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Sythe Users: What Do They Know? Do They Know Things?? Lets Find Out! #5

Can't until you provide set definitions. You've rejected existing definitions so you need to provide your own.

well, you know that 'N' = {0, 1, 2, 3, 4, ..., 'infinity'}

you know that 'N' is countable,

now you have to prove |R|*> |N| using mathematical proofs
 
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well, you know that 'N' = {0, 1, 2, 3, 4, ..., 'infinity'}

you know that 'N' is countable,

now you have to prove |R|*> |N| using mathematical proofs


they have no knowledge of the relationship between naturals and real numbers, therefore you can't just assume that one is a subset of another

Provide definitions for the sets.

If you can't assume one is a subset of another it must be because the definitions do not imply they are subsets of one another. Therefore most likely your definitions are impossible definitions.

Provide them.

The a priori or axiomatic reasoning you are asking for is implied by and contained completely within the definitions of the sets.
 
Provide definitions for the sets.

If you can't assume one is a subset of another it must be because the definitions do not imply they are subsets of one another. Therefore most likely your definitions are impossible definitions.

Provide them.

The a priori or axiomatic reasoning you are asking for is implied by and contained completely within the definitions of the sets.

alright, let's say that the overlords know that the Natural numbers are included in what is called the Real numbers, you still have to prove that one infinity is bigger than the other infinity with a valid proof.

You can't just assume that for example, because {0.1} is not in the Naturals, that |R| > |N|, because {-1} is not in the naturals, it's in the integers (Z), but 'Z' has as much elements as 'N' ( |Z| = |N| )


poor walter...
 
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alright, let's say that the overlords know that the Natural numbers are included in what is called the Real numbers, you still have to prove that one infinity is bigger than the other infinity with a valid proof

You're just asking for a proof that reals are uncountable set and naturals are a countable set but this is just down to definitions as well.

Posing an axiomatic proof question without definitions is like trying to wash your windscreen when you don't have a car.
 
You're just asking for a proof that reals are uncountable set and naturals are a countable set but this is just down to definitions as well.

Posing an axiomatic proof question without definitions is like trying to wash your windscreen when you don't have a car.

But the definitions didn't just appear, there are ways to prove them, just like you can't just assume the sun is not a spotlight

The overlords want walter to prove to them, through mathematical proofs, that |N| < |R|

i've edited OP to be more specific: Any valid mathematical proofs can be used to solve the problem
 
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But the definitions didn't just appear, there are ways to prove them and that is exactly what the overlords ask of Walter to spare his life

Walter needs to show them proof that |N| < |R|

Lol go on how do you derive a definition for natural numbers with pure mathematics.
 
Alright, walter has been beheaded. poor guy ...

here is how he could have proven to the overlords that the infinite set 'R' was greater in size than the infinite set 'N' :

given the fact that 'N' is countable

All he had to do was to show that |R| > |N|.

He could have done it in a few different ways, here is a way he could have used:

Step 1: he could have shown that |[0,1[|*> |N|.

To prove that he could have shown that there existed no surjective function 'f: N -> [0,1[' for any Naturals.

which by definition of surjective functions means the following: there is at least one element in the interval [0,1[ that is pointed by no element from the naturals

This is easily shown by the following:

Be 'a<n,i>' numbers from 0 to 9 that match the following

f(0) = 0.a<0,0>a<0,1>a<0,2> ...
f(1) = 0.a<1,0>a<1,1>a<1,2> ...
f(2) = 0.a<2,0>a<2,1>a<2,2> ...
.
.
.

these functions obviously represent any real numbers in the interval [0,1[.

now let's build a function that will never meet any numbers built by any of those functions

Be 'B(i)' , numbers from 0 to 8 assigned in the following order

B(i) = 4 <if a<i,i> != 4>
B(i) = 5 <if a<i,i> == 5>

we then have the real number:

b = 0.B(1)B(2)B(3)B(4)...

the real number 'B' is obviously inside the interval [0,1[ but will never be equal to any value obtained by any 'f' functions.

we then have 'B' != f(i) for any Natural numbers

so B is never part of the image of any f-function.

we then have that obviously, |[0,1[| >*|N| since there exist no surjective function f: N -> [0,1[.

Now, going from there to the next step is easy:

We know that [0,1[ is included inside the Real numbers.

so by definition we have, |[0,1[| <= |R|.


Since we found in Step 1 that |N| < |[0,1[|, we then have this relationship between the three sets:

|N| < |[0,1[| <= |R|

which by the transitivity of '<' is equal to:

|N| < |R|.

We then have it, |R| > |N| thus proving that the cardinality of the real numbers is greater than the cardinality of the naturals
 
Such an easy question. There is no one to one correlation, and so the infinite sets are unequal. Natural numbers are 1, 2, 3, 4, etc., real numbers include an infinite number of numbers between each integer as well.

Showing that there's not one-to-one correlation, it's a matter of Canton's diagonal proof. Essentially take a real number who's decimal stretches to infinity, as it's able to do due to the infinite nature of the set of real numbers. On the next line, copy the number, but change the second digit to anything but the same number on the previous line.

Example:

1.123456789...
1.223456789...
1.224456789...
1.234556789...

And so on. In this respect, you can produce an infinite number between 1 and 2.

Now, since there are infinite numbers between 1 and 2, there are infinitely more sets within real numbers than in natural numbers.

Whereas
.1...2...3...4...5...6...7...8...9
.&#8595;...&#8595;...&#8595;...&#8595;...&#8595;...&#8595;...&#8595;...&#8595;...&#8595;
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9


Observe that 1.5 = 1.50000...
Apply Cantor's diagonal theory

1.50000...
1.51000...
1.51100...

Etc. All of these numbers will fall on the number line between 1.5 and 1.6, but there's no number on the natural number line to correlate to these unique real numbers. Assuming the decimal goes on forever, as pi does (which is a real number), there can't be a natural number to correlate if the irrational number never terminates, because as per cantor's diagonal, a single change along the infinite string of numbers results in an entirely unique and unrepresented real number.
 
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Alright, walter has been beheaded. poor guy ...

here is how he could have proven to the overlords that the infinite set 'R' was greater in size than the infinite set 'N' :

given the fact that 'N' is countable

All he had to do was to show that |R| > |N|.

He could have done it in a few different ways, here is a way he could have used:

Step 1: he could have shown that |[0,1[|*> |N|.

To prove that he could have shown that there existed no surjective function 'f: N -> [0,1[' for any Naturals.

which by definition of surjective functions means the following: there is at least one element in the interval [0,1[ that is pointed by no element from the naturals

This is easily shown by the following:

Be 'a<n,i>' numbers from 0 to 9 that match the following

f(0) = 0.a<0,0>a<0,1>a<0,2> ...
f(1) = 0.a<1,0>a<1,1>a<1,2> ...
f(2) = 0.a<2,0>a<2,1>a<2,2> ...
.
.
.

these functions obviously represent any real numbers in the interval [0,1[.

now let's build a function that will never meet any numbers built by any of those functions

Be 'B(i)' , numbers from 0 to 8 assigned in the following order

B(i) = 4 <if a<i,i> != 4>
B(i) = 5 <if a<i,i> == 5>

we then have the real number:

b = 0.B(1)B(2)B(3)B(4)...

the real number 'B' is obviously inside the interval [0,1[ but will never be equal to any value obtained by any 'f' functions.

we then have 'B' != f(i) for any Natural numbers

so B is never part of the image of any f-function.

we then have that obviously, |[0,1[| >*|N| since there exist no surjective function f: N -> [0,1[.

Now, going from there to the next step is easy:

We know that [0,1[ is included inside the Real numbers.

so by definition we have, |[0,1[| <= |R|.


Since we found in Step 1 that |N| < |[0,1[|, we then have this relationship between the three sets:

|N| < |[0,1[| <= |R|

which by the transitivity of '<' is equal to:

|N| < |R|.

We then have it, |R| > |N| thus proving that the cardinality of the real numbers is greater than the cardinality of the naturals

According to your previous post there is no known relation between reals and naturals that you can depend on.

So for example when you said/copypastaed:
We know that [0,1[ is included inside the Real numbers.

How do they know that interval, defined from natural numbers, is in the real numbers?

Once you declined to define your sets whilst also rejecting the standard definitions of the sets the problem became impossible to answer.

Of course all you really wanted to do was repeat this proof from your first year text so ok.
 
According to your previous post there is no known relation between reals and naturals that you can depend on.

So for example when you said/copypastaed:


How do they know that interval, defined from natural numbers, is in the real numbers?

the [0,1[ interval is not actually a naturals interval, it is a real number interval [0,000000000000... , 0,999999999999...[

but seeing as how rational numbers act when dealing with infinite lines of 99999999999, the interval is writen as [0,1[.

In the proof, there is no link between reals and naturals that is being used other than the fact that they are both numbers
 
Such an easy question. There is no one to one correlation, and so the infinite sets are unequal. Natural numbers are 1, 2, 3, 4, etc., real numbers include an infinite number of numbers between each integer as well.

Showing that there's not one-to-one correlation, it's a matter of Canton's diagonal proof. Essentially take a real number who's decimal stretches to infinity, as it's able to do due to the infinite nature of the set of real numbers. On the next line, copy the number, but change the second digit to anything but the same number on the previous line.

Example:

1.123456789...
1.223456789...
1.224456789...
1.234556789...

And so on. In this respect, you can produce an infinite number between 1 and 2.

Now, since there are infinite numbers between 1 and 2, there are infinitely more sets within real numbers than in natural numbers.

Whereas
.1...2...3...4...5...6...7...8...9
.&#8595;...&#8595;...&#8595;...&#8595;...&#8595;...&#8595;...&#8595;...&#8595;...&#8595;
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9


Observe that 1.5 = 1.50000...
Apply Cantor's diagonal theory

1.50000...
1.51000...
1.51100...

Etc. All of these numbers will fall on the number line between 1.5 and 1.6, but there's no number on the natural number line to correlate to these unique real numbers. Assuming the decimal goes on forever, as pi does (which is a real number), there can't be a natural number to correlate if the irrational number never terminates, because as per cantor's diagonal, a single change along the infinite string of numbers results in an entirely unique and unrepresented real number.

That's basicly right, you create a number that switches digits on the diagonals to make sure that this number will never be equal to any numbers created by any functions thus making sure that there is never any surjective functions.

Once that is proven the proof is basicly done
 
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the [0,1[ interval is not actually a naturals interval, it is a real number interval [0,000000000000... , 0,999999999999...[

but seeing as how rational numbers act when dealing with infinite lines of 99999999999, the interval is writen as [0,1[.

lol you slimey butthole you even edited your post. Look at this:

The overlords believe in creationism, therefore they have no knowledge of the relationship between naturals and real numbers, therefore you can't just assume that one is a subset of another.


what you're induction proof tells the overlords is that 'N' would reach "infinity" 'faster' than "R" , but it doesn't prove that 'N' is of lesser size than 'R'


The overlords are starting to set-up the guillotine

You removed the underlined part after posting it.

Your proof is impossible given the changes you made to the problem, namely that you don't have a definition for reals or naturals.

Also there's no reason to disregard the inductive proof that one set is larger than the other. Your math lecturer probably told you not to use it. But it's valid. If it's true for any N arbitrarily large then this is true for an N infinitely large as well.
 
lol you slimey butthole you even edited your post. Look at this:



You removed the underlined part after posting it.

Your proof is impossible given the changes you made to the problem, namely that you don't have a definition for reals or naturals.

Also there's no reason to disregard the inductive proof that one set is larger than the other. Your math lecturer probably told you not to use it. But it's valid. If it's true for any N arbitrarily large then this is true for an N infinitely large as well.

I edited it and said I did so.


The syllogism I presented remains true irrespective of the number of elements in the set, provided the number of elements is two or greater.

For example if you allow 2 elements in your set of naturals then it is {0, 1};
Therefore the set of integers is {-1, 0, 1}. If the set of reals at least must contain the set of integers then {-1, 0, 1} is a subset of the reals and therefore the reals must contain more elements than the naturals.

This works for any number of naturals >= 2, even if you take the limit to infinity (as in the case of the true set of naturals.)

E.g. n=4 it would be count({0, 1, 2, 3}) < count({-3, -2, -1, 0, 1, 2, 3})

Your induction proof relies on the fact that |N| < |Z| but this statement is false as |N| = |Z| since it is possible to construct a bijective function f: N -> Z like so in extension:

f(n) = n/2 (if n is pair)
f(n) = -(n+1)/2 (if n is impair)


... f(9) f(7) f(5) f(3) f(1) f(0) f(2) f(4) f(6) f(8) f(10) ...
&#8595; &#8595; &#8595; &#8595; &#8595; &#8595; &#8595; &#8595; &#8595; &#8595; &#8595;
... -5 -4 -3 -2 -1 0 1 2 3 4 5 ...

The function f will be bijective and thus |Z| = |N|
 
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I edited it and said I did so.




Your induction proof relies on the fact that |N| < |Z| but this statement is false as |N| = |Z|

Can you prove |N| = |Z|?

The induction proof doesn't rely on the fact that |N| < |Z| it proves it.

Also just assuming ordinary definitions and set theory the only way for Z to be a strict subset of R and for N to be a strict subset of Z is for the numbers of elements in each of these to be different. This alone proves |R| > |Z| > |N|

Of course if you just assume |N| = |Z| = |R| then you can do anything you want. I haven't seen proof of these assertions however. I obviously didn't major in math so I'm interested to see what you've got.
 
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1. I lost it when you said i is an irrational number. It's imaginary. This thread shouldn't be taken seriously.
2. You've clearly never taken a proper Discrete Mathmatics class. This is a very basic proof and Richard got it correct right away.
3. Moh is also right.
4. Just a quick fun thing, you can literally represent the entirety of the set of natural numbers between any two real numbers. Edit: that's basically what Moh said
5. RIP Walter
 
Can you prove |N| = |Z|?

The induction proof doesn't rely on the fact that |N| < |Z| it proves it.

Also just assuming ordinary definitions and set theory the only way for Z to be a strict subset of R and for N to be a strict subset of Z is for the numbers of elements in each of these to be different. This alone proves |R| > |Z| > |N|

Of course if you just assume |N| = |Z| = |R| then you can do anything you want. I haven't seen proof of these assertions however. I obviously didn't major in math so I'm interested to see what you've got.

how can induction proof prove something that is not true?

|A| = |B| if there is a bijective function f: A -> B

Let's build such function,

be the function f: N -> Z
f(n) = n/2 (if n is pair)
f(n) = -(n+1)/2 (if n is impair)

the f-function is indeed bijective as for every natural, only one 'Z' satisfies f(a) = b

Which then means that |N| = |Z|
 
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