Re: The seven step series

From: Bruno Marchal <marchal.domain.name.hidden>
Date: Wed, 8 Jul 2009 19:31:00 +0200

On 08 Jul 2009, at 15:43, m.a. wrote:

> Second try:
>> (power {1, 2, 3}) = ? {{ }, {1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}}

This is far better! Not yet correct though.

I gave you the hint that there are 8 elements. Let us count:

The empty set { } ..................................1
Three singletons {1}, {2}, {3}................3
Two doubletons {1,2 }, {2,3 }................2
The biggest subset {1,2,3}..................1

1 + 03 + 02 + 01 = 7

A subset is missing! Can you see which one?


> And I give you a little subject research: if a set x has n elements,
> how many elements are in (power x)?
>
> How's this for a wild guess? I have a feeling that it's missing the
> accolades, but I have no idea where to put them
>
> (power {x}) = n(n-1) (n-2)...(n-x+1)
> x!
>

Good intuition that there is a problem with the accolades. Although
your expression is not *missing* accolades. Actually it has *too much*
accolades.

If x represents a set, for example the set {1,2}, it means that in the
formula x can be substituted by {1,2}. we could write that x = {1, 2},
the accolades are in x, if you want.
So on the left, you should have written (power x), like in the
enunciation of the subject research, actually.

If x is the set {1,2}, (power x) is (power {1, 2}). But (power {x}) is
(power {{1,2,3}}), i.e. the powerset of {{1,2,3}}, which is {{ },
{{1,2,3}}}.
(Power {x}) looks like the powerset of an indeterminate singleton, a
set with only one element.

You could have written this:

When x has n elements, then (power x) = n(n-1) ... (n-x+1) / x!

Let us see. After all you already computes the powerset of { }, which
is the set with 00 element, and you told me (power { }) = {{ }}. So it
has one element, and your formula should confirm this, and ... well,
your formula begins by n multiplied by something, if n = 0 then we
will get 0, because 0 times any number gives 0. But we have just seen
that (power { }) = {{ }}, which is a set with 1 element. So your
formula is already contradicted by the first example. Hmm...

May be that was bad luck, and sometimes in math the first example is
also the trickiest, so let us look for n = 1. Let us take a set with
one element, like {24}. Its power has 2 elements: {{ } {24}}, and you
can guess that all singletons (set with two elements) have the same
number of elements in their power set. So the answer is 2, in this
case. If x has 1 element, the powerset of x, (power x) has two
elements. Your formula should give 2, when n is equal to 1.
Let us see ...

it gives 1(1-1) ... (1 -

... but now, what could you mean by (n - x ...). ???


n is supposed to represent a number, x is supposed to represent a set,
how could I, or you, subtract a set from a number?

So I'm afraid that your formula is senseless, although I will perhaps
take the time, in some future, to explain why there *is* an atom of
truth in it! The correct formula is much simpler, though.

Morality: if you have a theory, or a formula, test it on what you
already know before submitting to publication!

To find the formula, you could try first the tedious brut force (if
this is english). (Few mathematicians admit to do that, but all
mathematicians do it!)

The number of sets included in { } = 1 (you have seen that). If x
has 0 elements, (power x) = 1.
The number of sets included in {a} = 2 (you have seen that). If x
has 1 elements, (power x) = 2.
The number of sets included in {a, b} = 04 (you have seen that). If x
has 2 elements (power x) = 4.
The number of sets included in {a,b,c} = 8 (cf the hint, exercise
above). If x has 3 elements (power x) = 8

Could you compute and/or guess the number of sets included in
{a,b,c,d} ?

And what about {a,b,c,d,e}?, and {a,b,c,d,e,f} ?, and ... ?

Bruno

PS Did I say that (power x) is called the powerset of x? It could be
better to write (powerset x) instead of (power x), to emphasize that
the powerset of x is the *set* of the sets included in x. OK? (Sorry).

http://iridia.ulb.ac.be/~marchal/




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Received on Wed Jul 08 2009 - 19:31:00 PDT

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