B How many numbers can you make with four 4s?

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The discussion revolves around the mathematical challenge of generating different numbers using exactly four 4s and various mathematical symbols. Participants explore methods to represent specific numbers, including 10 and π, while debating the validity of using functions and arbitrary constants in their solutions. There is a consensus that without explicit restrictions on the operations allowed, the problem can become trivial, as many solutions can be constructed. The conversation highlights the importance of defining the parameters of the problem to ensure meaningful engagement. Ultimately, the challenge remains an intriguing puzzle for mathematical exploration.
  • #31
mfb said:
Both at the same time with a total of 4 4s? Yes they can be expressed like that, but then there is no point in introducing integrals and whatever because you have expressed 10 via 4s already.
Thus one has to pick which method they follow ... (The primitive one is already in the video, so the question about 10 shouldn't have been asked in the first place ! ...)
 
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  • #32
Stavros Kiri said:
But back on track with the exact solution (not approximation), for any real number x (e.g. π) [integer, rational, or irrational ...].
A simple counting argument shows that this is not possible. The cardinality of the reals is larger than the cardinality of the set of finite expressions.
 
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  • #33
jbriggs444 said:
A simple counting argument shows that this is not possible. The cardinality of the reals is larger than the cardinality of the set of finite expressions.
Unless you use continues functions (in the variables and/or constants) to invert from the 4 4s numbers business. Then the cardinality can even be |R|, or |RxR| etc., if you fix appropriately variables and constants for the functions ... . But the point is that we say that this is not actually allowed (see previous posts about restrictions on arbitrariness and inversion, as well as the examples of continues functions there).

So basically, yes, I agree. Do you?
 
  • #34
Stavros Kiri said:
Unless you use continues functions (in the variables and/or constants) to invert from the 4 4s numbers business. Then the cardinality can even be |R|, or |RxR| etc., if you fix appropriately variables and constants for the functions ... . But the point is that we say that this is not actually allowed (see previous posts about restrictions on arbitrariness and inversion, as well as the examples of continues functions there).

So basically, yes, I agree. Do you?
I am not sure that I know what you mean by "use continuous functions". But if one is going to use a function there is a requirement that one identify the function. That is why the general approach given in post #2 (just use an arbitrary constant C that gives the desired result) is not acceptable.

It does not matter how rich the specification language is. The number of things that you can specify cannot exceed the number of valid specifications.

It does not matter that the set of not-necessarily-continuous functions has cardinality ##|\mathbb{R}^{|\mathbb{R}|}|##. The set of the ones you can finitely specify has cardinality no more than |N|.
 
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  • #35
jbriggs444 said:
I am not sure that I know what you mean by "use continuous functions"
Real continuous functions on the [also continuous] variables and constants, e.g. say (x, A) [continuous quantities (i.e. their domain is a continuous subset of R)], plus f(x;A) [or notation like fA(x)], for one variable and one constant etc.], continuous functions on their domain, a requirement for simplicity to give you easily every choice possible, that would make the puzzle trivial.

But as mentioned, and mostly agreed, none of that should be allowed, according to the video and OP. (Just the 4 4s, used only once [each], and basic step operations, basic functions [inverses I think are allowed*, if there are no arbitrary constants (that you can covenienty fix etc.)], mathematical symbols etc. ... ; no other numbers or arbitrary constants etc.)

So I think overall we probably agree, and for the rest of your post I completely agree and liked them both [posts of yours].

But although we cannot get all real numbers, we can try to get as many as we can, following:
Greg Bernhardt said:
How many different numbers can you make? It's best to think of a number and then try to make it

Thus the puzzle and challenge is not trivial, but a real and valid one (Note that in the video he deals only about 0, 1, 2 ... n ... ∞ , with "Dirac's busting" at the end ... , but the OP actually gave more challenge ... (as many numbers as we can, i.e. reals ...; π was one new accomplished etc.) ).

* e.g. ~as in mfb's solution for π (earlier), as both cos and arccos I think should be allowed (one is the inverse of the other, both basic functions ...).
 
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  • #36
Stavros Kiri said:
Real continuous functions on the [also continuous] variables and constants, e.g. say (x, A) [continuous quantities (i.e. their domain is a continuous subset of R)], plus f(x;A) [or notation like fA(x)], for one variable and one constant etc.], continuous functions on their domain, a requirement for simplicity to give you easily every choice possible, that would make the puzzle trivial.
I still do not understand this.

A constant function is continuous. But if you allow for arbitrary constant functions to be invoked (specified how?) then the problem becomes trivial. So one must have a limited roster of functions.

If you are allowed to include functions with no arguments (also known as zero-place functions or constants) then the problem becomes trivial. Given one such function (call it k), one can write ##\frac{k+k+k+k}{k+k+k}## or similar to encode any rational number. So we cannot allow zero-place functions . (Not even zero, see below).

If one is given a one-place monotone increasing function f and an unrelated one-place monotone decreasing function g then one can write expressions like f(f(g(g(f(g(x)))))) and use the pattern of f's and g's to encode values. So we must exercise care with the set of one-place functions that are permitted, otherwise the problem becomes trivial. [That was the point of post #26].

For two-place or greater functions, the fact that no constants (not even zero) are allowed is helpful. With only four constant terms (the four fours) to use, a maximum of three two-place functions can possibly be used. e.g. f(4,g(4,h(4,4))). Similarly, at most two three place functions or at most one four-place function. The point is that the user is prevented from chaining together an arbitrarily long string of function compositions.

So. Which functions do you have in mind, consistent with the above guidelines?

Plus, minus, times, divide, exponentiation and unary negation?
 
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  • #37
jbriggs444 said:
then the problem becomes trivial. So one must have a limited roster of functions.
That's what I also said.+Many interesting points in your reply.

+
jbriggs444 said:
Plus, minus, times, divide, exponentiation and unary negation?
In the video he uses more, e.g. concatenation, percentage and a bunch of special tricks ...
 
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