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You Can Make Any Number Out of Four 4s Because Math Is Amazing (youtube.com)
Andrew Moseman, writing for Popular Mechanics: Here's a fun math puzzle to brighten your day. Say you've got four 4s -- 4, 4, 4, 4 -- and you're allowed to place any normal math symbols around them. How many different numbers can you make? According to the fantastic YouTube channel Numberphile, you can make all of them. Really. You just have to have some fun and get creative. When you first start out, the problem seems pretty simple. So, for example, 4 - 4 + 4 - 4 = 0. To make 1, you can do 4 / 4 + 4 - 4. In fact, you can make all the numbers up to about 20 using only the basic arithmetic operations of addition, subtraction, multiplication, and division. But soon that's not enough. To start reaching bigger numbers, the video explains, you must pull in more sophisticated operations like square roots, exponents, factorials (4!, or 4 x 3 x 2 x 1), and concatenation (basically, turning 4 and 4 into 44).
"and concatenation"
No, that's not really maths,
deleting the extra space after periods so i can stay relevant, yeah.
arccos(-4*4/4/4)
If you want to be seen, stand up. If you want to be heard, speak up. If you want to be respected, sit down and shut up.
Also, you can't actually get "any number" since you'll max out at a number when using the highest performing operator which I think would be the power operator, so the highest number you can make is 3.4028236692093846346337460743177e+38 (i.e. 4^4^4^4). And you'll be unable to get most of the numbers on the way there since you run out of the 4's to use to fill in gaps.
My question would be, just how high can you get before you miss a whole number?
Variants of this problem have been along for a very long time. It was popular as a recreational game in the 1930s and is now used more with Middle School students as a way of getting them more familiar with different operations.
If one uses instead of 4 uses 1 and has a restricted operation set or the like then you can get some actually non-trivial math by asking how many 1s you need. For example, one can define the integer complexity of a number n, denoted by ||n|| as the minimum number of 1s needed to write n as a product or sum of 1s with any number of parentheses. Thus for example, 6=(1+1)(1+1+1) shows that ||n|| 1 one has 3log_3 n - a better result is actually known that what is in that post, and I'm writing it up now. The other person mentioned there is Harry Altman who probably has thought more about different notions of complexity of numbers than anyone else at this point (his dissertation was on the subject). We had a joint paper https://arxiv.org/abs/1207.4841 that is mostly readable.
Another neat variant of this is again looking at 1s and allowing just addition, but allow that once you have made a number you can use it again, and now you count how many operations you have used. So for example, in this framework, you can use 3 additions to get 6 because 3=1+1+1 and 6=3+3. This is the addition chain complexity https://en.wikipedia.org/wiki/Addition_chain and is closely related to how to quickly exponentiate objects (such as matrices, or specific numbers mod another group) https://en.wikipedia.org/wiki/Addition-chain_exponentiation which is very important for doing a lot of practical algorithms efficiently (such as RSA and Diffie-Hellman).
The problem in the original post is essentially silly but it connects to a lot of neat, serious math. (Also, Numberphile is in general great.)
Except 5318008, naturally...
Too funny. If you knew anything about the Numberphile channel, you'd know these are real mathematicians, not some BS. It's really just a math-related brain teaser. I really enjoy their videos. Even if you're not into math, they are sufficiently nerdy that I think many slashdotters would appreciate them. In fact a couple videos ago they had an interview with Ronald Rivest who was one of the inventors of RSA encryption. He's a down-to-earth, articulate person. He mentiones how he invented the MD5 hash which was later shown to be flawed.
Anyway, yes it turns out with just log, square root, and multiplication, you can assemble any whole number between 0 and infinity with just four fours. Fairly useless, but a neat puzzle.
Yes, they meant positive integer. And yes, one can. Watch the video. In fact one only needs logs, division and square root operations to do it.
log(sqrt(4)/4) (log(4)(sqrt(sqrt(sqrt(4))))) = 3
That's the same as
log(base 1/2) of (log(base 4) of 4^(1/8))
(log(base 4) of 4^(1/8)) = 1/8 (4 to what power = 4 to the power of 1/8?)
So you have
log(base 1/2) of 1/8
Which is 3. Just add another sqrt() to increment the result.
log(sqrt(4)/4) (log(4)(sqrt(sqrt(sqrt(sqrt(4)))))) = 4
That's the same as
log(base 1/2) of (log(base 4) of 4^(1/16))
(log(base 4) of 4^(1/16)) = 1/16 (4 to what power = 4 to the power of 1/16?)
So you have
log(base 1/2) of 1/16
Which is 4.
Well I think it's triangular, so there.
If you really want four fours, then do 4-4+4-4 and N times the increment operator.
Actually, you can, if you allow complex arithmetic - ln(-4/4)/sqrt(-4/4)
File under 'M' for 'Manic ranting'
Factorial already breaks the rule by using integers less than 4.
Except negative numbers.
You just have to wait for the Universe to roll over.
You seem to be confusing operators (or, more generally, functions) with constants here. arccos is just an operator, much like negation, addition, subtraction, etc. Pi is a constant, a numerical value, it has no other interpretation.
If you allow every function to be an "operator", the solution is pretty trivial, as to construct any number p you could just simply take the constant function with 4 input parameters that has value p. If you allow only a finite set of "operators" you won't be able to construct any real from it with only finitely many applications of your operators, as there are unaccountable many reals.
If you restrict yourself to rationals you can get everywhere by just writing / somewhere between the four's and applying increment/decrement on both sides often enough.
My bad... I forgot to parenthesize the i*pi in e^(i*pi)=-1, above... although it does not change the conclusion... I just did not correctly type the expression in that one equation.
File under 'M' for 'Manic ranting'
4^444 > 4^4^4^4
Hmmm... no. 4^4 = 256. 4^256 = 1.3407807929942597099574024998206e+154. So 4^that = too big to easily calculate.
It looks like my previous estimate of 4^4^4^4 got calculated as ((4^4)^4)^4, when it needs to be done as 4^(4^(4^4)) to give the correct biggest number with standard operators.
In the spirit of Brainf*ck, I will use this to create a programming language called "Fourtran".
Table-ized A.I.
Factorial already breaks the rule by using integers less than 4.
Okay, then how about Gamma(x+4/4)...
And I can make any number I want with just ones and a single zero (if I want negative numbers or zero). Big deal.
But you cannot make any integer using just 4 ones, and that's what makes this a bit more interesting.
The summary uses the phrase "any number", but that is wrong. What they should have said was "any integer".
If the summary had correctly said "any integer", I doubt I would have read any further, and probably half the comments would not have been posted.
It's kind of a neat trick. If you already know what a logarithm is, you only need to watch from the 8:00 minutes in the video to 8:40 see the method.
The method uses only log, square root, and division.
BTW, making a googleplex by adding a long string of ones takes about the same amount of paper as their method that uses only 4 fours, depending on your handwriting.