Every Number Expressed as Four 2s

[Karl]

A bit of a game, but needs more thought put into it to be placed in the Forum Games area, in my opinion.

Basically, the idea is to express each integer as an expression using only the number 2, and only using it 4 times.

Any functions may be used. Trigonometric functions must specify if they are in radians or degrees.

EDIT: Time to keep a tally.

1 = 22 / 22

2 = (2 / 2) + (2 / 2)

3 = (2 ^ 2) - (2 / 2)

4 = (2 ^ 2) * (2 / 2)

5 = (2 ^ 2) + (2 / 2)

6 = 2 * 2 * 2 - 2

7 = 2 + √(.2 ^ -2)

8 = 2 + 2 + 2 + 2

9 = (22 / 2) - 2

10 = 2 * 2 * 2 + 2

11 = (2 / .2) + (2 / 2)

12 = (2 + 2 + 2) * 2

13 = (22 / 2) + 2

14 = (2 ^ 2 ^ 2) - 2

15 = (.2 ^ -2) - (2 / .2)

16 = 2 * 2 * 2 * 2

17 = [[ (.2 ^ -2.2) / 2 ]]

18 = 22 - 2 - 2

19 = [[ 22 - 2.2 ]]

20 = (2 * 2)! - 2 - 2

21 = 22 - (2 / 2)

22 = 22 / (2 / 2)

23 = 22 + (2 / 2)

24 = (2 + 2)! + 2 - 2

25 = (.2 ^ -2) * (2 / 2)

26 = 22 + 2 + 2

27 = [[ (.2 ^ -2) + 2.2 ]]

28 = (2 * 2)! + 2 + 2

29 = (.2 ^ -2) + 2 + 2

30 = (2 + 2 + 2) / .2

31 =

32 = 2 * (2 ^ (2 + 2))

[Justin]

2 / 2 + 2 / 2

[e p]

Do exponents count? If so,

2^2 - (2\2)

[Karl]

(2^2) * (2 / 2)

Title your posts, maybe, so there's no confusion?

Also: Yes, any function may be used, as long as you are only using 2s.

[Syrito]

2^2 + 2/2 = 5

[Justin]

They better, otherwise we can't go higher than 2^4.

Also;

2 * 2 * 2 - 2

[Karl]

[[ (2 * 2 * 2) - sin(2) ]]

It works in Radians or Degrees, so I think it's valid.

Note: [[ ]] means to truncate the number.

[e p]

2 + 2 + 2 + 2

[Karl]

e is a constant unequal to 2, so, no.

(22 / 2) - 2

EDIT: Chejrw, are you removing posts? Makes me sound like a madman, you know.

[e p]

e is a constant unequal to 2, so, no.
Hmm?
edit: Screwed up on 10. I need to think on this one.

[¿CRAZYØ¿]

2*2*2 + 2

[Karl]

In that case, CrazyO would be correct, because you used five 2s in yours, evil penguin. Although, really his should be:

(2 * 2 * 2) + 2

[e p]

In that case, CrazyO would be correct, because you used five 2s in yours, evil penguin.

Yeah, I just screwed up. I edited my post.

[Karl]

[[ 22.2 / 2 ]]

This gets really fun once we reach the 20s or so.

[Justin]

22/2+2

[Spartan 1337]

2 + 2 + 2 * 2 = 12

[Karl]

(2 ^ 2 ^ 2) - 2

[Justin]

OK, we have 2 11's, then 13, then 12, then 14.

[Karl]

I count one of each, just out of order. You're the Admin here, you can fix it if it irks you.

[Justin]

eh, whatever. That was really my way of saying "I can't think of how to do 15"

[Karl]

Maybe try something with factorials? That's what I'm looking at right now, but it doesn't seem to be going anywhere.

EDIT: Never mind.

[[ (2 ^ 2 ^ 2) - sin(2) ]]

Same process as 7. It would be nice to find something without using trig functions, but I really can't think of anything.

[Justin]

Truncation... That's kinda cheap. Oh well.

2 * 2 * 2 * 2

[Serisium]

Here's another way to get 15:

[sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(22!)))))!)*2)^2]

I think I try too hard. . (http://www.google.com/search?hl=en&safe=off&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=BFU&q=sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(22!)))))!)*2) ^2&btnG=Search)

EDIT: Here's 17:

[sqrt(sqrt(sqrt(sqrt(22 !)))) - (2 * 2)]

[Karl]

Truncation... That's kinda cheap. Oh well.

2 * 2 * 2 * 2
Once you find a way to get 15 without it, we won't use it.

Also, Serisium, a sqrt() function implies a 2, since it is the 2nd root of some number.

So, each sqrt counts as a 2.

[Serisium]

In that case,

(2^2)^2+sin(2)

If we're allowed to do more than one at a time:

18: 22-2-2

[e p]

(2^2)^2 + 2

Hmm, apparently I'll need more than just a 10th grade education for some of these. Math is my worst subject, too. I already broke out my TI-84 for this.

[Justin]

In that case,

(2^2)^2+sin(2)

If we're allowed to do more than one at a time:

18: 22-2-2

16.02 =/= 17, even by the most liberal rounding rules.

[Serisium]

http://www.google.com/search?hl=en&safe=off&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&q=(2^2)^2%2Bsin(2)&btnG=Search

Google calculator is turning up with 16.909.

[Justin]

It must be using degrees, then.

Trig functions are ambiguous, to say the least.

[e p]

It must be using degrees, then.

Trig functions are ambiguous, to say the least.
Don't you mean radians? Under radians it's 16.90929743 on my TI-84.

On degrees it's 16.0348995

[Karl]

I may have needed to write down some rules first.

Trig functions are ambiguous, so we might want to not use them.

I still think truncation is valid, though. I doubt you'd get anywhere without it.

[Blixinator]

Should sqrt() not be allowed? Since it is really just ^.5
Also, are we typesetting our answers to integers because several of the answers aren't coming out to the exact number.

[Karl]

Should sqrt() not be allowed? Since it is really just ^.5
Also, are we typesetting our answers to integers because several of the answers aren't coming out to the exact number.
We are truncating some results, as in f(x) = [[x]]. It removes the decimal part of a number.

Also, as long we taking 2nd roots, we have no problem.

http://i3.photobucket.com/albums/y79/Naborr/root.png

If we disallow using 2nd roots because they can be expressed as x ^ (1/2) then we would have to disallow using 2/2 because they can be expressed as 1.

[Karl]

Because I don't like using trig functions:

[[ (2 ^ 2 ^ 2) - .2 ]]

Similarly,

7 = [[ (2 * 2 * 2) - .2 ]]

[e p]

Damn, e^(2) * 2 + 2 + .2 = 16.9781122

Does that count?

[Karl]

No, we're not rounding anything. Also, as I said before, e is a numerical constant not equal to 2, and it can't be used.

No matter how close 16.whatever is to 17, it's not 17.

The only exception I was going to make is truncation, and that's only until we can find another way without using it.

EDIT:

[[ 22 - 2.2 ]]

[e p]

No, we're not rounding anything. Also, as I said before, e is a numerical constant not equal to 2, and it can't be used.

No matter how close 16.whatever is to 17, it's not 17.

The only exception I was going to make is truncation, and that's only until we can find another way without using it.
Hmm, I've tried about 25 different combinations in the last 10-15 minutes. All of them revolved around e.

I guess I'll have to rethink.

[Joest]

(22 - 2) - 2

[Karl]

(22 - 2) - 2
18?

[Joest]

18?
Whoops.

[Karl]

Seems to start being repetitive, but:

[[ 22 - 2 + .2 ]]

[Serisium]

22-(2/2)

Also, here's 20 without having to round: (2*2)!-2-2

[Spartan 1337]

22 / (2/2) = 22

[Karl]

22-(2/2)

Also, here's 20 without having to round: (2*2)!-2-2
Very nice! Yeah, any chance to remove truncations is good.

[Rocinante]

[[(2+2)^2 + Ln(2)]]

Will this work or are you not allowed to use natural log.

[Spartan 1337]

22 + (2/2) = 23

[Karl]

[[(2+2)^2 + Ln(2)]]

Will this work or are you not allowed to use natural log.
Sorry, no, a natural log has a base of e. The only logarithm you could use would be Log2(2), which is 1.

EDIT: (2 + 2)! + 2 - 2

[Rocinante]

25 = .2^-2 *2/2
26 = 22 + 2 + 2
27 = [[.2^-2 + 2.2 ]]
17 = [[(.2^-2.2)/2]]
28 = (2 * 2)! + 2 + 2
29 = (.2 ^ -2 ) + 2 + 2

[Serisium]

22+2+2

Also, great solution for 25. I would've never seen that.

[Karl]

Ah whoever moved the thread is a jerk. =[

Probably warranted it, but still. It seems so much more legit in Irrelevance.

[e p]

Damnit!

I wanted to do 22 so I could do log2(2)^22.

I can't do 27. :/

[Rocinante]

15 = ( .2^-2) - (2/.2)

[Caesar Augustus]

17 = 22 - √(.2^(-2))

that same trick of getting the value 5 can be used to get 27

27 = 22 + √(.2^(-2))

[Karl]

Except that every root is counted as a 2, because it is the 2nd root.

So both your solutions for 17 and 27 use five 2s.

[Karl]

First post has been updated. I must say that Rocinante has been currently dubbed 'The Man' as it pertains to this thread.

[Rocinante]

7 = 2 + √(.2^(-2))

The only way i can think of getting to 30 involves trig functions , i think we are stuck unless trig functions are allowed.

[Serisium]

Here's 11 without truncating: (2/.2) + (2/2)

If square roots didn't count we could use this for 30: (.2^-2) + sqrt(.2^-2)

I think It's going to be impossible to continue without either square roots, or trig functions.

[Caesar Augustus]

The two is implied in a root without a number, but writing it isn't required, but whatever. Trying to work up a 30 here...

[Justin]

Yeah, a root is simply X^(1/Y), where Y is the base of the root (2 in the case of square roots).

[Karl]

So we're agreed that a root counts for one 2? It may be implied, but it's still necessary for it to be a square root.

As for trig functions, I don't like the idea unless it is true for both degrees and radians.

[Justin]

It should be fair enough to just specify: eg: tan(2 rad) or csc(2 degrees)

[Karl]

Fine, if it's specified. If we can simplify it to work without truncation or trigonometric functions, however, we should do so.

[Rocinante]

30 = [[csc (2 degrees ) +2 -.2 -.2 ]]
31 = [[csc (2 degrees ) + 2 * (2/2)]]
32 = [[csc (2 degrees ) + 2 + (2/2)]]

[Karl]

See this might be a dilemma. Cosecant implies a 1 / Sine. Do we count the 1?

Another thing: Tangent is Sine / Cosine. Would this count as one 2 or two?

This is another reason I wouldn't want to use trig functions. Besides, I'm fairly certain I've seen 30 done without trig functions in class. For the life of me I can't remember the process, but I'm pretty sure it was done.

[Rocinante]

Cosecant is the realtionship between hypotenuse side over the opposite side so unless we are not using trig functions at all , cosecant should be fine. Sine is not needed to find the Cosecant , the only real reason you need to know that Sine and Cosecant are inverses is because graphing calculators do not have a Cosecant button.

30 = (2+2+2)/.2
32 = 2 * 2 ^(2+2)

[Karl]

That's true.

It just seems like every time we say, "Oh, I don't see any way we could do this without using trig functions" we end up finding a way.

I was all proud of myself for figuring out 30, too, when I come to post and here you've already posted it.

[Blixinator]

Are the trig functions just being used as a cheap way of getting a 0 or 1 with only one 2?

If that's the case, I think we should try harder. Because I'm fairly certain that several of the numbers could be expressed without sin(2) ~ sin(0) ~ 0 or cos(2) ~ cos(0) ~ 1

I also think we could do without the [[]] tags. I'm pretty sure I saw the word integer in the first post.

Come on people!

Remeber, there aren't just + - * and /.

Try using some !s and %s

And I think I just found a way around using trig to get a 0 with just one 2.

Remember from calculus that the derivative of a constant is 0?

dy/dx(2) = 0

Now go back and get rid of some of those nasty [[]]s.

[Karl]

The idea is just to keep it simple. The simpler the solution, the better. Trading truncation for a derivative is just going from one complication to another.

[Justin]

Except d (2) / dx is exactly zero, unlike sin (2), which is only approximately zero.

[Blixinator]

Only derivatives are nice clean math. Type castings are dirty.

[denacioust]

Can we round numbers off? That'd be pretty handy.

[Karl]

All I'm saying is that ideally we would would keep it as simple as possible. By all means use what you need to get to the value, but if another solution comes along that is simpler, we'll switch them.

[Blixinator]

I'm pretty sure a more precise answer us better than a simpler one.

Or we'd all be going around willy-nilly dividing by zero and saying pi is exactly 3.

Also, difficulty is objective. Or is it subjective? I hate words.