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Four 4's

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Krabo
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28 May 2013, 3:03 pm

You are given a set of four numbers, they are all 4's. You are free to arrange them at will and also free to use any valid arithmetical operations, including parentheses and brackets, roots, what ever, but you are not allowed to include any extra numbers. Your goal is to construct expressions which yield all integers below a certain limit. (I'm not sure what this limit might be.)

Examples:

1 = 44/44
2 = 4/4 + 4/4
3 = 4 + 4/4 – √4
4 = 4 + 4·(4 – 4)

Please, continue. (Not necessarily in ascending order.)



Funkwelle
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28 May 2013, 3:49 pm

5 = √(4*4) + 4/4
5 = 4 + 4^(4-4)
6 = 4 + √(4) + 4 - 4
7 = 44/4 - 4
8 = 4 + 4 + 4 - 4
8 = 4 * √(4) + 4 - 4



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28 May 2013, 4:26 pm

9 = 4+4 + 4/4
10 = (44-4)/4


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Krabo
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28 May 2013, 4:59 pm

11 = 4!/√(4) – 4/4



Fnord
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28 May 2013, 5:06 pm

12 = 4+4+√4+√4



GGPViper
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28 May 2013, 5:09 pm

13=(4!/√(4))+(4/4)



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28 May 2013, 5:15 pm

14 = 4 × 4 - 4 + √4

Are you guys sure square roots aren’t cheating? You’re implicitly including a 2 or a 1/2 :)


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ruveyn
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28 May 2013, 5:16 pm

Krabo wrote:
You are given a set of four numbers, they are all 4's. You are free to arrange them at will and also free to use any valid arithmetical operations, including parentheses and brackets, roots, what ever, but you are not allowed to include any extra numbers. Your goal is to construct expressions which yield all integers below a certain limit. (I'm not sure what this limit might be.)

Examples:

1 = 44/44
2 = 4/4 + 4/4
3 = 4 + 4/4 – √4
4 = 4 + 4·(4 – 4)

Please, continue. (Not necessarily in ascending order.)


5 = (4*4 + 4)/4
6 = ( 4*4 + sqrt(4)*4) /4

7 = (4*4 + 4*4 - 4)/4

8 = (4*4 + 4*4)/4
9 = (4*4 + 4*4 + 4)/4

I will think of some more later on.

ruveyn



ruveyn
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28 May 2013, 5:19 pm

Spiderpig wrote:
14 = 4 × 4 - 4 + √4

Are you guys sure square roots aren’t cheating? You’re implicitly including a 2 or a 1/2 :)


Aha . That gives us 10 = 4*4 + sqrt(4) - 4 - 4

and 40 = 4*(4*4 + sqrt(4) - 4 -4 ).

or 40 = ( 4*4 + 4*4 + 4 + 4)



ruveyn
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28 May 2013, 5:19 pm

Spiderpig wrote:
14 = 4 × 4 - 4 + √4

Are you guys sure square roots aren’t cheating? You’re implicitly including a 2 or a 1/2 :)


Aha . That gives us 10 = 4*4 + sqrt(4) - 4 - 4

and 40 = 4*(4*4 + sqrt(4) - 4 -4 ).

or 40 = ( 4*4 + 4*4 + 4 + 4)



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28 May 2013, 5:20 pm

15 = 4x4 - (4/4)
16 = 4! -4 -√4 -√4
17 = 4x4 + (4/4)



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28 May 2013, 5:25 pm

16 = 4 × 4 - 4 + 4
17 = 4 × 4 + 4/4
18 = 4 × 4 + 4/√4
19 = 4! - 4 - 4/4
20 = 4! - 4 - 4 + 4
21 = 4! - 4 + 4/4
22 = 4! - 4 + 4/√4


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28 May 2013, 5:32 pm

23 = 4! +(4/4) - √4
24 = 4 + 4 + (4x4)
25 = 4! - (4/4) + √4
26 = 4! + (4!/4) - 4



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28 May 2013, 5:32 pm

23 = 4! - (√4+√4)/4
24 = 4 × 4 + 4 + 4
25 = 4! + (√4+√4)/4
26 = 4! + (4 + 4)/4
27 = 4! + √4 + 4/4
28 = 4! + 4 + 4 - 4
29 = 4! + 4 + 4/4
30 = 4! + 4 + 4/√4


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Krabo
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29 May 2013, 12:45 am

I think it ends here. Odd integers seem to require "4/4" and it's not possible to express either 30 or 32 with two 4's alone. Thank you all for your contribution.

We could, of course, define a function φ(x) = 7x and write
31 = φ(4) + 4 – 4/4
but this wouldn't be in the spirit of the problem.



Krabo
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29 May 2013, 2:21 am

We would get a long way forward if we could somehow express "1" by using one 4 alone. Truth values and calculator-style functions will do the job. For, consider FP (fractional part) with the logical ¬ ("not") thus,

1 = ¬[FP(4)]