### The Mizar article:

### Irrationality of $e$

**by****Freek Wiedijk**

- Received July 2, 1999
Copyright (c) 1999 Association of Mizar Users

- MML identifier: IRRAT_1
- [ MML identifier index ]

environ vocabulary INT_1, ARYTM, SEQ_1, FINSEQ_1, RAT_1, POWER, FILTER_0, SQUARE_1, ARYTM_3, FUNCT_1, ARYTM_1, NEWTON, QC_LANG1, RELAT_1, SEQ_2, ORDINAL2, ABSVALUE, SEQM_3, SUPINF_2, RLVECT_1, SERIES_1, RFINSEQ, FINSEQ_4, FINSEQ_2, GROUP_1, SIN_COS, IRRAT_1; notation XBOOLE_0, SUBSET_1, ORDINAL1, NUMBERS, XCMPLX_0, XREAL_0, INT_1, NAT_1, NEWTON, FUNCT_1, SEQ_1, SEQ_2, POWER, ABSVALUE, SEQM_3, FINSEQ_1, RVSUM_1, SERIES_1, FINSEQ_2, RFINSEQ, FINSEQ_4, SIN_COS, RAT_1, SQUARE_1, LIMFUNC1, NAT_LAT, INT_2, PEPIN; constructors NAT_1, REAL_1, SEQ_2, PARTFUN1, PREPOWER, RVSUM_1, SERIES_1, RFINSEQ, FINSEQ_4, SIN_COS, COMSEQ_3, LIMFUNC1, NAT_LAT, PEPIN, COMPLEX1, MEMBERED, ARYTM_0; clusters RELSET_1, XREAL_0, SEQ_1, INT_1, NEWTON, NAT_1, MEMBERED, ORDINAL2, NUMBERS; requirements REAL, NUMERALS, BOOLE, SUBSET, ARITHM; definitions SEQ_2; theorems SEQ_1, REAL_2, POWER, NEWTON, REAL_1, AXIOMS, NAT_1, SEQ_2, SEQ_4, ABSVALUE, SEQM_3, RVSUM_1, FINSEQ_2, SERIES_1, RFINSEQ, FINSEQ_3, FINSEQ_4, FINSEQ_5, FINSEQ_1, SIN_COS, SQUARE_1, INT_1, RAT_1, INT_2, NAT_LAT, FUNCT_2, XCMPLX_0, XCMPLX_1; 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 number; reserve seq, seq1, seq2 for Real_Sequence; reserve sq for FinSequence of REAL; definition let x; redefine attr x is rational; antonym x is irrational; end; definition let x, y; redefine func x to_power y; synonym x.^.y; end; theorem Th1: p is prime implies sqrt p is irrational proof assume A1: p is prime; then A2: p>1 by INT_2:def 5; then A3: p>0 by AXIOMS:22; assume sqrt p is rational; then consider i, n such that A4: n<>0 & sqrt p=i/n & for i1 being Integer, n1 being Nat st n1<>0 & sqrt p=i1/n1 holds n<=n1 by RAT_1:25; A5: i=sqrt p*n by A4,XCMPLX_1:88; sqrt p>=0 & n>=0 by A3,NAT_1:18,SQUARE_1:def 4; then i>=0 by A5,REAL_2:121; then reconsider m = i as Nat by INT_1:16; A6: m^2 = (sqrt p)^2*n^2 by A5,SQUARE_1:68 .= p*n^2 by A3,SQUARE_1:def 4; then p divides m^2 by NAT_1:def 3; then p divides m*m by SQUARE_1:def 3; then p divides m by A1,NAT_LAT:95; then consider m1 being Nat such that A7: m=p*m1 by NAT_1:def 3; n^2 = m^2/p by A3,A6,XCMPLX_1:90 .= p^2*m1^2/p by A7,SQUARE_1:68 .= p*p*m1^2/p by SQUARE_1:def 3 .= p*(p*m1^2)/p by XCMPLX_1:4 .= p*m1^2 by A3,XCMPLX_1:90; then p divides n^2 by NAT_1:def 3; then p divides n*n by SQUARE_1:def 3; then p divides n by A1,NAT_LAT:95; then consider n1 being Nat such that A8: n=p*n1 by NAT_1:def 3; A9: n1<>0 by A4,A8; then A10: n1>0 by NAT_1:19; A11: m1/n1 = sqrt p by A3,A4,A7,A8,XCMPLX_1:92; p*n1>1*n1 by A2,A10,REAL_2:199; hence contradiction by A4,A8,A9,A11; end; theorem ex x, y st x is irrational & y is irrational & x.^.y is rational proof set w = sqrt 2; w>0 by AXIOMS:22,SQUARE_1:84; then A1: (w.^.w).^.w = w.^.(w*w) by POWER:38 .= w.^.(w^2) by SQUARE_1:def 3 .= w.^.2 by SQUARE_1:def 4 .= w^2 by POWER:53 .= 2 by SQUARE_1:def 4; per cases; suppose A2: w.^.w is rational; take w, w; thus thesis by A2,Th1,INT_2:44; suppose A3: w.^.w is irrational; take w.^.w, w; thus thesis by A1,A3,Th1,INT_2:44; end; begin :: A proof that e = e scheme LambdaRealSeq{F(set)->real number}: (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 deffunc F1(set)=F($1); thus ex seq st for n holds seq.n=F1(n) from ExRealSeq; let seq1, seq2; assume A1: for n holds seq1.n=F(n); assume A2: for n holds seq2.n=F(n); now let n; thus seq1.n = F(n) by A1 .= seq2.n by A2; end; hence seq1 = seq2 by FUNCT_2:113; end; definition let k; 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; end; definition let k; 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; 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)) by XCMPLX_1:161 .= (n.^.(-k))/(n.^.1) by A1,POWER:34 .= (n.^.(-k))/n by POWER:30; end; theorem Th5: for x, y, z, v, w being real number holds x/(y*z*(v/w))=(w/z)*(x/(y*v)) proof let x, y, z, v, w be real number; thus x/(y*z*(v/w)) = x/(y*z*(v*w")) by XCMPLX_0:def 9 .= x/(z*(y*(v*w"))) by XCMPLX_1:4 .= x/(z*(y*v*w")) by XCMPLX_1:4 .= x/(z*((y*v)/w)) by XCMPLX_0:def 9 .= x/((y*v)/(w/z)) by XCMPLX_1:82 .= (w/z)*(x/(y*v)) by XCMPLX_1:82; end; theorem Th6: (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 Nat by INT_1:18; A2: l+1 = n-(k+1-1) by XCMPLX_1:37 .= n-k by XCMPLX_1:26; then reconsider l1 = n-k as Nat; k <= k+1 by NAT_1:29; then A3: k <= n by A1,AXIOMS:22; thus (n choose (k+1)) = n!/((k+1)!*(l!)) by A1,NEWTON:def 3 .= n!/((k!*(k+1))*(l!)) by NEWTON:21 .= n!/((k!*(k+1))*((l!)*(l+1)/(l+1))) by XCMPLX_1:90 .= n!/((k!*(k+1))*((l+1)!/(l+1))) by NEWTON:21 .= (l1/(k+1))*(n!/((k!)*(l1!))) by A2,Th5 .= ((n-k)/(k+1))*(n choose k) by A3,NEWTON:def 3; suppose A4: k+1>n & k<=n; then k>=n & k<=n by NAT_1:38; then k=n by AXIOMS:21; then n-k=0 by XCMPLX_1:14; hence (n choose (k+1)) = ((n-k)/(k+1))*(n choose k) by A4,NEWTON:def 3; suppose A5: 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 A5,NEWTON:def 3; end; theorem Th7: 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 Th6 .= ((n-k)/(k+1))*(n choose k)*((n.^.(-k))/n) by A1,Th4 .= (n-k)*(k+1)"*(n choose k)*((n.^.(-k))/n) by XCMPLX_0:def 9 .= (n-k)*(k+1)"*(n choose k)*((n.^.(-k))*n") by XCMPLX_0:def 9 .= (n-k)*(k+1)"*(n choose k)*(n.^.(-k))*n" by XCMPLX_1:4 .= (k+1)"*(n choose k)*(n-k)*(n.^.(-k))*n" by XCMPLX_1:4 .= (k+1)"*(n choose k)*(n.^.(-k))*(n-k)*n" by XCMPLX_1:4 .= (k+1)"*((n choose k)*(n.^.(-k)))*(n-k)*n" by XCMPLX_1:4 .= (k+1)"*((n choose k)*(n.^.(-k)))*((n-k)*n") by XCMPLX_1:4 .= (1/(k+1))*((n choose k)*(n.^.(-k)))*((n-k)*n") by XCMPLX_1:217 .= (1/(k+1))*((n choose k)*(n.^.(-k)))*((n-k)/n) by XCMPLX_0:def 9 .= (1/(k+1))*(bseq(k).n)*((n-k)/n) by Def2 .= (1/(k+1))*(bseq(k).n)*(aseq(k).n) by Def1; end; theorem Th8: 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) by XCMPLX_1:121 .= 1-(k/n) by A1,XCMPLX_1:60; end; theorem Th9: aseq(k) is convergent & lim(aseq(k))=1 proof A1: for eps be real number st 0<eps ex N st for n st N<=n holds abs(aseq(k).n-1)<eps proof let eps be real number; assume A2: eps>0; consider N such that A3: N>k/eps by SEQ_4:10; A4: k>=0*eps by NAT_1:18; take N; let n; assume A5: n>=N; then A6: n>0 by A2,A3,A4,REAL_2:177; k>=0*n by NAT_1:18; then A7: k/n>=0 by A6,REAL_2:177; A8: abs(aseq(k).n-1) = abs((1-(k/n))-1) by A6,Th8 .= abs((1+-(k/n))-1) by XCMPLX_0:def 8 .= abs(-(k/n)) by XCMPLX_1:26 .= abs(k/n) by ABSVALUE:17 .= k/n by A7,ABSVALUE:def 1; n>(k/eps) by A3,A5,AXIOMS:22; then (k/eps)*eps<n*eps by A2,REAL_1:70; then k<n*eps by A2,XCMPLX_1:88; hence thesis by A6,A8,REAL_2:178; end; thus aseq(k) is convergent proof take 1; thus thesis by A1; end; hence lim(aseq(k))=1 by A1,SEQ_2:def 7; end; theorem Th10: for seq st for n holds seq.n=x holds seq is convergent & lim(seq)=x proof let seq; assume A1: for n holds seq.n=x; then A2: seq is constant by SEQM_3:def 6; hence seq is convergent by SEQ_4:39; thus lim(seq) = seq.0 by A2,SEQ_4:41 .= x by A1; end; theorem Th11: 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:29 .= 1 by POWER:29; end; theorem Th12: (1/(k+1))*(1/(k!))=1/((k+1)!) proof thus (1/(k+1))*(1/(k!)) = 1/((k+1)*(k!)) by XCMPLX_1:103 .= 1/((k+1)!) by NEWTON:21; end; theorem Th13: 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: P[0] proof deffunc F(set)=1; consider bseq0 being Real_Sequence such that A2: for n holds bseq0.n=F(n) from ExRealSeq; A3: bseq0 is convergent & lim(bseq0)=1 by A2,Th10; A4: for n st n>=1 holds bseq(0).n=bseq0.n proof let n; assume n>=1; thus bseq(0).n = 1 by Th11 .= bseq0.n by A2; end; hence bseq(0) is convergent by A3,SEQ_4:31; thus lim(bseq(0)) = 1/(0!) by A3,A4,NEWTON:18,SEQ_4:32; end; A5: now let k; assume A6: P[k]; thus P[k+1] proof deffunc F(Nat)=(1/(k+1))*(bseq(k).$1); consider seq such that A7: for n holds seq.n = F(n) from ExRealSeq; A8: seq = (1/(k+1))(#)(bseq(k)) by A7,SEQ_1:13; A9: seq is convergent & lim(seq)=1/((k+1)!) proof thus seq is convergent by A6,A8,SEQ_2:21; thus lim(seq) = (1/(k+1))*(1/(k!)) by A6,A8,SEQ_2:22 .= 1/((k+1)!) by Th12; end; deffunc G(Nat)=seq.$1*(aseq(k).$1); consider seq1 such that A10: for n holds seq1.n=G(n) from ExRealSeq; A11: seq1=seq(#)(aseq(k)) by A10,SEQ_1:12; A12: seq1 is convergent & lim(seq1)=1/((k+1)!) proof A13: aseq(k) is convergent & lim(aseq(k))=1 by Th9; hence seq1 is convergent by A9,A11,SEQ_2:28; thus lim(seq1) = lim(seq)*lim(aseq(k)) by A9,A11,A13,SEQ_2:29 .= 1/((k+1)!) by A9,A13; end; A14: for n st n>=1 holds bseq(k+1).n=seq1.n proof let n; assume n>=1; then n>=0+1; then n>0 by NAT_1:38; hence bseq(k+1).n = (1/(k+1))*(bseq(k).n)*(aseq(k).n) by Th7 .= (seq.n)*(aseq(k).n) by A7 .= seq1.n by A10; end; hence bseq(k+1) is convergent by A12,SEQ_4:31; thus lim(bseq(k+1)) = 1/((k+1)!) by A12,A14,SEQ_4:32; end; end; for k holds P[k] from Ind(A1,A5); hence bseq(k) is convergent & lim(bseq(k))=1/(k!); hence lim(bseq(k))=eseq.k by Def5; end; :: 0 <= (n choose k)*(n.^.(-k))) <= 1/(k!) theorem Th14: k<n implies 0<aseq(k).n & aseq(k).n<=1 proof assume A1: k<n; A2: 0<=k by NAT_1:18; A3: aseq(k).n=(n-k)/n by Def1; n-k>0 by A1,SQUARE_1:11; hence aseq(k).n>0 by A1,A2,A3,REAL_2:127; n >= 0 by A1, A2, AXIOMS:22; then k/n>=0 by A2,REAL_2:125; then 1-(k/n)<=1-0 by REAL_2:106; hence aseq(k).n<=1 by A1,A2,Th8; end; theorem Th15: 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 assume A1: n>0; A2: bseq(k).n=(n choose k)*(n.^.(-k)) by Def2; (n choose k) is Nat by NEWTON:35; then A3: (n choose k)>=0 by NAT_1:18; n.^.(-k)>0 by A1,POWER:39; hence A4: 0<=bseq(k).n by A2,A3,REAL_2:121; defpred P[Nat] means bseq($1).n<=1/($1!); A5: P[0] by Th11,NEWTON:18; A6: now let k; assume A7: P[k]; thus P[k+1] proof A8: (k+1)!>0 by NEWTON:23; then A9: 1/((k+1)!)>0 by REAL_2:127; per cases; suppose A10: k<n; A11: bseq(k+1).n=(1/(k+1))*(bseq(k).n)*(aseq(k).n) by A1,Th7; k+1>0 by NAT_1:19; then 1/(k+1)>0 by REAL_2:127; then (1/(k+1))*(bseq(k).n)<=(1/(k+1))*(1/(k!)) by A7,AXIOMS:25; then A12: (1/(k+1))*(bseq(k).n)<=1/((k+1)!) by Th12; aseq(k).n>=0 by A10,Th14; then A13: (1/(k+1))*(bseq(k).n)*(aseq(k).n)<=(1/((k+1)!))*(aseq(k).n) by A12,AXIOMS:25; aseq(k).n<=1 by A10,Th14; then 1/((k+1)!)*(aseq(k).n)<=1/((k+1)!) by A9,REAL_2:147; hence thesis by A11,A13,AXIOMS:22; suppose k>=n; then A14: k+1>n by NAT_1:38; bseq(k+1).n = (n choose (k+1))*(n.^.(-(k+1))) by Def2 .= 0*(n.^.(-(k+1))) by A14,NEWTON:def 3 .= 0; hence thesis by A8,REAL_2:127; end; end; for k holds P[k] from Ind(A5,A6); hence A15: bseq(k).n<=1/(k!); hence bseq(k).n<=eseq.k by Def5; hence 0<=cseq(n).k & cseq(n).k<=1/(k!) & cseq(n).k<=eseq.k by A4,A15,Th3; end; :: n>0 implies (1+(1/n)).^.n = Sum (\k:(n choose k)*(n.^.(-k))) theorem Th16: for seq st seq^\1 is summable holds seq is summable & Sum(seq)=(seq.0)+Sum(seq^\1) proof let seq; assume seq^\1 is summable; hence seq is summable by SERIES_1:16; hence Sum(seq) = Partial_Sums(seq).0+Sum(seq^\(1+0)) by SERIES_1:18 .= (seq.0)+Sum(seq^\1) by SERIES_1:def 1; end; theorem Th17: for D being non empty set, sq being FinSequence of D st 1<=k & k<len sq holds (sq/^1).k=sq.(k+1) proof let D be non empty set, sq be FinSequence of D; assume A1: 1<=k & k<len sq; then A2: len(sq)>=1 by AXIOMS:22; then A3: len(sq/^1)=len sq-1 by RFINSEQ:def 2; A4: k+1<=len sq by A1,NAT_1:38; len sq=(len sq-1)+1 by XCMPLX_1:27; then k<=len sq-1 by A4,REAL_1:53; then k in dom(sq/^1) by A1,A3,FINSEQ_3:27; hence (sq/^1).k=sq.(k+1) by A2,RFINSEQ:def 2; end; theorem Th18: for sq st len(sq)>0 holds Sum(sq)=(sq.1)+Sum(sq/^1) proof let sq; assume A1: len sq>0; then 1<=1 & 0+1<=len sq by NAT_1:38; then A2: 1 in dom sq by FINSEQ_3:27; sq is non empty by A1,FINSEQ_1:25; hence Sum(sq) = Sum(<*sq/.1*>^(sq/^1)) by FINSEQ_5:32 .= Sum(<*sq.1*>^(sq/^1)) by A2,FINSEQ_4:def 4 .= (sq.1)+Sum(sq/^1) by RVSUM_1:106; end; theorem Th19: for n holds for seq, sq st len(sq)=n & (for k st k<n holds seq.k=sq.(k+1)) & (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 seq, sq; assume A1: len(sq)=0 & (for k st k<0 holds seq.k=sq.(k+1)) & (for k st k>=0 holds seq.k=0); then sq is Element of 0-tuples_on REAL by FINSEQ_2:110; then A2: Sum(sq)=0 by RVSUM_1:109; defpred P[Nat] means Partial_Sums(seq).$1=0; A3: now let k; k>=0 by NAT_1:18; hence seq.k = 0 by A1; end; then seq.0=0; then A4: P[0] by SERIES_1:def 1; A5: now let k; assume A6: P[k]; Partial_Sums(seq).(k+1) = (Partial_Sums(seq).k)+(seq.(k+1)) by SERIES_1:def 1; hence P[k+1] by A3,A6; end; for k holds P[k] from Ind(A4,A5); then A7: Partial_Sums(seq) is convergent & lim(Partial_Sums(seq))=0 by Th10; hence seq is summable by SERIES_1:def 2; thus Sum(seq) = Sum(sq) by A2,A7,SERIES_1:def 3; end; then A8: P[0]; now let n; assume A9: for seq, sq st len(sq)=n & (for k st k<n holds seq.k=sq.(k+1)) & (for k st k>=n holds seq.k=0) holds seq is summable & Sum(seq)=Sum(sq); let seq, sq; assume A10: len(sq)=n+1 & (for k st k<n+1 holds seq.k=sq.(k+1)) & (for k st k>=n+1 holds seq.k=0); A11: n>=0 by NAT_1:18; A12: n+1>0 by NAT_1:18; n+1>=0+1 by A11,AXIOMS:24; then A13: len(sq/^1) = len(sq)-1 by A10,RFINSEQ:def 2 .= n by A10,XCMPLX_1:26; A14: now let k; assume k<n; then A15: k+1<n+1 by REAL_1:53; k>=0 by NAT_1:18; then A16: k+1>=0+1 by AXIOMS:24; thus (seq^\1).k = seq.(k+1) by SEQM_3:def 9 .= sq.((k+1)+1) by A10,A15 .= (sq/^1).(k+1) by A10,A15,A16,Th17; end; now let k; assume k>=n; then A17: k+1>=n+1 by AXIOMS:24; thus (seq^\1).k = seq.(k+1) by SEQM_3:def 9 .= 0 by A10,A17; end; then A18: seq^\1 is summable & Sum(seq^\1)=Sum(sq/^1) by A9,A13,A14; hence seq is summable by Th16; thus Sum(seq) = (seq.0)+Sum(seq^\1) by A18,Th16 .= (sq.(0+1))+Sum(seq^\1) by A10,A12 .= Sum(sq) by A10,A12,A18,Th18; end; then A19: P[k] implies P[k+1]; thus P[n] from Ind(A8,A19); end; theorem Th20: x<>0 & y<>0 & k<=n implies ((x,y) In_Power n).(k+1)=(n choose k)*(x.^.(n-k))*(y.^.k) proof assume A1: x<>0 & y<>0 & k<=n; k>=0 by NAT_1:18; then A2: k+1>=0+1 by AXIOMS:24; then reconsider i1 = (k+1)-1 as Nat by INT_1:18; A3: i1=k by XCMPLX_1:26; then reconsider l = n-i1 as Nat by A1,INT_1:18; A4: len((x,y) In_Power n)=n+1 by NEWTON:def 4; 1<=k+1 & k+1<=n+1 by A1,A2,AXIOMS:24; then k+1 in dom((x,y) In_Power n) by A4,FINSEQ_3:27; hence ((x,y) In_Power n).(k+1) = (n choose i1)*(x|^l)*(y|^i1) by NEWTON:def 4 .= (n choose i1)*(x.^.l)*(y|^i1) by A1,POWER:48 .= (n choose k)*(x.^.(n-k))*(y.^.k) by A1,A3,POWER:48; end; theorem Th21: n>0 & k<=n implies cseq(n).k=((1,1/n) In_Power n).(k+1) proof assume A1: n>0 & k<=n; then 1/n>0 by REAL_2:127; hence ((1,1/n) In_Power n).(k+1) = (n choose k)*(1 .^. (n-k))*((1/n).^.k) by A1,Th20 .= (n choose k)*1*((1/n).^.k) by POWER:31 .= (n choose k)*(n.^.(-k)) by A1,POWER:37 .= cseq(n).k by Def3; end; theorem Th22: n>0 implies cseq(n) is summable & Sum(cseq(n))=(1+(1/n)).^.n & Sum(cseq(n))=dseq.n proof assume A1: n>0; then 1/n>0 by REAL_2:127; then 1+(1/n)>1+0 by REAL_1:53; then A2: 1+(1/n)>0 by AXIOMS:22; A3: len((1,1/n) In_Power n)=n+1 by NEWTON:def 4; A4: now let k; assume k<n+1; then k<=n by NAT_1:38; hence cseq(n).k=((1,1/n) In_Power n).(k+1) by A1,Th21; end; A5: now let k; assume k>=n+1; then A6: k>n by NAT_1:38; thus cseq(n).k = (n choose k)*(n.^.(-k)) by Def3 .= 0*(n.^.(-k)) by A6,NEWTON:def 3 .= 0; end; hence cseq(n) is summable by A3,A4,Th19; thus (1+(1/n)).^.n = (1+(1/n))|^n by A2,POWER:48 .= Sum((1,1/n) In_Power n) by NEWTON:41 .= Sum(cseq(n)) by A3,A4,A5,Th19; hence dseq.n = Sum(cseq(n)) by Def4; end; :: number_e = lim(\n:(1+(1/n)).^.n) theorem Th23: dseq is convergent & lim(dseq)=number_e proof A1: now let n; thus (dseq^\1).n = dseq.(n+1) by SEQM_3:def 9 .= (1+1/(n+1)).^.(n+1) by Def4; end; then A2: dseq^\1 is convergent by POWER:67; hence dseq is convergent by SEQ_4:35; number_e=lim(dseq^\1) by A1,POWER:def 4; hence number_e=lim(dseq) by A2,SEQ_4:36; end; :: exp(1) = Sum (\k:1/(k!)) theorem Th24: eseq is summable & Sum(eseq)=exp(1) proof now let k; thus eseq.k = 1/(k!) by Def5 .= (1 |^ k)/(k!) by NEWTON:15 .= (1 ExpSeq).k by SIN_COS:def 9; end; then A1: eseq=(1 ExpSeq) by FUNCT_2:113; hence eseq is summable by SIN_COS:49; thus exp(1) = exp.1 by SIN_COS:def 27 .= Sum(eseq) by A1,SIN_COS:def 26; end; :: lim(\n:(1+(1/n)).^.n) = Sum (\k:1/(k!)) theorem Th25: for K holds for dseqK being Real_Sequence st for n holds dseqK.n=Partial_Sums(cseq(n)).K holds dseqK is convergent & lim(dseqK)=Partial_Sums(eseq).K proof defpred P[Nat] means for dseqK being Real_Sequence st for n holds dseqK.n=Partial_Sums(cseq(n)).$1 holds dseqK is convergent & lim(dseqK)=Partial_Sums(eseq).$1; now let dseq0 be Real_Sequence; assume A1: for n holds dseq0.n=Partial_Sums(cseq(n)).0; now let n; thus dseq0.n = Partial_Sums(cseq(n)).0 by A1 .= cseq(n).0 by SERIES_1:def 1 .= bseq(0).n by Th3; end; then A2: dseq0 = bseq(0) by FUNCT_2:113; hence dseq0 is convergent by Th13; thus lim(dseq0) = eseq.0 by A2,Th13 .= Partial_Sums(eseq).0 by SERIES_1:def 1; end; then A3: P[0]; now let K; assume A4: for dseqK being Real_Sequence st for n holds dseqK.n=Partial_Sums(cseq(n)).K holds dseqK is convergent & lim(dseqK)=Partial_Sums(eseq).K; deffunc F(Nat)=Partial_Sums(cseq($1)).K; consider dseqK being Real_Sequence such that A5: for n holds dseqK.n=F(n) from ExRealSeq; A6: dseqK is convergent & lim(dseqK)=Partial_Sums(eseq).K by A4,A5; let dseqK1 be Real_Sequence; assume A7: for n holds dseqK1.n=Partial_Sums(cseq(n)).(K+1); now let n; thus dseqK1.n = Partial_Sums(cseq(n)).(K+1) by A7 .= Partial_Sums(cseq(n)).K+cseq(n).(K+1) by SERIES_1:def 1 .= dseqK.n+cseq(n).(K+1) by A5 .= dseqK.n+bseq(K+1).n by Th3 .= (dseqK+bseq(K+1)).n by SEQ_1:11; end; then A8: dseqK1=dseqK+bseq(K+1) by FUNCT_2:113; A9: bseq(K+1) is convergent by Th13; hence dseqK1 is convergent by A6,A8,SEQ_2:19; thus lim(dseqK1) = lim(dseqK)+lim(bseq(K+1)) by A6,A8,A9,SEQ_2:20 .= Partial_Sums(eseq).K+eseq.(K+1) by A6,Th13 .= Partial_Sums(eseq).(K+1) by SERIES_1:def 1; end; then A10: P[n] implies P[n+1]; thus P[n] from Ind(A3,A10); end; theorem Th26: seq is convergent & lim(seq)=x implies for eps st eps>0 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 abs(seq.n-x)<eps by A1,SEQ_2 :def 7; take N; let n; assume n>=N; then abs(seq.n-x)<eps by A2; then seq.n-x>-eps by SEQ_2:9; then (seq.n-x)+x>-eps+x by REAL_1:53; then seq.n>-eps+x by XCMPLX_1:27; hence seq.n>x-eps by XCMPLX_0:def 8; end; theorem Th27: (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 A1: (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); A2: for eps be real number st eps>0 ex N st for n st n>=N holds abs(seq.n-x)<eps proof let eps be real number; assume A3: eps>0; then consider N1 being Nat such that A4: for n st n>=N1 holds seq.n>x-eps by A1; consider N2 being Nat such that A5: for n st n>=N2 holds seq.n<=x by A1; take N = max(N1,N2); let n; assume n>=N; then n>=N1 & n>=N2 by SQUARE_1:50; then A6: seq.n>x-eps & seq.n<=x by A4,A5; then A7: seq.n-x>(x-eps)-x by REAL_1:54; A8: (x-eps)-x = (x+-eps)-x by XCMPLX_0:def 8 .= -eps by XCMPLX_1:26; x+eps>x+0 by A3,REAL_1:53; then seq.n<x+eps by A6,AXIOMS:22; then seq.n-x<eps by REAL_1:84; hence abs (seq.n-x)<eps by A7,A8,SEQ_2:9; end; hence seq is convergent by SEQ_2:def 6; hence lim(seq)=x by A2,SEQ_2:def 7; end; theorem Th28: seq is summable implies for eps st eps>0 holds ex K st Partial_Sums(seq).K>Sum(seq)-eps proof assume A1: seq is summable; let eps; assume A2: eps>0; Partial_Sums(seq) is convergent by A1,SERIES_1:def 2; then consider K such that A3: for k st k>=K holds Partial_Sums(seq).k>lim(Partial_Sums(seq))-eps by A2,Th26; take K; Sum(seq)=lim(Partial_Sums(seq)) by SERIES_1:def 3; hence Partial_Sums(seq).K>Sum(seq)-eps by A3; end; theorem Th29: n>=1 implies dseq.n<=Sum(eseq) proof assume n>=1; then n>=0+1; then A1: n>0 by NAT_1:38; then for k holds 0<=cseq(n).k & cseq(n).k<=eseq.k by Th15; then Sum(cseq(n))<=Sum(eseq) by Th24,SERIES_1:24; hence dseq.n<=Sum(eseq) by A1,Th22; end; theorem Th30: seq is summable & (for k holds seq.k>=0) implies Sum(seq)>=Partial_Sums(seq).K proof assume A1: seq is summable & (for k holds seq.k>=0); then A2: seq^\(K+1) is summable by SERIES_1:15; now let k; (seq^\(K+1)).k = seq.(K+1+k) by SEQM_3:def 9; hence (seq^\(K+1)).k>=0 by A1; end; then Sum(seq^\(K+1))>=0 by A2,SERIES_1:21; then Partial_Sums(seq).K+Sum(seq^\(K+1))>=Partial_Sums(seq).K+0 by AXIOMS:24; hence Sum(seq)>=Partial_Sums(seq).K by A1,SERIES_1:18; end; theorem Th31: 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 eps>0; then A1: eps/2>0 by REAL_2:127; then consider K such that A2: Partial_Sums(eseq).K>Sum(eseq)-eps/2 by Th24,Th28; deffunc F(Nat)=Partial_Sums(cseq($1)).K; consider dseqK being Real_Sequence such that A3: for n holds dseqK.n=F(n) from ExRealSeq; dseqK is convergent & lim(dseqK)=Partial_Sums(eseq).K by A3,Th25; then consider N such that A4: for n st n>=N holds dseqK.n>Partial_Sums(eseq).K-eps/2 by A1,Th26; take N1 = N+1; let n; assume A5: n>=N1; N+1>=N+0 by AXIOMS:24; then A6: n>=N by A5,AXIOMS:22; A7: n>0 by A5,NAT_1:18; then A8: for k holds cseq(n).k>=0 by Th15; cseq(n) is summable by A7,Th22; then Sum(cseq(n))>=Partial_Sums(cseq(n)).K by A8,Th30; then dseq.n>=Partial_Sums(cseq(n)).K by A7,Th22; then A9: dseq.n>=dseqK.n by A3; dseqK.n>Partial_Sums(eseq).K-eps/2 by A4,A6; then A10: dseq.n>Partial_Sums(eseq).K-eps/2 by A9,AXIOMS:22; A11: Sum(eseq)-eps/2-eps/2 = Sum(eseq)-(eps/2+eps/2) by XCMPLX_1:36 .= Sum(eseq)-eps by XCMPLX_1:66; Partial_Sums(eseq).K-eps/2>Sum(eseq)-eps/2-eps/2 by A2,REAL_1:54; hence dseq.n>Sum(eseq)-eps by A10,A11,AXIOMS:22; end; hence dseq is convergent & lim(dseq)=Sum(eseq) by Th27,Th29; end; :: number_e = exp(1) definition redefine func number_e equals :Def6: Sum(eseq); compatibility by Th23,Th31; end; definition redefine func number_e equals exp(1); compatibility by Def6,Th24; end; begin :: number_e is irrational theorem Th32: 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 & x=i/n by RAT_1:24; per cases; suppose n<1+1; then A2: n<=1 by NAT_1:38; n>0 by A1,NAT_1:19; then n>=0+1 by NAT_1:38; then n=1 by A2,AXIOMS:21; then reconsider x1 = x as Integer by A1; take n = 2; n!*x1 is integer; hence n>=2 & n!*x is integer; suppose A3: n>=2; take n; thus n>=2 by A3; n>=1 by A3,AXIOMS:22; then reconsider m = n-1 as Nat by INT_1:18; A4: n=m+1 by XCMPLX_1:27; then n!*x = (m+1)*(m!)*x by NEWTON:21 .= (n*x)*(m!) by A4,XCMPLX_1:4 .= i*(m!) by A1,XCMPLX_1:88; hence n!*x is integer; end; theorem Th33: n!*eseq.k = (n!)/(k!) proof thus n!*eseq.k = n!*(1/(k!)) by Def5 .= (n!)/(k!) by XCMPLX_1:100; end; theorem Th34: (n!)/(k!)>0 proof n!>0 & k!>0 by NEWTON:23; hence (n!)/(k!)>0 by REAL_2:127; end; theorem Th35: seq is summable & (for n holds seq.n>0) implies Sum(seq)>0 proof assume A1: seq is summable & (for n holds seq.n>0); then A2: Sum(seq)=(Partial_Sums(seq).0)+Sum(seq^\(0+1)) by SERIES_1:18 .= seq.0+Sum(seq^\1) by SERIES_1:def 1; A3: seq.0>0 by A1; A4: seq^\1 is summable by A1,SERIES_1:15; now let n; (seq^\1).n = seq.(1+n) by SEQM_3:def 9; hence (seq^\1).n>=0 by A1; end; then Sum(seq^\1)>=0 by A4,SERIES_1:21; then Sum(seq)>0+0 by A2,A3,REAL_1:67; hence Sum(seq)>0; end; theorem Th36: n!*Sum(eseq^\(n+1))>0 proof A1: n!>0 by NEWTON:23; A2: eseq^\(n+1) is summable by Th24,SERIES_1:15; now let k; A3: (eseq^\(n+1)).k = eseq.(n+1+k) by SEQM_3:def 9 .= 1/((n+1+k)!) by Def5; (n+1+k)!>0 by NEWTON:23; hence (eseq^\(n+1)).k>0 by A3,REAL_2:127; end; then Sum(eseq^\(n+1))>0 by A2,Th35; hence thesis by A1,REAL_2:122; end; theorem Th37: k<=n implies (n!)/(k!) is integer proof assume k<=n; then reconsider m = n-k as Nat by INT_1:18; defpred P[Nat] means (($1+k)!)/(k!) is integer; A1: n=m+k by XCMPLX_1:27; for m holds ((m+k)!)/(k!) is integer proof k!<>0 by NEWTON:23; then A2: P[0] by XCMPLX_1:60; now let m; assume ((m+k)!)/(k!) is integer; then reconsider i = ((m+k)!)/(k!) as Integer; A3: (m+k+1)*i is Integer; ((m+1)+k)! = (m+k+1)! by XCMPLX_1:1 .= (m+k+1)*((m+k)!) by NEWTON:21; then (((m+1)+k)!)/(k!) = (m+k+1)*((m+k)!)*(k!)" by XCMPLX_0:def 9 .= (m+k+1)*(((m+k)!)*(k!)") by XCMPLX_1:4 .= (m+k+1)*(((m+k)!)/(k!)) by XCMPLX_0:def 9; hence (((m+1)+k)!)/(k!) is integer by A3; end; then A4: for n being Nat holds P[n] implies P[n+1]; for n being Nat holds P[n] from Ind(A2,A4); hence thesis; end; hence (n!)/(k!) is integer by A1; end; theorem Th38: n!*Partial_Sums(eseq).n is integer proof defpred P[Nat] means $1<=n implies n!*Partial_Sums(eseq).$1 is integer; for k st k<=n holds n!*Partial_Sums(eseq).k is integer proof now assume A1: 0<=n; n!*Partial_Sums(eseq).0 = n!*eseq.0 by SERIES_1:def 1 .= (n!)/(0!) by Th33; hence n!*Partial_Sums(eseq).0 is integer by A1,Th37; end; then A2: P[0]; now let k; assume A3: k<=n implies n!*Partial_Sums(eseq).k is integer; assume A4: k+1<=n; k+0<=k+1 by AXIOMS:24; then reconsider i = n!*Partial_Sums(eseq).k as Integer by A3,A4,AXIOMS:22; n!*eseq.(k+1) = (n!)/((k+1)!) by Th33; then reconsider j = n!*eseq.(k+1) as Integer by A4,Th37; A5: 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) by XCMPLX_1:8; i+j is Integer; hence n!*Partial_Sums(eseq).(k+1) is integer by A5; end; then A6: P[k] implies P[k+1]; for k holds P[k] from Ind(A2,A6); hence thesis; end; hence thesis; end; theorem Th39: x=1/(n+1) implies (n!)/((n+k+1)!)<=x.^.(k+1) proof assume A1: x=1/(n+1); defpred P[Nat] means (n!)/((n+$1+1)!)<=x.^.($1+1); for k holds (n!)/((n+k+1)!)<=x.^.(k+1) proof A2: now A3: n+1<>0 & n!<>0 by NEWTON:23; (n!)/((n+1)!) = (n!)/((n+1)*(n!)) by NEWTON:21 .= (n!)*((n+1)*(n!))" by XCMPLX_0:def 9 .= (n!)*((n+1)"*(n!)") by XCMPLX_1:205 .= (n!)*(n!)"*(n+1)" by XCMPLX_1:4 .= 1*(n+1)" by A3,XCMPLX_0:def 7 .= x by A1,XCMPLX_1:217; hence P[0] by POWER:30; end; A4: now let k; assume A5: P[k]; A6: n+1>0 & n+(k+1)+1>0 & (n+k+1)!<>0 & (n+k)!<>0 by NAT_1:18,NEWTON:23; then A7: 1/(n+1)>0 by REAL_2:127; k+1>=0 by NAT_1:18; then n+(k+1)>=n+0 by AXIOMS:24; then n+(k+1)+1>=n+1 by AXIOMS:24; then 1/(n+(k+1)+1)<=1/(n+1) by A6,REAL_2:152; then A8: (n+(k+1)+1)"<=1/(n+1) by XCMPLX_1:217; A9: (n!)/((n+k+1)!)>0 by Th34; A10: (n+(k+1)+1)">0 by A6,REAL_1:72; A11: x.^.(k+1)*x = x.^.(k+1)*x.^.1 by POWER:30 .= x.^.((k+1)+1) by A1,A7,POWER:32; (n!)/((n+(k+1)+1)!) = (n!)*((n+(k+1)+1)!)" by XCMPLX_0:def 9 .= (n!)*((n+(k+1)+1)*((n+(k+1))!))" by NEWTON:21 .= (n!)*((n+(k+1)+1)*((n+k+1)!))" by XCMPLX_1:1 .= (n!)*((n+(k+1)+1)"*((n+k+1)!)") by XCMPLX_1:205 .= (n!)*((n+k+1)!)"*(n+(k+1)+1)" by XCMPLX_1:4 .= (n!)/((n+k+1)!)*(n+(k+1)+1)" by XCMPLX_0:def 9; hence P[k+1] by A1,A5,A8,A9,A10,A11,REAL_2:197; end; thus for n holds P[n] from Ind(A2,A4); end; hence thesis; end; theorem Th40: n>0 & x=1/(n+1) implies n!*Sum(eseq^\(n+1))<=x/(1-x) proof assume A1: n>0 & x=1/(n+1); then A2:n+1>0+1 by REAL_1:53; then A3: n+1>0 by AXIOMS:22; then A4:0<1/(n+1) by REAL_2:127; A5: 0<x & x<1 by A1,A2,A3,REAL_2:127,SQUARE_1:2; deffunc F(Nat)=x.^.($1+1); consider seq being Real_Sequence such that A6: for k holds seq.k=F(k) from ExRealSeq; A7: seq.0 = x.^.(0+1) by A6 .= x by POWER:30; A8: abs x=x by A1,A4,ABSVALUE:def 1; A9: abs x<1 by A5,ABSVALUE:def 1; A10: now let k; thus seq.(k+1) = x.^.(k+1+1) by A6 .= x.^.1*x.^.(k+1) by A1,A4,POWER:32 .= x*x.^.(k+1) by POWER:30 .= x*seq.k by A6; end; then A11: seq is summable & Sum(seq)=seq.0/(1-x) by A5,A8,SERIES_1:29; A12: Sum(seq)=x/(1-x) by A7,A9,A10,SERIES_1:29; A13: eseq^\(n+1) is summable by Th24,SERIES_1:15; now let k; A14: (n!(#)(eseq^\(n+1))).k = n!*((eseq^\(n+1)).k) by SEQ_1:13 .= n!*eseq.(n+1+k) by SEQM_3:def 9 .= n!*eseq.(n+k+1) by XCMPLX_1:1 .= n!*(1/((n+k+1)!)) by Def5 .= n!/((n+k+1)!) by XCMPLX_1:100; hence (n!(#)(eseq^\(n+1))).k>=0 by Th34; seq.k=x.^.(k+1) by A6; hence (n!(#)(eseq^\(n+1))).k<=seq.k by A1,A14,Th39; end; then Sum(n!(#)(eseq^\(n+1)))<=Sum(seq) by A11,SERIES_1:24; hence n!*Sum(eseq^\(n+1))<=x/(1-x) by A12,A13,SERIES_1:13; end; theorem Th41: for n be real number holds n>=2 & x=1/(n+1) implies x/(1-x)<1 proof let n be real number; A1: n+1 > n by REAL_2:174; assume A2: n>=2 & x=1/(n+1); then n>=0 by AXIOMS:22; then A3: x>0 by A1,A2,REAL_2:127; 2<n+1 by A1,A2,AXIOMS:22; then 2/(n+1)<1 by REAL_2:142; then 2*x<1 by A2,XCMPLX_1:100; then x+x<1 by XCMPLX_1:11; then x<1-x by REAL_1:86; hence x/(1-x)<1 by A3,REAL_2:142; end; theorem number_e is irrational proof assume number_e is rational; then consider n such that A1: n>=2 & n!*number_e is integer by Th32; A2: n!*number_e = n!*((Partial_Sums(eseq).n)+Sum(eseq^\(n+1))) by Def6,Th24,SERIES_1:18 .= n!*(Partial_Sums(eseq).n)+n!*Sum(eseq^\(n+1)) by XCMPLX_1:8; reconsider ne = n!*number_e as Integer by A1; reconsider ne1 = n!*Partial_Sums(eseq).n as Integer by Th38; A3: n!*Sum(eseq^\(n+1))=ne-ne1 by A2,XCMPLX_1:26; set x = 1/(n+1); A4: x/(1-x)<1 by A1,Th41; n>0 by A1,AXIOMS:22; then n!*Sum(eseq^\(n+1))<=x/(1-x) by Th40; then n!*Sum(eseq^\(n+1))<0+1 by A4,AXIOMS:22; then n!*Sum(eseq^\(n+1))<=0 by A3,INT_1:20; hence contradiction by Th36; end;

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