-5
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Make all positive integers, starting from $1$, using:

  1. Exactly four $4$'s (and no other digit), concatenation being allowed,
  2. Any standard mathematical symbol(s) and operation(s) not listed below.

But not using:

  1. Addition, subtraction, multiplication, division, modulo reduction,
  2. Logarithms, exponentiation, root extraction,
  3. Floor, ceiling, fractional part,
  4. Integral, differential, binomial coefficients,
  5. English, Greek, or Hebrew or any other language letters (Note that trig ratios require the use of English letters, and hence are not allowed),
  6. Any function or operation generated by you.

(And yes, it's possible.)


Enjoy! I'll be awaiting creative responses.

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closed as too broad by Deusovi, Gamow, JonMark Perry, CodeNewbie, Beastly Gerbil Jun 23 '16 at 20:01

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ 一,二,三,四,五,六,... $\endgroup$ – f'' Jun 23 '16 at 6:00
  • 6
    $\begingroup$ $|\{4444\}|$, $|\{4444,\varnothing\}|$, $|\{4444,\varnothing,\{\varnothing\}\}|$, $|\{4444,\varnothing,\{\varnothing\},\{\{\varnothing\}\}\}|$,... $\endgroup$ – f'' Jun 23 '16 at 6:30
  • $\begingroup$ @f'' chinese....well, I said letters and other digits are not allowed. The second solution is my solution too, with the $\varnothing$'s replaced by $\{\}$'s; would you like to post it as an answer? $\endgroup$ – Ankoganit Jun 23 '16 at 7:56
  • $\begingroup$ Letters and digits aren't allowed, but you never said anything about characters. $\endgroup$ – f'' Jun 23 '16 at 12:56
  • $\begingroup$ @f'' Hmm, looks like that was a bad puzzle to put up. Thanks anyway for your attention. :) $\endgroup$ – Ankoganit Jun 23 '16 at 12:59
6
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Easy :

$ \mathbb{N}^* \cup [4,444] $

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  • $\begingroup$ LOL what? N is a letter... $\endgroup$ – Ankoganit Jun 23 '16 at 12:40
  • 5
    $\begingroup$ @Ankoganit That, my pal, is not just a simple N. $\endgroup$ – ABcDexter Jun 23 '16 at 18:14
  • $\begingroup$ "Any standard mathematical symbol(s) and operation(s) not listed below." - is 5 meant to exclude $\Bbb{N}$? $\endgroup$ – Jonathan Allan Jul 3 '16 at 1:13
5
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Here's a solution:

The question asks for any positive integer, not every integer. Hence 4444 will do.

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  • 1
    $\begingroup$ And what about "starting from $1$"? +1 for the idea though. :) $\endgroup$ – Ankoganit Jun 23 '16 at 7:56
  • $\begingroup$ Now I have replaced "any" with "all", sorry for the inconvenience. $\endgroup$ – Ankoganit Jun 23 '16 at 8:40
  • $\begingroup$ Haha no inconvenience, just a bit of tongue-in-cheek. :-) $\endgroup$ – Shagnik Jun 23 '16 at 10:19
4
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There is a standard mathematical notation for all natural numbers using

$0$ $4$s and no other digit, it's called Church encoding

So all we need to do

is to replace $x$ with $44$ ($2$ $4$s concatenated)
in the standard lambda expansions of the Church numerals.

To yield

$1=\lambda f.\lambda44.f44$
$2=\lambda f.\lambda44.f(f44)$
$3=\lambda f.\lambda44.f(f(f44))$
$\vdots$
$n=\lambda f.\lambda44.f(\cdots f(f(f44))\cdots)\space\space\space$ (using $n$ $f$s)

This way, we can also

create zero
$0=\lambda f.\lambda44.44$


If we were only able to use $2a+1$ $4$s (some odd number of $4$s) we cannot do the above, but we can do

$n=|\{444\cdots4, \{\}, \{\}, \cdots, \{\}\}|\space\space\space$ (using $2a+1$ $4$s and $n-1$ $\{\}$s)
As
$\{x_1,x_2,\cdots,x_m\}$ is a set
$\{\}$ is the empty set, $\emptyset$
and $|X|$ is the cardinality of the set $X$


Going by the comment on Lord of dark's answer, I suppose $\lambda$s and $f$s are not allowed by not using #5 (any letters) - so the first solution is not a valid solution.

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3
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Let's see

!(44 ^ 44) = 1
4 >> ( !( (4 | 4) ^ 4) ) = 2
(!!4) | (4 >> ( !(4 ^ 4))) = 3
4 | 4 | 4 | 4 = 4
(4 | 4 | (!(4 ^ 4))) = 5
4 | (4 >> (!(4 ^ 4))) = 6
!!4 | (4 >> !!4) | 4 = 7
(4 & 4) << !(4 ^ 4) = 8
(4 << !(4 ^ 4)) | !!4 = 9
((4 << !!4)) | (4 >> !!4) = 10

And in case you don't think I am making up the things

see this.

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  • 2
    $\begingroup$ I Like the answer, but have a question: are coding operators 'standard mathematical operators? I.e. Do mathematicians use them like this on paper and in books? $\endgroup$ – BmyGuest Jun 23 '16 at 6:09
  • 1
    $\begingroup$ Logical operations is mathematics in IMHO. But I am not sure by the definition and scope of standard mathematical operators. $\endgroup$ – Mohit Jain Jun 23 '16 at 6:10
  • $\begingroup$ Logical operations, yes. Bitwise operations, I'm not so sure. $\endgroup$ – Gareth McCaughan Jun 23 '16 at 10:14
  • $\begingroup$ if they do, those are not the symbols they use. $\endgroup$ – Jasen Jul 22 '16 at 5:22

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