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

[Accounts Shop]

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

1. Yea my bad on that one, i mix those two sometimes as i haven't worked much with them
2. Richard merely took the definition of the two sets and used a false induction proof to prove it
3. Moh is right, Cantor did prove that the set of real numbers in uncountable

 

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|

Alternatively how can a bijective function prove something that is not true?

For any finite subset of naturals fed into this function you will have a subset of integers with cardinal exactly half less one of the subset of naturals. And this is true for any arbitrarily large subset of naturals.

When you allow infinity to become a number instead of a concept you produce a paradox (i.e. a contradiction.) Basically you are setting the number of elements in the input subset equal to the number in the output subset which of course leads to a contradiction.

From my understanding they introduced this countable uncountable terminology to sort of sweep the problem under the rug.

Edit: Lol reading about this now actually: https://en.wikipedia.org/wiki/Transfinite_number

You silly mathematicians

 

I've got one for you Accounts Shop:

Let's build a bijective function.

Here it is:
f: Z -> R {
f(z) = z * (infinitesimal)
}

infinitesimal = 1/infinity

If you're allowed to put infinity as a number into the function I can use infinity as a number in the function.

Now the reals and the integers have the same cardinality.

inb4 just broke mathematics

 

[Accounts Shop]

I've got one for you Accounts Shop:

Let's build a bijective function.

Here it is:
f: Z -> R {
f(z) = z * (infinitesimal)
}

infinitesimal = 1/infinity

If you're allowed to put infinity as a number into the function I can use infinity as a number in the function.

Now the reals and the integers have the same cardinality.

inb4 just broke mathematics

notice how i never used infinity as a number when i defined the function f?

i merely gave the general term of the f-function that would allow you to calculate any natural number and show it would be assigned to only one value of 'Z'

also, the f-function you defined is not even a bijective function.

the function is only injective (each value of Z is assigned to a value of R) but not surjective (every value of R can be represented by a value of Z)

for example:

lets say "infinitesimal = 0.005" <we can use a finite value here because although 'infinity' doesn't have a value, it is used as a constant in your definition of the f-function>

so we have
f(-3) = -3/0.005 = -600
f(-2) = -2/0.005 = -400
f(-1) = -1/0.005 = -200
f(0) = 0/0.005 = 0
f(1) = 1/0.005 = 200
f(2) = 2/0.005 = 400
f(3) = 3/0.005 = 600

since 'Z' is the set of integers, the f-function does not represent every 'R' and thus f is not surjective

however, to be bijective, a function needs to be Injective and Surjective.

Thus, |Z| != |R| which makes sense, because 'Z' is countable whereas 'R' is not

let's do the same test with the f-function i built:

f(a) = a/2 <if 'a ' is pair>
f(a) = -(a+1)/2 <if 'a' is impair>

f(0) = 0/2 = 0
f(1) = -(1+1)/2 = -1
f(2) = 2/2 = 1
f(3) = -(3+1)/2 = -2
f(4) = 4/2 = 2
f(5) = -(5+1)/2 = -3
f(6) = 6/2 = 3
...

if we continue to infinity, each value of 'a' will have been assigned to one and only one value of 'Z', and each value of 'Z' will be assigned to one and only one 'a' thus proving that the f-function is in fact a bijective function which then means that |N| = |Z|

 

[MohtasaUnique]

I've got one for you Accounts Shop:

Let's build a bijective function.

Here it is:
f: Z -> R {
f(z) = z * (infinitesimal)
}

infinitesimal = 1/infinity

If you're allowed to put infinity as a number into the function I can use infinity as a number in the function.

Now the reals and the integers have the same cardinality.

inb4 just broke mathematics

The notation you're looking for is

Infinity can't be represented as a number, only an idea. Therefore you can claim N will approach infinity, but it cannot be equal to infinity as a number itself.

To rewrite the equation it would be
f: Z -> R {
f(z) = z * lim 1/n (as n approaches infinity)
}

and in this case, the cardinality is still not the same because N will approach infinity, but still can never reach it.

 

notice how i never used infinity as a number when i defined the function f?

i merely gave the general term of the f-function that would allow you to calculate any natural number and show it would be assigned to only one value of 'Z'

Yeah it's clever, you don't use infinity in your function definition but then your function definition isn't the whole story. You then place it in a counting function whose input is 0 and +infinity.

also, the f-function you defined is not even a bijective function.

the function is only injective and not injective + surjection (which is needed to be bijective)

Interesting. Can you show it's not bijective?

 

[Accounts Shop]

Yeah it's clever, you don't use infinity in your function definition but then your function definition isn't the whole story. You then place it in a counting function whose input is 0 and +infinity.



Interesting. Can you show it's not bijective?

if i can show your f-function is not bijective? i did in my post showing the first few values of it in extension, i was editing my post when you quoted it it seems

 

The notation you're looking for is

Infinity can't be represented as a number, only an idea.

Actually no, this isn't what modern mathematics puts forward.
https://en.wikipedia.org/wiki/Transfinite_number

Therefore you can claim N will approach infinity, but it cannot be equal to infinity as a number itself.

To rewrite the equation it would be
f: Z -> R {
f(z) = z * lim 1/n (as n approaches infinity)
}

No still wrong. If infinity can be a number then an infinitesimal can be a number.

and in this case, the cardinality is still not the same because N will approach infinity, but still can never reach it.

That's not how limits work. The Lim 1/n as n approaches infinity is 0. http://www.wolframalpha.com/input/?i=lim+n->infinity+1/n

 

if i can show your f-function is not bijective? i did in my post showing the first few values of it in extension, i was editing my post when you quoted it it seems

for example:

lets say "infinitesimal = 0.005" <we can use a finite value here because although 'infinity' doesn't have a value, it is used as a constant in your definition of the f-function>

so we have
f(-3) = -3/0.005 = -600
f(-2) = -2/0.005 = -400
f(-1) = -1/0.005 = -200
f(0) = 0/0.005 = 0
f(1) = 1/0.005 = 200
f(2) = 2/0.005 = 400
f(3) = 3/0.005 = 600

since 'Z' is the set of integers, the f-function does not represent every 'R' and thus f is not surjective

however, to be bijective, a function needs to be Injective and Surjective.

Thus, |Z| != |R| which makes sense, because 'Z' is countable whereas 'R' is not

No it's a fallacy.

You can no more assume that my infinitesimal is a finite constant than I can assume your infinite set has a finite number of elements.

You're assuming the initial point. The proof fails.

 

[Accounts Shop]

No it's a fallacy.

You can no more assume that my infinitesimal is a finite constant than I can assume your infinite set has a finite number of elements.

You're assuming the initial point. The proof fails.

the value of 'infinity' in your function is not dependant on the value of 'z' you put in your function , ergo it acts as a constant, even if we don't know the value of it, it will have the same value for all the integers you put in your function.

I could do a proper proof of the surjectivity of my function and the non-surjectivity of yours but the formatting of the comments would make it unreadable.

here is what you need to prove to verify that a function is surjective:

R is surjective <=> (forall 'b' member-of 'T' | (there-is 'a' member-of 'S' | a R b) )

it's easy to verify that my function is surjective for all 'N' whereas your function is not surjective for all 'Z'

 

the value of 'infinity' in your function is not dependant on the value of 'z' you put in your function , ergo it acts as a constant, even if we don't know the value of it, it will have the same value for all the integers you put in your function

I said it's not a finite constant. I didn't say it wasn't a constant.

I must say as defender of the ZFC status quo you're not doing a very good job.

The problem that you have is because of the nature of an infinitesimal it necessarily WILL fill every gap in the R set. If the gap between any two arbitrarily close together but non equal real numbers contains an infinite number of points, then my infinitesimal will happily fill that range when you multiply it by the set of the naturals.

 

[Accounts Shop]

I said it's not a finite constant. I didn't say it wasn't a constant.

I must say as defender of the ZFC status quo you're not doing a very good job.

The problem that you have is because of the nature of an infinitesimal it necessarily WILL fill every gap in the R set. If the gap between any two arbitrarily close together but non equal real numbers contains an infinite number of points, then my infinitesimal will happily fill that range when you multiply it by the set of the naturals.

no it wont ... 1/infinity will be a fraction of 1, a very small one.

if you divide an 'integer' by a fraction, the result will never be a fraction per the arithmetic properties.

so your function would only represent integer values in 'R'

 

[MohtasaUnique]

Actually no, this isn't what modern mathematics puts forward.
https://en.wikipedia.org/wiki/Transfinite_number




No still wrong. If infinity can be a number then an infinitesimal can be a number.


That's not how limits work. The Lim 1/n as n approaches infinity is 0. http://www.wolframalpha.com/input/?i=lim+n->infinity+1/n

You were saying infinitesimally small right? Isn't that essentially the same thing as lim 1/n? The difference is n is approaching infinity, but the result is still an infinitesimal number.

 

no it wont ... 1/infinity will be a fraction of 1, a very small one.

if you divide an 'integer' by a fraction will never be a fraction per the arithmetic properties.

so your function would only represent integer values in 'R'

No no no. This is not how infinitesimals work.

It's an infinitely small number that's still greater than zero. It is the smallest number. Literally every possible (real) number is a multiple of it.

 

You were saying infinitesimally small right? Isn't that essentially the same thing as lim 1/n? The difference is n is approaching infinity, but the result is still an infinitesimal number.

You're using infinity as a concept which is imho correct, and certainly the way we work with infinity in physics and computer science.

Mathematics doesn't treat it as a concept but rather an actual number like 0 or 1. It has a special significance however.

But interestingly ZFC doesn't appear to include infinitesimals strongly in their model. An infinitesimal is just the inverse concept to infinity. If infinity is an actual number then an infinitesimal is an actual number defined as 1/infinity.

We're arguing essentially about which proof is correct. Does induction prove that the cardinality of the reals must be larger than the naturals or does the bijective function prove that the cardinality of the reals is equal to the cardinality of the naturals.

As can be plainly seen it comes down to what model of mathematics you use. It's certainly interesting I've never stopped to research it before.

Also this is my race horse: https://en.wikipedia.org/wiki/Law_of_Continuity support for my main man Leibniz

 

[Accounts Shop]

No no no. This is not how infinitesimals work.

It's an infinitely small number that's still greater than zero. It is the smallest number. Literally every possible (real) number is a multiple of it.

infinitisimal = 1/infinity

it's safe to assume that infinity is > 1

thus making infinitissimal ]0,1[

so it will be "z / (1/infinity)" which is equal to "z * (infinity/1)"

edit: so you're infinitesimals would be what i know as "0+" and "0-" ?

basically my proof relies on the axiom of choices, if you don't "accept" it, then there isn't really any way to tell which infinite set has a greater cardinality that I know of

 

infinitisimal = 1/infinity

it's safe to assume that infinity is > 1

thus making infinitissimal ]0,1[

so it will be "z / (1/infinity)" which is equal to "z * (infinity/1)"

edit: so you're infinitesimals would be what i know as "0+" and "0-" ?

basically my proof relies on the axiom of choices, if you don't "accept" it, then there isn't really any way to tell which infinite set has a greater cardinality that I know of

You've not worked with infinitesimals before? Have you taken calculus?

Infinitesimal is hard to pin down. To be quite fair and because you have various different types of infinity in your mathematical model it is more correct to define it not as 1/infinity, rather it is strictly the smallest possible number which is not zero.

That said it is often written as 1/infinity which is what I mean to say earlier.

 

[Accounts Shop]

You've not worked with infinitesimals before? Have you taken calculus?

Infinitesimal is hard to pin down. To be quite fair and because you have various different types of infinity in your mathematical model it is more correct to define it not as 1/infinity, rather it is strictly the smallest possible number which is not zero.

That said it is often written as 1/infinity which is what I mean to say earlier.

i've had calculus, it's probably what we referred to as "0+" and "0-" but it was only used when wanting to calculate limits (as n approaches 0+) rather than limit (as n approches 0)

we basically used it to get infinity, so yea that is it aha ...

1/0+ = +infinity
1/0- = -infinity

I had never used these 'numbers' outside the context of limits

 

i've had calculus, it's probably what we referred to as "0+" and "0-" but it was only used when wanting to calculate limits (as n approaches 0+) rather than limit (as n approches 0)

we basically used it to get infinity, so yea that is it aha ...

1/0+ = +infinity
1/0- = -infinity

I had never used these 'numbers' outside the context of limits

Right you could probably say 0+ less 0 is an infinitesimal and therefore 0+ is an infinitesimal.


I strongly suspect they are incompatible with the axiom of choice. But honestly this is the limit of my knowledge at this point. I'll get back to you on that.

 

[Accounts Shop]

Right you could probably say 0+ less 0 is an infinitesimal and therefore 0+ is an infinitesimal.


I strongly suspect they are incompatible with the axiom of choice. But honestly this is the limit of my knowledge at this point. I'll get back to you on that.

Pretty sure they are incompatible because if not, there could be a bijective function from Z -> R which would make |Z| = |R|

Pretty sure that the 0+ /0- things i was taugh to only use when i wanted to know the behavior of a function as it approaches 0