# Let's discuss Mathematics

Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.

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You can call M(n) the number of people recruited on step n. That function is a^n. Then write N using that new M function.

2. ### Lohrenswald老仁森林

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Thanks for the jumpkick lol, I've figured the answer (by using your hint):

N(n)=(1-a^(n+1))/(1-a)

3. ### SamsonDeity

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I have a population of things, say genes. I think some of them will have a feature, say involved in metabolic syndrome. If I do a frequentist test for this feature I expect the ones that do not have this feature to produce a p value with the uniform distribution, and the ones that have this feature to produce a p value skewed towards zero.

If I am interested in selecting a subset of this population to reject the null hypothesis with a certain confidence I could do something like the Benjamini Hochberg procedure. If however I am interested in the characteristics of the whole population, such as how many have this feature or even what is their effect size distribution, what techniques could I use?

I have tried searching for this, but I really do not know what search terms to use and can find nothing. I have thought about it and can imagine assuming the true effect size follows a beta distribution in the cases where the null hypothesis is false and estimating the parameters of the distribution with monte carlo bayesian inference, perhaps with BUGS or STAN, feeding in as observations p values and effect size estimates. I feel someone must have thought about this before.

4. ### onejayhawkAfflicted with reason

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What about the proposition that 0.999... =/= 1.000...? The proof is trivial. 0.999... is an element of [0,1) and 1.000 is not.# QED

J

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false

L

6. ### warpusIn pork I trust

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Basically if you could prove that you would take down modern mathematics and we'd have to start from scratch

7. ### onejayhawkAfflicted with reason

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Demonstrate.

Not really. Calculus is built on the principle of close enough. They are definitely that.

J

8. ### Lohrenswald老仁森林

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since 0,999...=1 it's not in [0, 1)

you don't have a better arguement for why this isn't the case

9. ### onejayhawkAfflicted with reason

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I see, wave a magic wand.

You have stated one truth, 1 is not an element of [0,1). 0.999... is an element of the set {0.9, 0.99, 0.999...} which is a subset of [0, 1). Hence 0.999... is an element of [0,1). This is a contradiction and your assumption that 0,999...=1 is disproven.

J

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It's certainly NOT an element of the set {0.9, 0.99, 0.999...}. Or you could point out which one it is.

11. ### SamsonDeity

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The proposition can be writen as when x = 0.999... then x <> 1.
If you put [0,1) into wolfram alpha it translates it to :

[EDIT] Wolfram alpha images are not permanent. It says:
Inequality: 0 <= x < 1
x < 1 therefore x<> 1 does not seem a very strong argument.

Last edited: Nov 29, 2018

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J's argument can also be applied to the limit of 2^n when n-> infinity. All numbers in this set are in [0;+infinity[ (an interval that exludes +infinity), therefore the limit (ie +infinity) is also in [0;+infinity[

13. ### onejayhawkAfflicted with reason

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Why? That's like saying you can point to an infinite number. You are just saying that the set is not finite, which we already knew.

J

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Ok so according to you +infinity is an element of the set {1, 2, 4, 8, 16, 32...). Good for you I guess, but you're the only person in the world who believes that.

15. ### Lohrenswald老仁森林

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I'll try to explain why 1=0,999... as best as I understand it
In , the "number density is infinite". If two numbers a and b are different and b>a, then there's an infinite amount of numbers r that's a<r<b
so if 0,999... and 1 where different numbers, you should be able to produce at least one number that fits between them, like the average
however, you can't

I'm sure there's a much more vigurous "version" of this proof, but this is roughly how I remember learning about it

and speaking of me learning about it: I thought initialy that 0,999... and 1 were different, however I learned about it, saw it was true, accepted it and then moved on with my life

like, what do you want to achieve by this inanity, hawkboy?

J

16. ### onejayhawkAfflicted with reason

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No, and don't be absurdist.

What I claim is that your set contains all numbers of the form 2^n. You cannot point to the largest element of either set.

This is a difficulty with irrational numbers--you cannot write them out. If you were to try to define the highest number less than 1, 0.999... is a candidate. Instead, we absurdly state that these sets have no highest number and that open segments have no endpoints. Like Schroedinger's cat, things only resolve on inspection. Regardless, nothing in your post contradicts any part of my proof.

As to the value, abstract mathematics needs no defined purpose. They are thought experiments (also like Schroedinger's cat).

J

Last edited: Nov 30, 2018

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The set {0.9, 0.99, 0.999...} contains all the numbers in that form with a finite number of 9s. That's what you don't seem to understand.

Abstract mathematics are not a thought experiment, it's not just a mental game. Some things in abstract mathematics are true, others are false, and your assertion is 100% false. The proof for 0.999999... = 1 was given to you several times in the other thread about this.

18. ### onejayhawkAfflicted with reason

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False. You know better. It is a bounded, infinite set with an accumulation point that lies outside the set.

Say rather attempted proofs--failed attempts. Regardless, the discussion is profitable, or it can be. No one can make you stretch your mind.

J

19. ### formerdc81Chieftain

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Mathematics are indeed arbitrary in the sense that the axioms can be chosen however you wish. HOWEVER, once a set of axioms is chosen, all the logical consequences of it follow and it is not possible to then wave your hands and say "it's just a theoretical construct and I can believe whatever I like" towards results you personally disagree with.

IN your case, you are making a basic mistake. If S is a bounded sequence that can be contained in an open set X, then the sup and inf of S lie NOT in X, but in the closure of X.
Proof: (I'll just show it for the supremum since the proof is the same for the other extrema.)
Since S is a bounded sequence, sup S is not infinite. By definition if s is an element of S, s is less than or equal to sup S.
If s = sup S, this is trivial because sup S is in X, and thus, is in the closure of X.
If s < sup S, S is an infinite sequence. More importantly, by the definition of supremum, for any r > 0, we can always find an n such that s_n is the nth element of S and sup S - s_n < r. However, since the above is true for any r, sup S by definition exists in the closure of X.

Now back to your problem. The closure of the set X [0.9, 1) is [0.9, 1]. Thus, if S = (0.9, 0.99, 0.999, ....), then S is in X and sup S = lim n->infinity of s_n is in the set [0.9, 1]. Since clearly sup S is not in X, sup S must be in the set cl(X) - X = {1}. Hence, 0.999.... = 1.

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20. ### onejayhawkAfflicted with reason

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Interesting approach though you seem to assume your conclusion. I would like to see a more definitive explanation of why the closure of S is relevant, since C(S) = [0.9,1.0) union 1. This seems equivalent of lim (0.9+0.09+0.009+...) = 1.0 which is inadequate for reasons given previously.

On a silly level, I ran across this. Assume you have ordered a sausage and mushroom from your local Italian restaurant. If the radius of the pie = z and the thickness = a, then the volume of your repast is V = Pi*z*z*a.

J

Last edited: Jan 9, 2019