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Theorem List for Metamath Proof Explorer - 12501-12600   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremrpnnen2lem7 12501* Lemma for rpnnen2 12506. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
 |-  F  =  ( x  e.  ~P NN  |->  ( n  e.  NN  |->  if ( n  e.  x ,  ( ( 1  / 
 3 ) ^ n ) ,  0 )
 ) )   =>    |-  ( ( A  C_  B  /\  B  C_  NN  /\  M  e.  NN )  -> 
 sum_ k  e.  ( ZZ>=
 `  M ) ( ( F `  A ) `  k )  <_  sum_ k  e.  ( ZZ>= `  M ) ( ( F `  B ) `
  k ) )
 
Theoremrpnnen2lem8 12502* Lemma for rpnnen2 12506. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
 |-  F  =  ( x  e.  ~P NN  |->  ( n  e.  NN  |->  if ( n  e.  x ,  ( ( 1  / 
 3 ) ^ n ) ,  0 )
 ) )   =>    |-  ( ( A  C_  NN  /\  M  e.  NN )  ->  sum_ k  e.  NN  ( ( F `  A ) `  k
 )  =  ( sum_ k  e.  ( 1 ... ( M  -  1
 ) ) ( ( F `  A ) `
  k )  +  sum_
 k  e.  ( ZZ>= `  M ) ( ( F `  A ) `
  k ) ) )
 
Theoremrpnnen2lem9 12503* Lemma for rpnnen2 12506. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
 |-  F  =  ( x  e.  ~P NN  |->  ( n  e.  NN  |->  if ( n  e.  x ,  ( ( 1  / 
 3 ) ^ n ) ,  0 )
 ) )   =>    |-  ( M  e.  NN  -> 
 sum_ k  e.  ( ZZ>=
 `  M ) ( ( F `  ( NN  \  { M }
 ) ) `  k
 )  =  ( 0  +  ( ( ( 1  /  3 ) ^ ( M  +  1 ) )  /  ( 1  -  (
 1  /  3 )
 ) ) ) )
 
Theoremrpnnen2lem10 12504* Lemma for rpnnen2 12506. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
 |-  F  =  ( x  e.  ~P NN  |->  ( n  e.  NN  |->  if ( n  e.  x ,  ( ( 1  / 
 3 ) ^ n ) ,  0 )
 ) )   &    |-  ( ph  ->  A 
 C_  NN )   &    |-  ( ph  ->  B 
 C_  NN )   &    |-  ( ph  ->  m  e.  ( A  \  B ) )   &    |-  ( ph  ->  A. n  e.  NN  ( n  <  m  ->  ( n  e.  A  <->  n  e.  B ) ) )   &    |-  ( ps  <->  sum_ k  e.  NN  ( ( F `  A ) `  k
 )  =  sum_ k  e.  NN  ( ( F `
  B ) `  k ) )   =>    |-  ( ( ph  /\ 
 ps )  ->  sum_ k  e.  ( ZZ>= `  m )
 ( ( F `  A ) `  k
 )  =  sum_ k  e.  ( ZZ>= `  m )
 ( ( F `  B ) `  k
 ) )
 
Theoremrpnnen2lem11 12505* Lemma for rpnnen2 12506. (Contributed by Mario Carneiro, 13-May-2013.)
 |-  F  =  ( x  e.  ~P NN  |->  ( n  e.  NN  |->  if ( n  e.  x ,  ( ( 1  / 
 3 ) ^ n ) ,  0 )
 ) )   &    |-  ( ph  ->  A 
 C_  NN )   &    |-  ( ph  ->  B 
 C_  NN )   &    |-  ( ph  ->  m  e.  ( A  \  B ) )   &    |-  ( ph  ->  A. n  e.  NN  ( n  <  m  ->  ( n  e.  A  <->  n  e.  B ) ) )   &    |-  ( ps  <->  sum_ k  e.  NN  ( ( F `  A ) `  k
 )  =  sum_ k  e.  NN  ( ( F `
  B ) `  k ) )   =>    |-  ( ph  ->  -. 
 ps )
 
Theoremrpnnen2 12506* The other half of rpnnen 12507, where we show an injection from sets of natural numbers to real numbers. The obvious choice for this is binary expansion, but it has the unfortunate property that it does not produce an injection on numbers which end with all 0's or all 1's (the more well-known decimal version of this is 0.999... 12339). Instead, we opt for a ternary expansion, which produces (a scaled version of) the Cantor set. Since the Cantor set is riddled with gaps, we can show that any two sequences that are not equal must differ somewhere, and when they do, they are placed a finite distance apart, thus ensuring that the map is injective.

Our map assigns to each subset  A of the natural numbers the number  sum_ k  e.  A ( 3 ^
-u k )  = 
sum_ k  e.  NN ( ( F `  A ) `  k
), where  ( ( F `  A ) `  k )  =  if ( k  e.  A ,  ( 3 ^
-u k ) ,  0 ) ) (rpnnen2lem1 12495). This is an infinite sum of real numbers (rpnnen2lem2 12496), and since  A 
C_  B implies  ( F `  A )  <_  ( F `  B ) (rpnnen2lem4 12498) and  ( F `  NN ) converges to  1  /  2 (rpnnen2lem3 12497) by geoisum1 12337, the sum is convergent to some real (rpnnen2lem5 12499 and rpnnen2lem6 12500) by the comparison test for convergence cvgcmp 12276. The comparison test also tells us that  A  C_  B implies  sum_ ( F `  A )  <_ 
sum_ ( F `  B ) (rpnnen2lem7 12501).

Putting it all together, if we have two sets  x  =/=  y, there must differ somewhere, and so there must be an  m such that  A. n  < 
m ( n  e.  x  <->  n  e.  y
) but  m  e.  ( x  \  y ) or vice versa. In this case, we split off the first  m  -  1 terms (rpnnen2lem8 12502) and cancel them (rpnnen2lem10 12504), since these are the same for both sets. For the remaining terms, we use the subset property to establish that  sum_ ( F `
 y )  <_  sum_ ( F `  ( NN  \  { m }
) ) and  sum_ ( F `
 { m }
)  <_  sum_ ( F `
 x ) (where these sums are only over  ( ZZ>= `  m
)), and since  sum_ ( F `
 ( NN  \  { m } ) )  =  ( 3 ^ -u m )  /  2 (rpnnen2lem9 12503) and  sum_ ( F `  { m } )  =  ( 3 ^
-u m ), we establish that  sum_ ( F `
 y )  <  sum_ ( F `  x
) (rpnnen2lem11 12505) so that they must be different. By contraposition, we find that this map is an injection. (Contributed by Mario Carneiro, 13-May-2013.) (Proof shortened by Mario Carneiro, 30-Apr-2014.)

 |-  F  =  ( x  e.  ~P NN  |->  ( n  e.  NN  |->  if ( n  e.  x ,  ( ( 1  / 
 3 ) ^ n ) ,  0 )
 ) )   =>    |- 
 ~P NN  ~<_  ( 0 [,] 1 )
 
Theoremrpnnen 12507 The cardinality of the continuum is the same as the powerset of  om. This is a stronger statement than ruc 12523, which only asserts that  RR is uncountable, i.e. has a cardinality larger than  om. The main proof is in two parts, rpnnen1 10349 and rpnnen2 12506, each showing an injection in one direction, and this last part uses sbth 6983 to prove that the sets are equinumerous. By constructing explicit injections, we avoid the use of AC. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 23-Aug-2014.)
 |- 
 RR  ~~  ~P NN
 
Theoremrexpen 12508 The real numbers are equinumerous to their own cross product, even though it is not necessarily true that  RR is well-orderable (so we cannot use infxpidm2 7646 directly). (Contributed by NM, 30-Jul-2004.) (Revised by Mario Carneiro, 16-Jun-2013.)
 |-  ( RR  X.  RR )  ~~  RR
 
Theoremcpnnen 12509 The complex numbers are equinumerous to the powerset of the natural numbers. (Contributed by Mario Carneiro, 16-Jun-2013.)
 |- 
 CC  ~~  ~P NN
 
TheoremrucALT 12510 The set of natural numbers is strictly dominated by the set of real numbers, i.e. the real numbers are uncountable. This proof is a simple corollary of rpnnen 12507, which determines the exact cardinality of the reals. For an alternate proof discussed at http://us.metamath.org/mpeuni/mmcomplex.html#uncountable, see ruc 12523. (Contributed by NM, 13-Oct-2004.) (Revised by Mario Carneiro, 13-May-2013.) (Proof modification is discouraged.) (New usage is discouraged.)
 |- 
 NN  ~<  RR
 
Theoremruclem1 12511* Lemma for ruc 12523 (the reals are uncountable). Substitutions for the function  D. (Contributed by Mario Carneiro, 28-May-2014.) (Revised by Fan Zheng, 6-Jun-2016.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  ( ph  ->  A  e.  RR )   &    |-  ( ph  ->  B  e.  RR )   &    |-  ( ph  ->  M  e.  RR )   &    |-  X  =  ( 1st `  ( <. A ,  B >. D M ) )   &    |-  Y  =  ( 2nd `  ( <. A ,  B >. D M ) )   =>    |-  ( ph  ->  ( ( <. A ,  B >. D M )  e.  ( RR  X.  RR )  /\  X  =  if (
 ( ( A  +  B )  /  2
 )  <  M ,  A ,  ( (
 ( ( A  +  B )  /  2
 )  +  B ) 
 /  2 ) ) 
 /\  Y  =  if ( ( ( A  +  B )  / 
 2 )  <  M ,  ( ( A  +  B )  /  2
 ) ,  B ) ) )
 
Theoremruclem2 12512* Lemma for ruc 12523. Ordering property for the input to  D. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  ( ph  ->  A  e.  RR )   &    |-  ( ph  ->  B  e.  RR )   &    |-  ( ph  ->  M  e.  RR )   &    |-  X  =  ( 1st `  ( <. A ,  B >. D M ) )   &    |-  Y  =  ( 2nd `  ( <. A ,  B >. D M ) )   &    |-  ( ph  ->  A  <  B )   =>    |-  ( ph  ->  ( A  <_  X  /\  X  <  Y  /\  Y  <_  B ) )
 
Theoremruclem3 12513* Lemma for ruc 12523. The constructed interval  [ X ,  Y ] always excludes  M. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  ( ph  ->  A  e.  RR )   &    |-  ( ph  ->  B  e.  RR )   &    |-  ( ph  ->  M  e.  RR )   &    |-  X  =  ( 1st `  ( <. A ,  B >. D M ) )   &    |-  Y  =  ( 2nd `  ( <. A ,  B >. D M ) )   &    |-  ( ph  ->  A  <  B )   =>    |-  ( ph  ->  ( M  <  X  \/  Y  <  M ) )
 
Theoremruclem4 12514* Lemma for ruc 12523. Initial value of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  C  =  ( { <. 0 ,  <. 0 ,  1 >. >. }  u.  F )   &    |-  G  =  seq  0 ( D ,  C )   =>    |-  ( ph  ->  ( G `  0 )  = 
 <. 0 ,  1 >.
 )
 
Theoremruclem6 12515* Lemma for ruc 12523. Domain and range of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  C  =  ( { <. 0 ,  <. 0 ,  1 >. >. }  u.  F )   &    |-  G  =  seq  0 ( D ,  C )   =>    |-  ( ph  ->  G : NN0 --> ( RR  X.  RR ) )
 
Theoremruclem7 12516* Lemma for ruc 12523. Successor value for the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  C  =  ( { <. 0 ,  <. 0 ,  1 >. >. }  u.  F )   &    |-  G  =  seq  0 ( D ,  C )   =>    |-  ( ( ph  /\  N  e.  NN0 )  ->  ( G `  ( N  +  1 ) )  =  ( ( G `  N ) D ( F `  ( N  +  1 ) ) ) )
 
Theoremruclem8 12517* Lemma for ruc 12523. The intervals of the  G sequence are all nonempty. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  C  =  ( { <. 0 ,  <. 0 ,  1 >. >. }  u.  F )   &    |-  G  =  seq  0 ( D ,  C )   =>    |-  ( ( ph  /\  N  e.  NN0 )  ->  ( 1st `  ( G `  N ) )  < 
 ( 2nd `  ( G `  N ) ) )
 
Theoremruclem9 12518* Lemma for ruc 12523. The first components of the  G sequence are increasing, and the second components are decreasing. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  C  =  ( { <. 0 ,  <. 0 ,  1 >. >. }  u.  F )   &    |-  G  =  seq  0 ( D ,  C )   &    |-  ( ph  ->  M  e.  NN0 )   &    |-  ( ph  ->  N  e.  ( ZZ>= `  M ) )   =>    |-  ( ph  ->  (
 ( 1st `  ( G `  M ) )  <_  ( 1st `  ( G `  N ) )  /\  ( 2nd `  ( G `  N ) )  <_  ( 2nd `  ( G `  M ) ) ) )
 
Theoremruclem10 12519* Lemma for ruc 12523. Every first component of the  G sequence is less than every second component. That is, the sequences form a chain a1  < a2 
<...  < b2  < b1, where ai are the first components and bi are the second components. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  C  =  ( { <. 0 ,  <. 0 ,  1 >. >. }  u.  F )   &    |-  G  =  seq  0 ( D ,  C )   &    |-  ( ph  ->  M  e.  NN0 )   &    |-  ( ph  ->  N  e.  NN0 )   =>    |-  ( ph  ->  ( 1st `  ( G `  M ) )  < 
 ( 2nd `  ( G `  N ) ) )
 
Theoremruclem11 12520* Lemma for ruc 12523. Closure lemmas for supremum. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  C  =  ( { <. 0 ,  <. 0 ,  1 >. >. }  u.  F )   &    |-  G  =  seq  0 ( D ,  C )   =>    |-  ( ph  ->  ( ran  ( 1st  o.  G )  C_  RR  /\  ran  ( 1st  o.  G )  =/=  (/)  /\  A. z  e. 
 ran  ( 1st  o.  G ) z  <_ 
 1 ) )
 
Theoremruclem12 12521* Lemma for ruc 12523. The supremum of the increasing sequence  1st  o.  G is a real number that is not in the range of  F. (Contributed by Mario Carneiro, 28-May-2014.)
 |-  ( ph  ->  F : NN --> RR )   &    |-  ( ph  ->  D  =  ( x  e.  ( RR 
 X.  RR ) ,  y  e.  RR  |->  [_ ( ( ( 1st `  x )  +  ( 2nd `  x ) )  /  2
 )  /  m ]_ if ( m  <  y , 
 <. ( 1st `  x ) ,  m >. , 
 <. ( ( m  +  ( 2nd `  x )
 )  /  2 ) ,  ( 2nd `  x ) >. ) ) )   &    |-  C  =  ( { <. 0 ,  <. 0 ,  1 >. >. }  u.  F )   &    |-  G  =  seq  0 ( D ,  C )   &    |-  S  =  sup ( ran  ( 1st  o.  G ) ,  RR ,  <  )   =>    |-  ( ph  ->  S  e.  ( RR  \  ran  F ) )
 
Theoremruclem13 12522 Lemma for ruc 12523. There is no function that maps  NN onto  RR. (Use nex 1544 if you want this in the form  -.  E. f
f : NN -onto-> RR.) (Contributed by NM, 14-Oct-2004.) (Proof shortened by Fan Zheng, 6-Jun-2016.)
 |- 
 -.  F : NN -onto-> RR
 
Theoremruc 12523 The set of natural numbers is strictly dominated by the set of real numbers, i.e. the real numbers are uncountable. The proof consists of lemmas ruclem1 12511 through ruclem13 12522 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 12522 for the function existence version of this theorem. For an informal discussion of this proof, see http://us.metamath.org/mpeuni/mmcomplex.html#uncountable. For an alternate proof see rucALT 12510. (Contributed by NM, 13-Oct-2004.)
 |- 
 NN  ~<  RR
 
Theoremresdomq 12524 The set of rationals is strictly less equinumerous than the set of reals ( RR strictly dominates  QQ). (Contributed by NM, 18-Dec-2004.)
 |- 
 QQ  ~<  RR
 
Theoremaleph1re 12525 There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.)
 |-  ( aleph `  1o )  ~<_  RR
 
Theoremaleph1irr 12526 There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.)
 |-  ( aleph `  1o )  ~<_  ( RR  \  QQ )
 
Theoremcnso 12527 The complex numbers can be linearly ordered. (Contributed by Stefan O'Rear, 16-Nov-2014.)
 |- 
 E. x  x  Or  CC
 
PART 6  ELEMENTARY NUMBER THEORY

Here we introduce elementary number theory, in particular the elementary properties of divisibility and elementary prime number theory.

 
6.1  Elementary properties of divisibility
 
6.1.1  Irrationality of square root of 2
 
Theoremsqr2irrlem 12528 Lemma for irrationality of square root of 2. The core of the proof - if  A  /  B  =  sqr ( 2 ), then 
A and  B are even, so  A  /  2 and  B  /  2 are smaller representatives, which is absurd by the method of infinite descent (here implemented by strong induction). (Contributed by NM, 20-Aug-2001.) (Revised by Mario Carneiro, 12-Sep-2015.)
 |-  ( ph  ->  A  e.  ZZ )   &    |-  ( ph  ->  B  e.  NN )   &    |-  ( ph  ->  ( sqr `  2
 )  =  ( A 
 /  B ) )   =>    |-  ( ph  ->  ( ( A  /  2 )  e. 
 ZZ  /\  ( B  /  2 )  e.  NN ) )
 
Theoremsqr2irr 12529 The square root of 2 is irrational. See zsqrelqelz 12831 for a generalization to all non-square integers. The proof's core is proven in sqr2irrlem 12528, which shows that if  A  /  B  =  sqr ( 2 ), then  A and  B are even, so  A  /  2 and  B  /  2 are smaller representatives, which is absurd. An older version of this proof was included in The Seventeen Provers of the World compiled by Freek Wiedijk. It is also the first "top 100" mathematical theorems whose formalization is tracked by Freek Wiedijk on his Formalizing 100 Theorems page at http://www.cs.ru.nl/~freek/100/. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
 |-  ( sqr `  2
 )  e/  QQ
 
Theoremsqr2re 12530 The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.)
 |-  ( sqr `  2
 )  e.  RR
 
6.1.2  Some Number sets are chains of proper subsets
 
Theoremnthruc 12531 The sequence  NN,  ZZ,  QQ,  RR, and  CC forms a chain of proper subsets. In each case the proper subset relationship is shown by demonstrating a number that belongs to one set but not the other. We show that zero belongs to  ZZ but not  NN, one-half belongs to  QQ but not  ZZ, the square root of 2 belongs to  RR but not  QQ, and finally that the imaginary number  _i belongs to  CC but not  RR. See nthruz 12532 for a further refinement. (Contributed by NM, 12-Jan-2002.)
 |-  ( ( NN  C.  ZZ  /\  ZZ  C.  QQ )  /\  ( QQ  C.  RR  /\  RR  C.  CC ) )
 
Theoremnthruz 12532 The sequence  NN,  NN0, and  ZZ forms a chain of proper subsets. In each case the proper subset relationship is shown by demonstrating a number that belongs to one set but not the other. We show that zero belongs to 
NN0 but not  NN and minus one belongs to  ZZ but not  NN0. This theorem refines the chain of proper subsets nthruc 12531. (Contributed by NM, 9-May-2004.)
 |-  ( NN  C.  NN0  /\ 
 NN0  C.  ZZ )
 
6.1.3  The divides relation
 
Syntaxcdivides 12533 Extend the definition of a class to include the divides relation. See df-dvds 12534.
 class  ||
 
Definitiondf-dvds 12534* Define the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ||  =  { <. x ,  y >.  |  ( ( x  e.  ZZ  /\  y  e.  ZZ )  /\  E. n  e.  ZZ  ( n  x.  x )  =  y ) }
 
Theoremdivides 12535* Define the divides relation.  M  ||  N means  M divides into  N with no remainder. For example,  3  ||  6 (ex-dvds 20837). As proven in dvdsval3 12537, 
M  ||  N  <->  ( N  mod  M )  =  0. See divides 12535 and dvdsval2 12536 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <-> 
 E. n  e.  ZZ  ( n  x.  M )  =  N )
 )
 
Theoremdvdsval2 12536 One nonzero integer divides another integer if and only if their quotient is an integer. (Contributed by Jeff Hankins, 29-Sep-2013.)
 |-  ( ( M  e.  ZZ  /\  M  =/=  0  /\  N  e.  ZZ )  ->  ( M  ||  N  <->  ( N  /  M )  e.  ZZ ) )
 
Theoremdvdsval3 12537 One nonzero integer divides another integer if and only if the remainder upon division is zero. (Contributed by Mario Carneiro, 22-Feb-2014.) (Revised by Mario Carneiro, 15-Jul-2014.)
 |-  ( ( M  e.  NN  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <->  ( N  mod  M )  =  0 )
 )
 
Theoremdvdszrcl 12538 Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.)
 |-  ( X  ||  Y  ->  ( X  e.  ZZ  /\  Y  e.  ZZ )
 )
 
Theoremnndivdvds 12539 Strong form of dvdsval2 12536 for natural numbers. (Contributed by Stefan O'Rear, 13-Sep-2014.)
 |-  ( ( A  e.  NN  /\  B  e.  NN )  ->  ( B  ||  A 
 <->  ( A  /  B )  e.  NN )
 )
 
Theoremmoddvds 12540 Two ways to say  A  ==  B
(  mod  N ). (Contributed by Mario Carneiro, 18-Feb-2014.)
 |-  ( ( N  e.  NN  /\  A  e.  ZZ  /\  B  e.  ZZ )  ->  ( ( A  mod  N )  =  ( B 
 mod  N )  <->  N  ||  ( A  -  B ) ) )
 
Theoremdvds0lem 12541 A lemma to assist theorems of 
|| with no antecedents. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  /\  ( K  x.  M )  =  N )  ->  M  ||  N )
 
Theoremdvds1lem 12542* A lemma to assist theorems of 
|| with one antecedent. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ph  ->  ( J  e.  ZZ  /\  K  e.  ZZ ) )   &    |-  ( ph  ->  ( M  e.  ZZ  /\  N  e.  ZZ ) )   &    |-  ( ( ph  /\  x  e.  ZZ )  ->  Z  e.  ZZ )   &    |-  (
 ( ph  /\  x  e. 
 ZZ )  ->  (
 ( x  x.  J )  =  K  ->  ( Z  x.  M )  =  N ) )   =>    |-  ( ph  ->  ( J  ||  K  ->  M  ||  N ) )
 
Theoremdvds2lem 12543* A lemma to assist theorems of 
|| with two antecedents. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ph  ->  ( I  e.  ZZ  /\  J  e.  ZZ ) )   &    |-  ( ph  ->  ( K  e.  ZZ  /\  L  e.  ZZ ) )   &    |-  ( ph  ->  ( M  e.  ZZ  /\  N  e.  ZZ )
 )   &    |-  ( ( ph  /\  ( x  e.  ZZ  /\  y  e.  ZZ ) )  ->  Z  e.  ZZ )   &    |-  (
 ( ph  /\  ( x  e.  ZZ  /\  y  e.  ZZ ) )  ->  ( ( ( x  x.  I )  =  J  /\  ( y  x.  K )  =  L )  ->  ( Z  x.  M )  =  N ) )   =>    |-  ( ph  ->  ( ( I  ||  J  /\  K  ||  L )  ->  M  ||  N )
 )
 
Theoremiddvds 12544 An integer divides itself. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( N  e.  ZZ  ->  N  ||  N )
 
Theorem1dvds 12545 1 divides any integer. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( N  e.  ZZ  ->  1  ||  N )
 
Theoremdvds0 12546 Any integer divides 0. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( N  e.  ZZ  ->  N  ||  0 )
 
Theoremnegdvdsb 12547 An integer divides another iff its negation does. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <->  -u M  ||  N ) )
 
Theoremdvdsnegb 12548 An integer divides another iff it divides its negation. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <->  M  ||  -u N ) )
 
Theoremabsdvdsb 12549 An integer divides another iff its absolute value does. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <->  ( abs `  M )  ||  N ) )
 
Theoremdvdsabsb 12550 An integer divides another iff it divides its absolute value. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <->  M  ||  ( abs `  N ) ) )
 
Theorem0dvds 12551 Only 0 is divisible by 0 . (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( N  e.  ZZ  ->  ( 0  ||  N  <->  N  =  0 ) )
 
Theoremdvdsmul1 12552 An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  M  ||  ( M  x.  N ) )
 
Theoremdvdsmul2 12553 An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  N  ||  ( M  x.  N ) )
 
Theoremiddvdsexp 12554 An integer divides a positive integer power of itself. (Contributed by Paul Chapman, 26-Oct-2012.)
 |-  ( ( M  e.  ZZ  /\  N  e.  NN )  ->  M  ||  ( M ^ N ) )
 
Theoremmuldvds1 12555 If a product divides an integer, so does one of its factors. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  ->  ( ( K  x.  M )  ||  N  ->  K 
 ||  N ) )
 
Theoremmuldvds2 12556 If a product divides an integer, so does one of its factors. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  ->  ( ( K  x.  M )  ||  N  ->  M 
 ||  N ) )
 
Theoremdvdscmul 12557 Multiplication by a constant maintains the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ  /\  K  e.  ZZ )  ->  ( M  ||  N  ->  ( K  x.  M )  ||  ( K  x.  N ) ) )
 
Theoremdvdsmulc 12558 Multiplication by a constant maintains the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ  /\  K  e.  ZZ )  ->  ( M  ||  N  ->  ( M  x.  K )  ||  ( N  x.  K ) ) )
 
Theoremdvdscmulr 12559 Cancellation law for the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ  /\  ( K  e.  ZZ  /\  K  =/=  0 ) )  ->  ( ( K  x.  M )  ||  ( K  x.  N ) 
 <->  M  ||  N )
 )
 
Theoremdvdsmulcr 12560 Cancellation law for the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ  /\  ( K  e.  ZZ  /\  K  =/=  0 ) )  ->  ( ( M  x.  K )  ||  ( N  x.  K ) 
 <->  M  ||  N )
 )
 
Theoremdvds2ln 12561 If an integer divides each of two other integers, it divides any linear combination of them. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( ( I  e.  ZZ  /\  J  e.  ZZ )  /\  ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ ) )  ->  (
 ( K  ||  M  /\  K  ||  N )  ->  K  ||  ( ( I  x.  M )  +  ( J  x.  N ) ) ) )
 
Theoremdvds2add 12562 If an integer divides each of two other integers, it divides their sum. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  ->  ( ( K  ||  M  /\  K  ||  N )  ->  K  ||  ( M  +  N )
 ) )
 
Theoremdvds2sub 12563 If an integer divides each of two other integers, it divides their difference. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  ->  ( ( K  ||  M  /\  K  ||  N )  ->  K  ||  ( M  -  N ) ) )
 
Theoremdvdstr 12564 The divides relation is transitive. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  ->  ( ( K  ||  M  /\  M  ||  N )  ->  K  ||  N ) )
 
Theoremdvdsmultr1 12565 If an integer divides another, it divides a multiple of it. (Contributed by Paul Chapman, 17-Nov-2012.)
 |-  ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  ->  ( K  ||  M  ->  K  ||  ( M  x.  N ) ) )
 
Theoremdvdsmultr2 12566 If an integer divides another, it divides a multiple of it. (Contributed by Paul Chapman, 17-Nov-2012.)
 |-  ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  ->  ( K  ||  N  ->  K  ||  ( M  x.  N ) ) )
 
Theoremordvdsmul 12567 If an integer divides either of two others, it divides their product. (Contributed by Paul Chapman, 17-Nov-2012.) (Proof shortened by Mario Carneiro, 17-Jul-2014.)
 |-  ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  ->  ( ( K  ||  M  \/  K  ||  N )  ->  K  ||  ( M  x.  N ) ) )
 
Theoremdvdssub2 12568 If an integer divides a difference, then it divides one term iff it divides the other. (Contributed by Mario Carneiro, 13-Jul-2014.)
 |-  ( ( ( K  e.  ZZ  /\  M  e.  ZZ  /\  N  e.  ZZ )  /\  K  ||  ( M  -  N ) )  ->  ( K 
 ||  M  <->  K  ||  N ) )
 
Theoremdvdsadd 12569 An integer divides another iff it divides their sum. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 13-Jul-2014.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <->  M  ||  ( M  +  N ) ) )
 
Theoremdvdsaddr 12570 An integer divides another iff it divides their sum. (Contributed by Paul Chapman, 31-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <->  M  ||  ( N  +  M ) ) )
 
Theoremdvdssub 12571 An integer divides another iff it divides their difference. (Contributed by Paul Chapman, 31-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <->  M  ||  ( M  -  N ) ) )
 
Theoremdvdssubr 12572 An integer divides another iff it divides their difference. (Contributed by Paul Chapman, 31-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ )  ->  ( M  ||  N 
 <->  M  ||  ( N  -  M ) ) )
 
Theoremdvdsadd2b 12573 Adding a multiple of the base does not affect divisibility. (Contributed by Stefan O'Rear, 23-Sep-2014.)
 |-  ( ( A  e.  ZZ  /\  B  e.  ZZ  /\  ( C  e.  ZZ  /\  A  ||  C )
 )  ->  ( A  ||  B  <->  A  ||  ( C  +  B ) ) )
 
Theoremfsumdvds 12574* If every term in a sum is divisible by  N, then so is the sum. (Contributed by Mario Carneiro, 17-Jan-2015.)
 |-  ( ph  ->  A  e.  Fin )   &    |-  ( ph  ->  N  e.  ZZ )   &    |-  (
 ( ph  /\  k  e.  A )  ->  B  e.  ZZ )   &    |-  ( ( ph  /\  k  e.  A ) 
 ->  N  ||  B )   =>    |-  ( ph  ->  N  ||  sum_ k  e.  A  B )
 
Theoremdvdslelem 12575 Lemma for dvdsle 12576. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  M  e.  ZZ   &    |-  N  e.  NN   &    |-  K  e.  ZZ   =>    |-  ( N  <  M  ->  ( K  x.  M )  =/= 
 N )
 
Theoremdvdsle 12576 The divisors of a positive integer are bounded by it. The proof does not use  /. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( ( M  e.  ZZ  /\  N  e.  NN )  ->  ( M  ||  N  ->  M  <_  N ) )
 
Theoremdvdsleabs 12577 The divisors of a nonzero integer are bounded by its absolute value. (Contributed by Paul Chapman, 21-Mar-2011.) (Proof shortened by Fan Zheng, 3-Jul-2016.)
 |-  ( ( M  e.  ZZ  /\  N  e.  ZZ  /\  N  =/=  0 ) 
 ->  ( M  ||  N  ->  M  <_  ( abs `  N ) ) )
 
Theoremdvdseq 12578 If two integers divide each other, they must be equal, up to a difference in sign. (Contributed by Mario Carneiro, 30-May-2014.)
 |-  ( ( ( M  e.  NN0  /\  N  e.  NN0 )  /\  ( M 
 ||  N  /\  N  ||  M ) )  ->  M  =  N )
 
Theoremdvds1 12579 The only nonnegative integer that divides 1 is 1. (Contributed by Mario Carneiro, 2-Jul-2015.)
 |-  ( M  e.  NN0  ->  ( M  ||  1  <->  M  =  1
 ) )
 
Theoremalzdvds 12580* Only 0 is divisible by all integers. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  ( N  e.  ZZ  ->  ( A. x  e. 
 ZZ  x  ||  N  <->  N  =  0 ) )
 
Theoremdvdsext 12581* Poset extensionality for division. (Contributed by Stefan O'Rear, 6-Sep-2015.)
 |-  ( ( A  e.  NN0  /\  B  e.  NN0 )  ->  ( A  =  B  <->  A. x  e.  NN0  ( A  ||  x  <->  B  ||  x ) ) )
 
Theoremfzm1ndvds 12582 No number between  1 and  M  - 
1 divides  M. (Contributed by Mario Carneiro, 24-Jan-2015.)
 |-  ( ( M  e.  NN  /\  N  e.  (
 1 ... ( M  -  1 ) ) ) 
 ->  -.  M  ||  N )
 
Theoremfzo0dvdseq 12583 Zero is the only one of the first 
A nonnegative integers that is divisible by  A. (Contributed by Stefan O'Rear, 6-Sep-2015.)
 |-  ( B  e.  (
 0..^ A )  ->  ( A  ||  B  <->  B  =  0
 ) )
 
Theoremfzocongeq 12584 Two different elements of a half-open range are not congruent mod its length. (Contributed by Stefan O'Rear, 6-Sep-2015.)
 |-  ( ( A  e.  ( C..^ D )  /\  B  e.  ( C..^ D ) )  ->  ( ( D  -  C )  ||  ( A  -  B )  <->  A  =  B ) )
 
Theoremdvdsfac 12585 A positive integer divides any greater factorial. (Contributed by Paul Chapman, 28-Nov-2012.)
 |-  ( ( K  e.  NN  /\  N  e.  ( ZZ>=
 `  K ) ) 
 ->  K  ||  ( ! `  N ) )
 
Theoremdvdsexp 12586 A power divides a power with a greater exponent. (Contributed by Mario Carneiro, 23-Feb-2014.)
 |-  ( ( A  e.  ZZ  /\  M  e.  NN0  /\  N  e.  ( ZZ>= `  M ) )  ->  ( A ^ M ) 
 ||  ( A ^ N ) )
 
Theoremdvdsmod 12587 Any number  K whose mod base  N is divisible by a divisor  P of the base is also divisible by  P. This means that primes will also be relatively prime to the base when reduced  mod  N for any base. (Contributed by Mario Carneiro, 13-Mar-2014.)
 |-  ( ( ( P  e.  NN  /\  N  e.  NN  /\  K  e.  ZZ )  /\  P  ||  N )  ->  ( P 
 ||  ( K  mod  N )  <->  P  ||  K ) )
 
Theoremodd2np1lem 12588* Lemma for odd2np1 12589. (Contributed by Scott Fenton, 3-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
 |-  ( N  e.  NN0  ->  ( E. n  e.  ZZ  ( ( 2  x.  n )  +  1 )  =  N  \/  E. k  e.  ZZ  (
 k  x.  2 )  =  N ) )
 
Theoremodd2np1 12589* An integer is odd iff it is one plus twice another integer. (Contributed by Scott Fenton, 3-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
 |-  ( N  e.  ZZ  ->  ( -.  2  ||  N 
 <-> 
 E. n  e.  ZZ  ( ( 2  x.  n )  +  1 )  =  N ) )
 
Theoremoddm1even 12590 An integer is odd iff its predecessor is even. (Contributed by Mario Carneiro, 5-Sep-2016.)
 |-  ( N  e.  ZZ  ->  ( -.  2  ||  N 
 <->  2  ||  ( N  -  1 ) ) )
 
Theoremoddp1even 12591 An integer is odd iff its successor is even. (Contributed by Mario Carneiro, 5-Sep-2016.)
 |-  ( N  e.  ZZ  ->  ( -.  2  ||  N 
 <->  2  ||  ( N  +  1 ) ) )
 
Theoremoexpneg 12592 The exponential of the negative of a number, when the exponent is odd. (Contributed by Mario Carneiro, 25-Apr-2015.)
 |-  ( ( A  e.  CC  /\  N  e.  NN  /\ 
 -.  2  ||  N )  ->  ( -u A ^ N )  =  -u ( A ^ N ) )
 
Theorem3dvds 12593* A rule for divisibility by 3 of a number written in base 10. (Contributed by Mario Carneiro, 14-Jul-2014.) (Revised by Mario Carneiro, 17-Jan-2015.)
 |-  ( ( N  e.  NN0  /\  F : ( 0
 ... N ) --> ZZ )  ->  ( 3  ||  sum_ k  e.  ( 0 ... N ) ( ( F `
  k )  x.  ( 10 ^ k
 ) )  <->  3  ||  sum_ k  e.  ( 0 ... N ) ( F `  k ) ) )
 
6.1.4  The division algorithm
 
Theoremdivalglem0 12594 Lemma for divalg 12604. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  N  e.  ZZ   &    |-  D  e.  ZZ   =>    |-  ( ( R  e.  ZZ  /\  K  e.  ZZ )  ->  ( D  ||  ( N  -  R )  ->  D  ||  ( N  -  ( R  -  ( K  x.  ( abs `  D ) ) ) ) ) )
 
Theoremdivalglem1 12595 Lemma for divalg 12604. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  N  e.  ZZ   &    |-  D  e.  ZZ   &    |-  D  =/=  0   =>    |-  0  <_  ( N  +  ( abs `  ( N  x.  D ) ) )
 
Theoremdivalglem2 12596* Lemma for divalg 12604. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  N  e.  ZZ   &    |-  D  e.  ZZ   &    |-  D  =/=  0   &    |-  S  =  { r  e.  NN0  |  D  ||  ( N  -  r ) }   =>    |-  sup ( S ,  RR ,  `'  <  )  e.  S
 
Theoremdivalglem4 12597* Lemma for divalg 12604. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  N  e.  ZZ   &    |-  D  e.  ZZ   &    |-  D  =/=  0   &    |-  S  =  { r  e.  NN0  |  D  ||  ( N  -  r ) }   =>    |-  S  =  {
 r  e.  NN0  |  E. q  e.  ZZ  N  =  ( (
 q  x.  D )  +  r ) }
 
Theoremdivalglem5 12598* Lemma for divalg 12604. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  N  e.  ZZ   &    |-  D  e.  ZZ   &    |-  D  =/=  0   &    |-  S  =  { r  e.  NN0  |  D  ||  ( N  -  r ) }   &    |-  R  =  sup ( S ,  RR ,  `'  <  )   =>    |-  (
 0  <_  R  /\  R  <  ( abs `  D ) )
 
Theoremdivalglem6 12599 Lemma for divalg 12604. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  A  e.  NN   &    |-  X  e.  ( 0 ... ( A  -  1 ) )   &    |-  K  e.  ZZ   =>    |-  ( K  =/=  0  ->  -.  ( X  +  ( K  x.  A ) )  e.  (
 0 ... ( A  -  1 ) ) )
 
Theoremdivalglem7 12600 Lemma for divalg 12604. (Contributed by Paul Chapman, 21-Mar-2011.)
 |-  D  e.  ZZ   &    |-  D  =/=  0   =>    |-  ( ( X  e.  ( 0 ... (
 ( abs `  D )  -  1 ) ) 
 /\  K  e.  ZZ )  ->  ( K  =/=  0  ->  -.  ( X  +  ( K  x.  ( abs `  D ) ) )  e.  ( 0
 ... ( ( abs `  D )  -  1
 ) ) ) )
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