>tfw non-constructive existence proof
Those proofs are often quite fun desu.
>>7749114
I hate that shit.
>Thing exists because if it doesn't then it means our assumptions are wrong.
>>7749178
>equivalence
That's the identity predicate you're talking about there you greaseball.
>having a problem with this
pigeonhole principle is non-constructive too
>>7749114
Theorem: there exists irrational numbers when raised to a irrational power are rational
Proof: ∛2^√3 is either rational or irrational, if irrational then (∛2^√3)^√3=2 ∎
>>7749248
irrational numbers are not well defined.
>>7749251
Aw come on, there are some fun measure theory arguments in that vein. I saw Bishop use it to prove this sweet lemma that any compact set in R^n contains a sort of generalized cantor set of arbitrarily less dimension and you can dial the separation of nested segments up as much as you want.
>>7749258
Suck my dick, they are
>>7749258
>irrationals are not well defined
top kek
>>7749114
>Proof. It is an easy exercise.
>>7749271
>arbitrarily less
should be arbitrarily little less
>>7749114
Those are the best, because you don't have to construct it. You did the fun part (proving it), without having to potentially do tedious shit like applying it.
>>7749114
>>7750079
They're easy to write and practically useless. How is this "the best" again?
>intuitionistic logic
why cant I construct the reals like this to show that they are countably infinite?
>>7751154
good question
>>7751127
Generalizes classical logic and lets you talk about more general branches of math (especially if you include anticlassical axioms).
>>7751154
Because this method only gives you strings of finite length. You could use it to show that all the rationals are countably infinite (because every rational would appear somewhere on your tree) but reals would have to be binary strings of infinite length.
>>7751154
because they aren't
>>7751166
>Because this method only gives you strings of finite length. You could use it to show that all the rationals are countably infinite (because every rational would appear somewhere on your tree) but reals would have to be binary strings of infinite length.
I thought about that too but then again a real number could be interpreted as the results of all iterations of some function that generates it, all these iterations are in that list and therefore all real numbers must be, but there is no order.
>>7751182
no. you are generating all finite numbers. what you're saying is that an infinite sequence of elements in your tree converges to any real number, and yes, this is a property of the rationals. you're building rationals.
>>7751182
Thing is that then you're equating these as sequences of elements on your list. This set of sequences ends up being larger.
For reference, take a finite set and call it an "alphabet". Because it is finite it has an ordering. Then create sequences of elements in your alphabet and call them "words", including the empty word. So for example if your alphabet is {a,b} then you have words like (b), (a,a,b,a,b), (), and so on. Consider the set of all words in your alphabet, note that it is countable. First you have the words of zero length, then the words of length 1, then the words of length two, and so on, each of which has finite length and an order (given by the alphabetical order). However while your set of words contains all words with integer length for all integers (arbitrarily large) it does not contain words of infinite length.
Next, you can call a set of words a language. Each language is thus a subset of the set of all words. So the set of all languages is uncountable (it is the power set of the set of all words).
Here you have done something similar, your set of sequences is not exactly the power set of "all items on your list" but it's still an uncountably large subset of that power set. There's some handwaving here but the argument is ugly in full formality.
(cont.)
>>7751221
(cont.)
There's this logical subtlety that many books and courses just outright fail to mention (handwave away) whenever you're using counting arguments and diagonal arguments. Namely that
>When you iteratively construct a list of elements your list only contains elements of finite length (length is an integer). You can only claim that set containing the elements of the list is countable. It does not infinite length elements (length is not an integer) and a set of subsets of your list is not necessarily countable.
>When you're using a diagonal argument you have this extra assumption that says all of your elements have finite representation. This means that when you construct the element that differs for each element on the diagonal, you get two things. Either the constructed element has infinite representation or the constructed element does not appear on the list. This is why you cannot apply the diagonal argument to integers and thus prove they are uncountable.
>>7751234
>When you're using a diagonal argument you have this extra assumption that says all of your elements have finite representation. This means that when you construct the element that differs for each element on the diagonal, you get two things. Either the constructed element has infinite representation or the constructed element does not appear on the list.
Thanks for the answer. If thats so then why is cantors Diagonal argument valid? There is no base in wich you could write down his argument in wich every faction has a finite representation (f.e. 2/3 = 0,666... in base 10), therefore it has elements of infinite length.
I assume if we want to have a list with infinite long elements we need an algorithm representing each element, this means a list that has all real numbers simply needs all possible permuations of "infinite-sum-of-fractions algorithms", right?
>>7751323
>Thanks for the answer. If thats so then why is cantors Diagonal argument valid? There is no base in wich you could write down his argument in wich every faction has a finite representation (f.e. 2/3 = 0,666... in base 10), therefore it has elements of infinite length.
Well there's two things. For counting the rationals we don't deal with the rationals in decimal form. That technique of representing numbers originates from Stevin and it's often avoided in formal mathematics because, well, it's hard to deal with formally. Instead we use the fraction representation of rationals (and just count some numbers multiple times).
https://en.wikipedia.org/wiki/Simon_Stevin#Decimal_fractions
However, with regards to the diagonal argument being used to prove uncountability of the reals in decimal representation, there's formally a bunch of extra assumptions that are also often handwaved away. Unfortunately I've never seen a truly satisfactory detailed exposition of this so I won't attempt to convince you of something I don't completely believe.
>I assume if we want to have a list with infinite long elements we need an algorithm representing each element, this means a list that has all real numbers simply needs all possible permuations of "infinite-sum-of-fractions algorithms", right?
I'm not sure this works either. The reason being that the formal way to deal with algorithms is as computable functions (then your sequences are recursively enumerable sets enumerated in some way). However, a standard result in computability theory is that the set of all computable functions is countable.
In other words, in order to describe your set of functions (which you may attempt to do by using some version of the axiom of choice) then you'll automatically be leaving the set of computable functions (you'll be describing functions for which no [math]finite[/math] algorithm exists).
>>7749248
Irrational, algebraic numbers (such as cube root of 2) raised to an algebraic irrational, by the Gelfond–Schneider theorem, are transcendental, meaning that you're wrong.
In case you don't know what an algebraic number is, it's the solution to a finite polynomial with rational coefficients (x^5-x-1=0, x is algebraic but can't be expressed in regular square or fifth roots or anything like that)
>>7752010
I'm not that guy but you're wrong.
So, suppose that this number is irrational.
∛2^√3
Note that √3 is irrational and then compute
(∛2^√3)^√3
=∛2^(√3*√3)
=∛2^(3)
=2
Thus we have an irrational raised to another irrational producing an integer. This is a fairly well known result and a sort of poster-child for nonconstructive proofs though normally this proof is done with square roots of 2.
>>7752010
But [math] 2^{ \sqrt{3} } [/math] is transcendental, funnily we know this by the Gelfond-Schneider theorem. So [math] \left( 2^{ \sqrt{3} } \right) ^{ \sqrt {3} [/math] is a transcendental to the power of an algebraic, and so isn't covered by the theorem.
>>7749114
>tfw proof by "drawing a fucking picture"
>>7751154
At which step would you count 0001 0001 ... ,0001 0001... ?
>>7749258
that's why I come back to /sci/
I made a small mistake and i fixed it
>>7752222
>At which step would you count 0001 0001 ... ,0001 0001... ?
Well lets look at what you are actually saying, you say there is a function that generates the sequence 0001 0001 ... ,0001 0001... , I assume what you mean is a function like this:
1) 0001,0001
2) 00010001,00010001
3) 000100010001,000100010001
x) 0001... x times , 0001.... x times
What I can tell you, because im permutating all possible combinations of 0 and 1s, is that every number this function generates is in my list. But the function itself, wich you might call a "number" isnt.
>>7752386
forgot le pic
>>7752386
you're not gonna construct reals, only a subset of rationals
>>7752405
ok, this is something I finally understand
