Infinity (Topic)

Relatively ancient and inactive

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Fine, let me prove that the set of integers and the set of integers that are multiples of four is the same size.

Proof:
Define Z to be the set of integers.
Define F to be the set of integers that are multiples of four.

I define a bijection between Z and F:
f(x) = 4x.
This is a bijection because it is both surjective and injective:
Surjective:
y = 4x. Hence, x = y/4. Thus f^-1(y) = x exists.
Therefore, f(x) is surjective.

Injective:
Define a and b to be in domain. Let f(a) = f(b). Then we see that 4a=4b. Thus, a = b.
Therefore, f(x) is injective.

Since f(x) is both surjective and injective, it is bijective. Since I have found a bijective function that relates the set of integers to the set of integers of multiples of four, the two sets are equal in size.
Therefore, |Z| = |F|. (|Z| is the size of Z and |F| is the size of F)

I'll admit, I'm not particularly used to group theory proofs - however, I'm fairly certain that it's not bijective, because not every value in the group with multiples of four has a corresponding value in the group with all integers. The vice versa is true, but it seems that to prove equality you have to make sure it works both ways.

None.

Quote from

Centreri

Quote from

cheeze

Fine, let me prove that the set of integers and the set of integers that are multiples of four is the same size.

Proof:
Define Z to be the set of integers.
Define F to be the set of integers that are multiples of four.

I define a bijection between Z and F:
f(x) = 4x.
This is a bijection because it is both surjective and injective:
Surjective:
y = 4x. Hence, x = y/4. Thus f^-1(y) = x exists.
Therefore, f(x) is surjective.

Injective:
Define a and b to be in domain. Let f(a) = f(b). Then we see that 4a=4b. Thus, a = b.
Therefore, f(x) is injective.

Since f(x) is both surjective and injective, it is bijective. Since I have found a bijective function that relates the set of integers to the set of integers of multiples of four, the two sets are equal in size.
Therefore, |Z| = |F|. (|Z| is the size of Z and |F| is the size of F)

I'll admit, I'm not particularly used to group theory proofs - however, I'm fairly certain that it's not bijective, because not every value in the group with multiples of four has a corresponding value in the group with all integers. The vice versa is true, but it seems that to prove equality you have to make sure it works both ways.

1 - 4
2 - 8
3 - 12
4 - 16
5 - 20
...

There's exactly one element in the set of the multiples of 4 for each element in the natural numbers set. Also there's exactly one element in the natural numbers set for each element in the set of the multiples of 4. So we've paired all of them: both sets have the same size.

Also you're refering to set theory and not group theory.

?????

Just to be sure. Lets just say infinity is not a quantity. Because it behaves nothing like quantitable units.

Quote

If you had x = 0
1x=x,
divide both sides by x, (supposing we are allowed to divide by 0 and it will act like a real number)
1(x/x) = (x/x)
1*1 = 1
1 = 1,
How does that say anything about infinity? 1x=x is just x=x.

You can't have x/x because x is 0. You cannot have a divisor that is 0. For plain obvious reasons explained very much alive above. Thats not how you explain infinity. And 0/0 is not 1. 0/0 is not defined. What you've done is not math but bullshit. Man how do you take 0 and divide it by 0 and get 1.

None.

Quote from

BeDazed

Just to be sure. Lets just say infinity is not a quantity. Because it behaves nothing like quantitable units.

As I said before, there are two different concepts of infinity which are not related. The ∞ one, which represents the behaviour of a function and is by no way related to a quantity, and the א₀, א₁, ..., which measure the size of a set and are, in fact, quantities. However they are not very useful for anything and thus you probably have never seen them used in any operation. This kind of numbers are called transfinite numbers.

If any of you is interested in further reading I'd suggest searching something on google. This result looks quite good: http://mathforum.org/library/drmath/view/51472.html

?????

Relatively ancient and inactive

Quote from

Clokr_

Quote from

Centreri

Quote from

cheeze

Fine, let me prove that the set of integers and the set of integers that are multiples of four is the same size.

Proof:
Define Z to be the set of integers.
Define F to be the set of integers that are multiples of four.

I define a bijection between Z and F:
f(x) = 4x.
This is a bijection because it is both surjective and injective:
Surjective:
y = 4x. Hence, x = y/4. Thus f^-1(y) = x exists.
Therefore, f(x) is surjective.

Injective:
Define a and b to be in domain. Let f(a) = f(b). Then we see that 4a=4b. Thus, a = b.
Therefore, f(x) is injective.

Since f(x) is both surjective and injective, it is bijective. Since I have found a bijective function that relates the set of integers to the set of integers of multiples of four, the two sets are equal in size.
Therefore, |Z| = |F|. (|Z| is the size of Z and |F| is the size of F)

I'll admit, I'm not particularly used to group theory proofs - however, I'm fairly certain that it's not bijective, because not every value in the group with multiples of four has a corresponding value in the group with all integers. The vice versa is true, but it seems that to prove equality you have to make sure it works both ways.

1 - 4
2 - 8
3 - 12
4 - 16
5 - 20
...

There's exactly one element in the set of the multiples of 4 for each element in the natural numbers set. Also there's exactly one element in the natural numbers set for each element in the set of the multiples of 4. So we've paired all of them: both sets have the same size.

Also you're refering to set theory and not group theory.

Ah. Okay. Thank you, oh lord Cheeze, for your proof - Although, that's in terms of sets... you'll notice I was actually talking about infinity as an actual length, insofar as you can take any number in both sets and the amount of numbers fitting would be different. So, I still rather think that's infinity is natively contradictory

.

Yes, I just don't give up.

None.

Quote

insofar as you can take any number in both sets and the amount of numbers fitting would be different

Huh? This statement makes no logical sense.

Quote

Proof:
Define Z to be the set of integers.
Define F to be the set of integers that are multiples of four.

I define a bijection between Z and F:
f(x) = 4x.
This is a bijection because it is both surjective and injective:
Surjective:
y = 4x. Hence, x = y/4. Thus f^-1(y) = x exists.
Therefore, f(x) is surjective.

Injective:
Define a and b to be in domain. Let f(a) = f(b). Then we see that 4a=4b. Thus, a = b.
Therefore, f(x) is injective.

Since f(x) is both surjective and injective, it is bijective. Since I have found a bijective function that relates the set of integers to the set of integers of multiples of four, the two sets are equal in size.

For formality:

Call the set of integers that are an even multiple of 4 a set F = {x ∈ ℤ | x/4 ∈ ℤ}. (ℤ is by convention the set of integers).

Define a function
f : ℤ → F
x → 4x

f is well-defined:
Choose z ∈ ℤ. f(z) = 4z. 4z/4 = z ∈ ℤ, thus 4z ∈ F.
Choose a, b ∈ ℤ s.t. a = b. f(a) = 4a = 4b = f(b), thus f is well-defined.

Injective:
Choose a, b ∈ ℤ s.t. f(a) = f(b). ⇒ 4a = 4b ⇒ a = b. Thus, f is injective.

Surjective:
Choose a ∈ F. Consider a/4. By the definition of set F, a/4 ∈ ℤ. f(a/4) = 4(a/4) = a. Thus ∀ a ∈ F ∃ z ∈ ℤ s.t. f(z) = a, and f is surjective.

f is both surjective and injective; thus it is bijective. A bijection exists between F and ℤ, therefore they have equal cardinality, by definition of the equality of cardinality.

Post has been edited 5 time(s), last time on Jan 15 2009, 3:09 am by DT_Battlekruser.

None.