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

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

 

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|

 

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.

 

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 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'

 

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'

 

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.

 

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

 

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

 

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