Conference rusure::math

    >(4) Is i greater than 0 or less than 0 ? (i is the square root of -1).
    
    this one is easy.
    
    since the class of complex numbers is larger than class of integers
    we can represent both as complex number:
    
    let  a= 0+ I*0
    let  b= 0+ I*1
    
      so   a-b = (0+ I*0) - ( 0 + I*1)
               = 0 - I 
    
      if a-b >= 0+I*0  then 
         a>= b  
      else a < b 
    
      but  a-b < 0 + 0*I
      so   a <b
    
      i.e. 0+ 0*I < 0+ 1*I   
    
      i.e. 0 is less than I 
    
    /naser
     
    
      

>      but  a-b < 0 + 0*I
    
    What is the justification for this line?
    
      John
    
    p.s., This really is a joke, right?

    well john, 
    if iam wrong , then this is a joke, if iam right, then iam very serious!
    
    how do you compare 2 complex numbers? if I use R(length of vector) and 
    Phi (angle) on the complex plan for comparing the 'values', then
    
    R for (0+ 1*I) = 0+1 = 1
    R for (0+ 0*I) = 0
    
    and both angles are equal.
    
    then  0+ 0*I < 0 + 1*I
    
    actually my first reaction was to say that you cant compare apples to
    oranges, but if express both as complex numbers you could.
    
    /naser

        An ordered field is a field in which there is a strict
        linear order < that satisfies
        
        	for all a,b,c: a < b ==> a + c < b + c
        	for all a,b,c: a < b and c positive ==> ac < bc
        
        ("c positive" means 0 < c)
        
        The reals are an ordered field using the usual order. 
        There is no order relation on the complex plane that will
        make the complex numbers into an ordered field.
        
        For example, if 0 < i, then using the second rule above
        with a=0, b=c=i yields 0i < i^2 or 0 < -1.  Using the
        second rule again with a=0, b=c=-1 gives 0 < 1.  But
        using the first rule with a=0,b=-1,c=1 gives 1 < 0.  This
        contradicts that < is a strict linear ordering. 
        Likewise, if i < 0, then adding -i to both sides by the
        first rule gives 0 < -i, and multiplying both sides by -i
        by the second rule again gives 0 < -1 which leads to the
        same contradiction.
        
        So you can invent linear orderings of the complex
        numbers, but not in a way that makes them an ordered
        field with the usual arithmetic operations.
        
        Dan

    ok, cant we invent our own ordering relation on complex numbers that
    will satisfy the above conditions? make our own algebra.
    
     assume length of vector was the order. (ignore angle)
    
     then that takes us back to the real field. 
    
     so a=0 has length 0
        b=I has length 1
    
    so a<b  i.e. 0<I in this agebra.
     
    

>(6) What are the last 4 digits of 5 to the 7777th power ?
>    (YOu are not allowed to use a calculator. Anyone who uses a
>    calculator will be expelled, their reputation tarnished, their
>    future ruined and their children left to fend for themselves
>    in a cold and hostile world.)
    
    No calculators?  Hmmm.  I guess computers are OK then? 
    
    No?  Well then, I would say that the answer was  "3125".
    
    --  Barry

    ref .-1 (Barry)
    
    He Is RIGHT ! 
    
    below is the proof 
    
    this is my calcuations, took me 43 hours and plenty of coffee to compute :-)
    I wonder How many hours did it take you ?
    
    77575345986817035105916821854909000686574093573482823257759620783761750
    45338203402426097968752443686951907337901773428893988313694611944602011
    81663460084338132578291386091259472280222217103195299791412438868155591
    08254636447252475290124153970444621379125204280090556574030047515119001
    89799970847099490736990370155655726921359883288044357560195198891994572
    11038165339686247774160774643493704422324621366849330101473417449103368
    01662777367030229330626847843415401670592730198618034589067672394436878
    70748380935320941047794404087793703184074972527540962356127847576179466
    13895453600834067692732725779815636343869416117693596340524674810397396
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    41609383632236488641361075301092059931880305384058137864079117211842710
    37265059478350840819538310156651510756030401524255776364072545142463115
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    39266334102586994333657230207450840957185083262354317661895912370669555
    91902983019886169214612465751343574848165414536478096766933068808744404
    45703729717295154552887060732749412914425753074325462388795143436935926
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    14425753074325462388795143436935926414967734674126485916182237872216775
    74379193271867347451976010151175516663236812827904225116466671845779606
    61489321801119881144806571225919955856183619075544056479058065505586678
    27325568777103699193956697843685632680703635939279917934930319975601140
    14284033578042646187483494942373784048746930426833774294704448527664800
    56413830164847005676513995937061201492404818455574450688675827484816942
    64563112305226766172027790062611843780508132227973295801559398200158511
    62274602717667458268706209404067145965592231711200429636084958091080815
    56116782451767384948206617143017690099098759324086313075995341458003926
    29470260967487803257401159157623432646552603384578605812485879289228920
    02444542487344758329238525537611507850103333849636139412947214657829120
    49915417544224808028344099967080485938142442234828211976925770759013577
    08142064047423490975134545128954988325127556112674517938471995061968576
    72196684351274201893868597800223586152812416116830718237906694412231445
    3125
    ^^^^
    

    re. 1431.7

    I'm lost for words!

    After a few jottings with pencil and paper (cue old joke about
    constipated mathematician) form the hypothesis that -

    5 ** (4n+1) ends in 3125

    n=1
    ---

    5 ** 5 = 3125

    n=k+1
    -----

    Assume true for n=k. Then

    5 ** (4k+5) = 5 ** (4k+1) * 625

    5 ** (4k+1) = m*10000 + 3125 for some integer m

    so 5 ** (4k+5) = 3125 * 625 + 625m*10000

                   = 1953125 + 625m*10000

    therefore hypothesis is true for k+1.
    Therefore hypothesis is true for all integer n > 0.

    7777 = 4*1944 + 1

    Therefore 5 ** 7777 ends in 3125.

    Brian

        Right.  It's easy to see that the powers of 5 mod 10000
        go 5^0 = 1, 5, 25, 125, 625, 3125, 5625, 8125, 625, ...
                                 ^
                                 |                      |
                                 ------------------------
        
        Dan

    ref .8
    
    Yes, but if you do it this way, you dont get tp experience the thrill
    of long multiplications :-)
    

    Re .7:
    
    I was wondering if anyone would ask how long it took me.  ;-)
    For the record, with no calculator or computing it took me about 60
    seconds.
    
    As previous replies indicate, a little insight (ie. analysis) goes a 
    long way with this problem.  The irony is that the method that I used
    works well (with only a little bit more work) for much more 'difficult'
    problems that do not have the obvious cycles that 5^X does.
    
    For instance, what are the last four digits of 7^7777 ?
    
    
    --  Barry

    
    
    ref .-1 (Barry)
     
    I've got the last 2 digits. is this good enough ?
    
    I did few multiplications and saw the pattern repeate for the
    right most 2 digits, and so I figured since this is a determinstic world
    and GOD does't play dice, the pattern will repeate forever :
    
                                07  ---
                                49    |  pattern  07,49,43,01 repeat
                               343    |  
                              2401  ---
                             16807
                            117649
                            823543
                           5764801
                                 
    since it repeates every 4 rows, and we have 7777/4 sets of 4 rows
    i.e 1944.25 sets of 4 rows, .25 of 4 is one, so we last row will be
    the one with last 2 digits  "07"
    
    I wanted to do the proof by induction, but I love VLMM more, plus
    for induction method, one needs to start with a hypothises, and cant
    see one here.
    
    so are you going to tell us you great method?
    
    /naser
    VLMM = Very Long Multiplication Methods
    it took me 25 minutes, including break time to make some tea.
    

    Well it's sort of like a magic trick in that once you know how
    to do it, you will not think that it is that great.  But if you do not
    know it it seems miraculous.  (Remember, I used to dabble in Lightning
    Calculator type stuff).
    
    And really, the previous replies are about 90% of the way there. 
    Someone else will probably figure it out by the end of the day (it
    really isn't very hard).
    
    Well, just for the record here it is, work and all:
    
    1:33am
    
    7^1		= 7
    7^2		= 49
    7^3		= 343
    7^4		= 2401
    7^5		= 16807
    		= k*10^4 + 6807		where k= any non-negative integer
    
    7^10	= k*k*10^8 + 2*k*10^4 + (6807)^2
    					47649
    				      52456
    				     40842
    				-------------
    				     46135249
    
    		= 	k*10^4     +     5249
    
    
    7^20	=			47241
    				       20996
    				      10498
    				     26245
    				-------------
    				     27553001
    
    		= 	k*10^4     +     3001		<-- Lucky break!
    
    7^40	= 	k*10^4     +     6001
    7^80	= 	k*10^4     +     2001
    7^160	= 	k*10^4     +     4001
    7^320	= 	k*10^4     +     8001
    7^640	= 	k*10^4     +     6001
    7^1280	= 	k*10^4     +     2001
    7^2560	= 	k*10^4     +     4001
    7^5120	= 	k*10^4     +     8001
    
    7^7680 = 7^5120 * 7^2560 = k*10^4 + ( 8001 * 4001 )
    			     = k*10^4 + 2001
    
    7^7760 = 7^7680 * 7^80   = k*10^4 + ( 2001 * 2001 )
    			     = k*10^4 + 4001
    
    7^7770 = 7^7760 * 7^10   = k*10^4 + ( 2001 * 5249 )
    					     10498
    					     --------
    					     10503249
    			     = k*10^4 + 3249
    
    7^7775 = 7^7770 * 7^5    = k*10^4 + (3249 * 6807)
    					       61263
                                              27228
                                             13614
    					    20421
    					    --------
                                            22115943
    			     = k*10^4 + 5943
    
    7^7777 = 7^7775 * 7^2    = k*10^4 + ( 5943 * 49 )
    				         53487
                                        23772
                                        ------
                                        291207
    			     
                             = k*10^4 + 1207
    
    Therefore, the answer is "1207".
    
    --  Barry

    I did not check this by the way.  I just did it as I typed it in.
    
    Someone else (MAPLE?) may want to check it.  I haven't done long
    multiplication in years and I may well have messed up.
    
    The principle still applies, however.
    
    --  Barry

	You don't have to multiply anything out to the power of 7777.

	Suppose that you're interested in the last k digits of n^a.   This
is n^a mod 10^k.   Let's assume for simplicity that (10,n) = 1, although the
argument develops in a similar way (actually simpler) if (10,n) = 2,5 or 10.

	n^a mod 10^k is defined by n^a mod 5^k and n^a mod 2^k
So let's pursue these independently.   The first thing
to note is that the group of numbers prime mod p^k is cyclic, of order
(p-1)*p^(k-1), *unless* p=2, in which case the group has a cyclic subgroup of
index 2.   Thus we know immediately, that as a increases, the last
k digits of n^a will repeat themselves with a period which will divide:
	4*5^(k-1) & 2^max(1,k-2)
which is:
	5^(k-1) * 2^max(2,k-2)

	In the case of k=4, this means that the period of n^a divides 500.
So in particular, 7^7777 == 7^277 mod 10^4.   In fact, 7^100 == 1 mod 5^4,
and 7^2 == 1 mod 16, so "all you need to find" are the last 4 digits of 7^77, 
since 7^a has period 100.

	This is similar to the way in which 5^a has period 4, for k >= 4,
which Dan showed.

Regards,
Andrew.

>(1) Any positive real number can be represented as an infinite
>   decimal (e.g.3.14159265358979323846...), possibly ending in
>   all zeroes or all nines. 

No, it can't. It takes too long. You can *begin* to represent one, but you 
can't do it. By the way, you failed to do it, *even for a well-known case*,
in the statement above. :-)

>			     We teach students how to add decimals.
>   How do we add positive real numbers represented as infinitely
>   long decimals ? How do we subtract or multiply or divide them ?

We don't. In general, real numbers are not representable unless they are
functions of integers. Only if the functional representation happens to 
exist can you find a functional representation for arithmetic operations on 
them.

As Lynn said, in practice no one ever uses anything but rationals 
for +-*/ and ^.  However, the rest of the reals are needed for theoretical
reasons, e.g., to make calculations about the circumference of a circle in units 
of its radius, and to prove theorems to keep mathematicians employed.

There is a theorem that says something to the effect that a field that 
is **** (I can't remember the word) like the rationals can be *completed* by
defining reals to be equivalence classes of converging (in the Cauchy sense)
sequences of rationals.  Then it's easy to show that the usual arithmetic 
operations still follow the piano axioms (is "axia" the plural of axiom?).

For example, if r = a/b, where a & b are integers, then

x^r = (x^a)^(1/b) (assuming whatever is needed to make this make sense).

If the irrational r is r = [a1/b1, a2/b2, a3/b3, ...], then

x^r = [x^(a1/b1), x^(a2/b2), x^(a3/b3), ...].   

And, in fact, any continuous function on an irrational is defined in the
same manner.

Avery

        re .19,
        
>> There is a theorem that says something to the effect that a field that 
>> is **** (I can't remember the word) like the rationals can be *completed* by
>> defining reals to be equivalence classes of converging (in the Cauchy sense)
>> sequences of rationals.  Then it's easy to show that the usual arithmetic 
>> operations still follow the piano axioms (is "axia" the plural of axiom?).
        
        "Peano axioms" (it's a person's name).
        
        "Complete" and "Cauchy sequence" are terms that apply to
        metric spaces.  So what you need is an absolute value
        function on the field.
        
        Dan

>       "Peano axioms" (it's a person's name).
 
	An unfortunate typo.  Your right about this.
	Oh well, their are lots more important things to worry about. Its 
	to bad I didn't sea that won.
       
>        "Complete" and "Cauchy sequence" are terms that apply to
>        metric spaces.  So what you need is an absolute value
>        function on the field.
        
	Right.  The fact that you are in a metric space is essential.  In 
	fact, depending on what you are doing, it doesn't necessarily have to 
	be a field to be extended continuously, as long as it has a metric.

	-avery
	

    Bummer.

    My critical error in .15 was that 7^20 should have had a residue of
    2001, *not* 3001.
    
    --  Barry

    I thought I'll add this here, since it seems a 'subtle' problem
    (to me at least);
    
    we all know the proof that 2^.5 is irrational:
    
    let c= 2^.5 = m/n to lowest terms
       
        c^2= m^2/n^2
        2 n^2= m^2
      so 2 n^2 is even, so m^2 is even , so m is even, let m= 2x
      so 2 n^2 = 4 x^2
      so n^2 = 2 x^2
      so n^2 is even, i.e n is even
      but then m and n are even, contradicting the assumption that m/n are
     to lowest common terms.
    
    now to show that othe radicals are irrationals might not be so easy
    example  ------------------
            /      ------------
           /      /      ----
       7  /    4 /      /
        \/ 6 - \/ 5 + \/ 11
                                                              ---
     is an irrational, to proof it , i started similarly to \/2
     and let the above beast = m/n, then take power 7th, etc..
    
     my question, do you see a short cut to show above is irrational?
     is the method must go along the lines of 2^.5 ?
     the above one got really messy in expansion even with maple help.
    
    thank you,
    /naser

>    now to show that other radicals are irrationals might not be so easy
>    example  ------------------
>            /      ------------
>           /      /      ----
>       7  /    4 /      /
>        \/ 6 - \/ 5 + \/ 11

	Call this chap x.   ((6 - x^7)^4 - 5)^2 - 11 = 0

	This is a *monic* polynomial. ("monic" means the term of highest 
degree (56) has coefficient 1.)   Thus x is what is known as an "algebraic 
integer".

	There is an easy theorem which says that the only rational, algebraic
integers are the integers themselves.   Now x can't be an integer, because
if it were, then so is _/11.   Thus x is irrational.

	Incidentally, I've always felt that the term "algebraic integer" is
very misleading, since it does *not* mean "an integer which is algebraic".
All integers are algebraic.   

	Algebraic integers are very important in group representation theory.
All entries in the character table of a finite group over the complex field
must be algebraic integers.   The fact that such entries cannot be non-integer
rationals is, I seem to recall, a critical step of Burnside's p-alpha-q-beta
lemma, which shows that all groups with order p-alpha-q-beta, where p & q
are prime, must be soluble.

Regards,
Andrew.

>	There is an easy theorem which says that the only rational, algebraic
>integers are the integers themselves.

	It occured to me that I should show the theorem, since it's here
that you actually see the generalization from the _/2 irrationality proof.
Also, I didn't make clear that the monic polynomials I was referring to
must have coefficients drawn from the integers.

	Suppose we have a monic poly over the integers:

	t^n + a_(n-1)*t^(n-1) + ... + a_1*t + a_0

and that a rational written b/c in lowest terms is a zero of this poly.

	Then multiplying through by c^n, we've got:

	b^n + a_(n-1)*b^(n-1)*c + ... + a_1*b*c^(n-1) + a_0*c^n = 0

Take this mod p, where p is some prime dividing c:

	b^n == 0 mod p.

Yet (b,c) = 1, contradiction.   So no such p exists so c is 1.   End.


	Another neat fact is that the set of algebraic integers is closed 
under addition, subtraction and multiplication.

Cheers,
Andrew.

PS: Thanks Avery for correcting a typo in an earlier version of this.

re .-1:

>	Another neat fact is that the set of algebraic integers is closed 
>under addition, subtraction and multiplication.

So it's a ring, in fact, a division ring.  Is it a field?  That is, if 
a is a solution of p() and b is a zero of q(), with p and q monic 
polynomials with integer coefficients, what can you say about a/b?  
Nothing comes to mind ...


avery

        It's not a field ... 1/2 is not an algebraic integer, but
        1 and 2 are.  What's a division ring?
        
        The algebraic integers are just the algebraic numbers
        that are zeroes of monic polynomials (coefficient of
        highest degree term is 1).  1/2 is the zero of 2x - 1.
        
        Dan

>        It's not a field ... 1/2 is not an algebraic integer, but
>        1 and 2 are.  What's a division ring?
 
	That's what you just proved in .28.  Sorry, I wasn't thinking when
	I asked the question.

	A division ring is one in which cancellation works, even though
	multiplicative inverses aren't defined.  That is,
	
	a*b = a*c  implies b = c.

	avery

    Another easy subtle problem for you number theory folks:
    
    a*b = c
    a,b are relatively prime, c is a cubic number , prove that a,b must
    then each be cubic numbers.
    
    thanks,
    /naser

	The subtlety, I suppose, is that this works over the integers, and not
just over the naturals.

Regards,
Andrew.

        re .31
        
>>	A division ring is one in which cancellation works, even though
>>	multiplicative inverses aren't defined.  That is,
>>	
>>	a*b = a*c  implies b = c.
        
        That must be assuming a is not 0.
        
        Aha.  Any ring between the integers and the complex
        numbers has no zero dividers (nonzero elements the
        product of which is zero) and so for such a ring
        
        	ab = ac ==> ab - ac = 0
        		==> a(b - c) = 0
        		==> a = 0 or b - c = 0
        		==> a = 0 or b = c
        
        Likewise if ab = ac implies b = c for nonzero a, then
        for nonzero a you have
        
        	ab = 0	==> ab = a0
        		==> b = 0
        
        So it sounds like division rings are rings without zero
        dividers, although other conditions may apply.
        
        I tend to see rings classified with the following three
        features, with the most general case being not having the
        feature:
        
        	commutative
        	with unity [i.e., a multiplicative identity]
        	with no zero dividers
        
        If you have all three the ring is called an integral
        domain.
        
        Are these three conditions independent of each other
        (i.e. for each of the eight combinations of having/not
        having that property does such a ring exist)?
        
        Dan

    ref .33 (Andrew)
    Thanks, but how does proof this? proof by induction?

	It's curious that I was just reading the other day an article about
intuitionist math which asked as an exercise for a proof that _/2 was irrational
which was not a reductio ad absurdum proof.   At the time, I couldn't think of
one...

Regards,
Andrew

    ref .-1 (Topher)
    That's a neat solution!
    
    Oh well..I guess I will always have chess to fall back to...
    
    Iam going to look for something to hit my head with, may be this
    will help fix the problem.
    
    Thanks..
    /naser

    The following is a comment on style of proof, NOT on the validity
    of .28, which is obviously well done.

    I had a math professor who discouraged us from using indirect 
    (reductio ad absurdum) proofs if it could be avoided. Many 
    indirect proofs assume that an object is of a certain form, then
    show it can't be  because it is of a different form.  These proofs
    can be rewritten in a  positive manner by simply removing the initial
    assumption and deriving the form in the proof.

    For example, in .28, the last line in
  
>Take this mod p, where p is some prime dividing c:
>
>	b^n == 0 mod p.
>
>Yet (b,c) = 1, contradiction.   So no such p exists so c is 1.   End.

    can be re written as 

    Since (b,c) = 1, the only prime that divides c is 1, so the 
    rational root b/c of the polynomial is an integer.

    That is, it is as easy to show that if the root is rational, it
    must be an integer, as it is to assume a rational root that is not
    an integer and arrive at a contradiction because it IS a integer.  
    They are both logically equivalent, but the former is more 
    acceptable to the constructionists, and in some cases easier 
    to follow.

    Avery

	Re -.1, yes the rewriting possibility crossed my mind.   This was why
I implicitly alleged a few replies ago that the proof I gave essentially did not 
involve RAA.

	More generally, I do not have a good "intuitive " :-) feel for what
kinds of things are possible in constructivist mathematics, compared to the
vanilla variety.   It could be for instance that some results that we assumed
to get the non-RAA proof of the rationality of _/2 rely on RAA themselves.
It could be, but I don't think it is.   Until I actually understand the
intuitionist arguments and implications, it would be mere superstition of me 
to avoid RAA.

	The aesthetic issue is a different one.   I agree with Avery that
other things being equal, a constructive proof is simpler to state.   An RAA
proof is like using a sentence with a double negative:  although one's first
draft of a sentence may include a double negative because that's the way that
the thought occured to one, later editing should iron out the wrinkle.   However,
one doesn't always have time to do this...

Cheers,
Andrew.

    Srinivasa Ramanujan was described (among other things) as having a great
    mathematical 'intuition' .

re Intuitionism:
  I (used to) feel rather convinced by intuitionistic criticisms of
  classical logic.

  The basis of the difference is in what you hold to be knowing the
  meaning of a mathematical proposition. The intuitionistic view
  is  that knowing the meaning of a statement means knowing what would
  count as a proof of it. The classical view is that it is knowing what
  would be the case were it true. In a slogan, it is "proof conditions
  versus truth conditions".

  If you write out the rules of intuitionistic logic in a natural deduction
  style, the introduction rules for the logical constants are the formal
  counterpart of the proof conditions. It is rather striking that rules
  such  as reductio ad absurdum, or excluded middle, which must be added
  to get classical logic, stick out like a sore thumb when the rules are
  formulated in this way. (I'd explain this if challenged.)

  Real mathematics doesn't really get carried out with delicate attention
  to logical subtleties like these. But in practice any real mathematical
  result has an intuitionistically provable counterpart. There was a book
  by a constructive mathematician caller Erret Bishop (a measure theorist)
  which showed how to what measure  theory (and other things) looked like
  when done constructively. There are also formal results, which show how
  how to translate deductions in classical calculi so as to expose their
  constructive content. I used to find this all rather fascinating. Perhaps
  I still do.

    David Hilbert had a great line about the intuitionists.  The
    intuitionists were let by some guy (Brower?) in France.  Hilbert was
    initially neutral about this debate.  Then he solved Gordon's problem
    with some very clever non-constructive argument (When Gordon first saw
    that proof, he said "This is not mathematics.  This is theology"
    because the theorem is about the existance of some basis.  Rather than
    constructing such a basis as Gordon and other mathematicians of the
    time were trying to do, Hilbert showed that such a basis must exist).
    Well, after Hilbert solved Gordon's problem he naturally joined the
    side against the intuitionists, and his great argument against the
    intuitionists was an ontological one (the same stuff that was used to
    prove the existence of God in the middle ages):
    
    When a French mathematician was visiting Goettingen, and got into a
    debate on constructive mathematics.  In the heat of the debate, Hilbert 
    said, 
    
    It is obvious that we the non-intuitive mathematicians can prove more 
    theorems than the intuitionists.  Given the choice of more theorems or 
    less theorems, I say we choose to have more theorem.  
    
    This was how low the debate went, but I also think Hilbert had a point
    in what he said.  That is there isn't really any point debating it 
    since the nature of the debate is a philosophical one not a mathematical 
    one, and one might as well use the ontological argument for that.
    
    Eugene

Brouwer was Dutch.
He was a Nazi collaborator in the war (I hope I'm not repeating a smear).
He was a brilliant topologist: I think he is most famous for something
called Brouwers fixed point theorem (which he proved by a non-constructive
argument: I  heard that disatisfaction with his proof is what led him
into the foundations of mathematics).

Another mathematician whose name is somewhat linked with the intuitionists
is Kolmogorov, the Russian who laid the foundations for probability theory.
It was he who first explained the  semantics of the intuitionistic logical
connectives. (In his formulation, proposition = problem, proof = solution.)
I'm not sure  whether he can be described as an intuitionist.

I know of one (living) mathematician with some reputation who is a serious
constructivist: the Swedish statistician Per Martin-Lof (whose name is
connected with  the theory  of randomness [see Knuth], and was a
student of Kolmogorov). When working  as a statistician, he was perfectly
content to publish non-constructive arguments. I think he held that by
suitable reformulations (which would interest  logicians more than
statisticians), constructively  impeccable results could be obtained: at
any rate, in mathematics that anyone could apply in practice.

$!
$!	Here's an example of a recursive function in DCL, which prints
$!	fascinating Hilbert curves.  IT REQUIRES A REGIS TERMINAL, SUCH AS
$!	VT240.
$!
$!	Save this procedure as HILBERT.COM.  Then, run it with
$!
$!		$ @HILBERT 4
$!
$!	Then, you can experiment with other small integers.
$!
$!	Author: Eric Osman 12/12/86
$!
$ all_options = p3 + p4 + p5 + p6 + p7 + p8
$ debug_flag = all_options .nes. ""
$ if .not. debug_flag then on control_y then goto leave
$ esc[0,8] = 27
$ if debug_flag then esc = "`"
$ s = p2
$ if s .eqs. "" then s = 15
$ level = 0
$ h0 = 180
$ hx180 = -1
$ hy180 = 0
$ h1 = 90
$ hx90 = 0
$ hy90 = 1
$ h2 = 0
$ hx0 = 1
$ hy0 = 0
$ h3 = 270
$ hx270 = 0
$ hy270 = -1
$ width = 800
$ height = 500
$ h = 270
$ !if .not. debug_flag then -
$ write sys$output "''esc'[2J''"
$ x = 0
$ y = 22 * height / 24
$ write sys$output "''esc'P1pP[''x',''y']"
$ whereto'level = "m0"
$ newarg = p1
$ if newarg .eqs. "" then newarg = 5
$ goto recursing
$m0: goto leave
$!
$recursing: level = level + 1
$ n'level = newarg
$ if n'level .eq. 0 then goto is0
$ a'level = 90 - (n'level .lt. 0) * 180
$ m'level = n'level + (n'level .lt. 0) * 2 - 1
$ h = h + a'level		! turn: a
$ newarg = 0 - m'level
$ whereto'level = "tag1"
$ goto recursing		! hilbert: 0 - m side: s
$tag1: h = h + a'level		! turn: a
$ h = h - h/360*360
$ if h .lt. 0 then h = h + 360
$ x = x + s*hx'h
$ y = y + s*hy'h
$ write sys$output "V[''x',''y']"
$ newarg = m'level
$ whereto'level = "tag2"
$ goto recursing		! hilbert: m side: s
$tag2: h = h - a'level		! turn: 0 - a
$ h = h - h/360*360
$ if h .lt. 0 then h = h + 360
$ x = x + s*hx'h
$ y = y + s*hy'h
$ write sys$output "V[''x',''y']"
$ h = h - a'level		! turn: 0 - a
$ newarg = m'level
$ whereto'level = "tag3"
$ goto recursing		! hilbert: m side: s
$tag3: h = h - h/360*360
$ if h .lt. 0 then h = h + 360
$ x = x + s*hx'h
$ y = y + s*hy'h
$ write sys$output "V[''x',''y']"
$ h = h + a'level		! turn: a
$ newarg = 0 - m'level
$ whereto'level = "tag4"
$ goto recursing		! hilbert: 0 - m side: s
$tag4: h = h + a'level		! turn: a
$ goto exit_please

$is0: h = h + 180			! ^turn: 180
$exit_please: level = level - 1
$ tag = whereto'level
$ goto 'tag
$!
$!
$leave: write sys$output ";''esc'/''esc'[24;1f''esc'[K''esc'[A"
$ exit

I was intrigued by numbers that equal the sum of powers of their digits.
So far, I've searched thru the 14th powers.  The list of numbers appears
below.  Through the 10th power, these were also published in .8.

power	numbers

   3	1, 153, 370, 371, and 407.	(0 is excluded by fiat)
   4	1, 1634, 8208, and 9474.
   5	1, 4150, 4151, 54748, 92727, 93084, and 194979.
   6	1, and 548834.
   7	1, 1741725, 4210818, 9800817, 9926315, and 14459929.
   8	1, 24678050, 24678051, and 88593477.
   9	1, 146511208, 472335975, 534494836, and 912985153.
  10	1, and 4679307774.
  11	1, 40028394225, 44708635679, 49388550606, 82693916578, and 94204591914.
  12	1.
  13	1, and 564240140138.
  14	1. 

> I was intrigued by numbers that equal the sum of powers of their digits.
> So far, I've searched thru the 14th powers.  
    
    How far have you searched within a power?  Is there a natural upper limit?

>> Is there a natural upper limit?
        
        If you assume k-th powers, then an n-digit number must be
        at least 10^(n-1) but its digits only sum to at most n9^k.
        That limits how high n can go for a given k.  An earlier
        reply showed the bounds for a small k.
        
        Dan