)" by NEWTON:15
.= (m+k+1)*(((m+k)!)*(k!)")
.= (m+k+1)*(((m+k)!)/(k!));
assume ((m+k)!)/(k!) is integer;
then reconsider i = ((m+k)!)/(k!) as Integer;
(m+k+1)*i is Integer;
hence (((m+1)+k)!)/(k!) is integer by A2;
end;
then
A3: for n being Nat holds P[n] implies P[n+1];
A4: P[0] by XCMPLX_1:60;
for n being Nat holds P[n] from NAT_1:sch 2(A4,A3);
hence thesis by A1;
end;
theorem Th37:
n!*Partial_Sums(eseq).n is integer
proof
defpred P[Nat] means
$1<=n implies n!*Partial_Sums(eseq).$1 is integer;
now
let k;
assume
A1: k<=n implies n!*Partial_Sums(eseq).k is integer;
assume
A2: k+1<=n;
k+0<=k+1 by XREAL_1:6;
then reconsider i = n!*Partial_Sums(eseq).k as Integer by A1,A2,XXREAL_0:2;
n!*eseq.(k+1) = (n!)/((k+1)!) by Th32;
then reconsider j = n!*eseq.(k+1) as Integer by A2,Th36;
A3: i+j is Integer;
n!*Partial_Sums(eseq).(k+1) = n!*(Partial_Sums(eseq).k+eseq.(k+1)) by
SERIES_1:def 1
.= n!*Partial_Sums(eseq).k+n!*eseq.(k+1);
hence n!*Partial_Sums(eseq).(k+1) is integer by A3;
end;
then
A4: P[k] implies P[k+1];
now
assume 0<=n;
n!*Partial_Sums(eseq).0 = n!*eseq.0 by SERIES_1:def 1
.= (n!)/(0!) by Th32;
hence n!*Partial_Sums(eseq).0 is integer by Th36;
end;
then
A5: P[0];
for k holds P[k] from NAT_1:sch 2(A5,A4);
hence thesis;
end;
theorem Th38:
x=1/(n+1) implies (n!)/((n+k+1)!)<=x ^ (k+1)
proof
defpred P[Nat] means (n!)/((n+$1+1)!)<=x ^ ($1+1);
assume
A1: x=1/(n+1);
A2: now
let k;
assume
A3: P[k];
A4: (n!)/((n+(k+1)+1)!) = (n!)*((n+(k+1)+1)!)"
.= (n!)*((n+(k+1)+1)*((n+k+1)!))" by NEWTON:15
.= (n!)*((n+(k+1)+1)"*((n+k+1)!)") by XCMPLX_1:204
.= (n!)*((n+k+1)!)"*(n+(k+1)+1)"
.= (n!)/((n+k+1)!)*(n+(k+1)+1)";
n+(k+1)>=n+0 by XREAL_1:6;
then n+(k+1)+1>=n+1 by XREAL_1:6;
then
A5: (n+(k+1)+1)"<=1/(n+1) by XREAL_1:85;
x ^ (k+1)*x = x ^ (k+1)*(x^1)
.= x ^ ((k+1)+1) by A1,POWER:27;
hence P[k+1] by A1,A3,A5,A4,XREAL_1:66;
end;
(n!)/((n+1)!) = (n!)/((n+1)*(n!)) by NEWTON:15
.= (n!)*((n+1)*(n!))"
.= (n!)*((n+1)"*(n!)") by XCMPLX_1:204
.= (n!)*(n!)"*(n+1)"
.= 1*(n+1)" by XCMPLX_0:def 7
.= x by A1;
then
A6: P[0];
for n holds P[n] from NAT_1:sch 2(A6,A2);
hence thesis;
end;
theorem Th39:
n>0 & x=1/(n+1) implies n!*Sum(eseq^\(n+1))<=x/(1-x)
proof
assume that
A1: n>0 and
A2: x=1/(n+1);
deffunc F(Nat)=x ^ ($1+1);
consider seq being Real_Sequence such that
A3: for k holds seq.k=F(k) from SEQ_1:sch 1;
A4: now
let k;
A5: (n!(#)(eseq^\(n+1))).k = n!*((eseq^\(n+1)).k) by SEQ_1:9
.= n!*eseq.(n+1+k) by NAT_1:def 3
.= n!*(1/((n+k+1)!)) by Def5
.= n!/((n+k+1)!);
hence (n!(#)(eseq^\(n+1))).k>=0;
seq.k=x ^ (k+1) by A3;
hence (n!(#)(eseq^\(n+1))).k<=seq.k by A2,A5,Th38;
end;
A6: seq.0 = x ^ (0+1) by A3
.= x;
A7: eseq^\(n+1) is summable by Th23,SERIES_1:12;
n+1>0+1 by A1,XREAL_1:6;
then
A8: x<1 by A2,XREAL_1:212;
A9: now
let k;
thus seq.(k+1) = x ^ (k+1+1) by A3
.= x ^ 1*(x^(k+1)) by A2,POWER:27
.= x*(x^(k+1))
.= x*seq.k by A3;
end;
|.x.|=x by A2,ABSVALUE:def 1;
then seq is summable by A8,A9,SERIES_1:25;
then
A10: Sum(n!(#)(eseq^\(n+1)))<=Sum(seq) by A4,SERIES_1:20;
|.x.|<1 by A2,A8,ABSVALUE:def 1;
then Sum(seq)=x/(1-x) by A6,A9,SERIES_1:25;
hence thesis by A7,A10,SERIES_1:10;
end;
theorem Th40:
for n be Real st n>=2 & x=1/(n+1) holds x/(1-x)<1
proof
let n be Real;
assume that
A1: n>=2 and
A2: x=1/(n+1);
n+1 > n by XREAL_1:29;
then 2=2 and
A2: n!*number_e is integer by Th31;
reconsider ne = n!*number_e as Integer by A2;
set x = 1/(n+1);
reconsider ne1 = n!*Partial_Sums(eseq).n as Integer by Th37;
n!*number_e = n!*((Partial_Sums(eseq).n)+Sum(eseq^\(n+1))) by Th23,
SERIES_1:15
.= n!*(Partial_Sums(eseq).n)+n!*Sum(eseq^\(n+1));
then
A3: n!*Sum(eseq^\(n+1))=ne-ne1;
x/(1-x)<1 & n!*Sum(eseq^\(n+1))<=x/(1-x) by A1,Th39,Th40;
then n!*Sum(eseq^\(n+1))<0+1 by XXREAL_0:2;
hence contradiction by A3,Th35,INT_1:7;
end;