Question

2+2=?

Answer 1

1

Answer 2

2 + 2 = 22

Answer 3

Uhm * Holds up fingers and starts counting * Its.... Uhm.... 7

Answer 4

correctamundo

Answer 5

lemme go get my AP Cal book and i will ask a real question :P

Answer 6

it`s 4

Answer 7

i don't believe you... lolz

Answer 8

glad you waited a while. We first begin by constructing the Real numbers, The synthetic approach axiomatically defines the real number system as a complete ordered field. Precisely, this means the following. A model for the real number system consists of a set R, two distinct elements 0 and 1 of R, two binary operations + and * on R (called addition and multiplication, resp.), a binary relation ≤ on R, satisfying the following properties. 1. (R, +, *) forms a field. In other words, For all x, y, and z in R, x + (y + z) = (x + y) + z and x * (y * z) = (x * y) * z. (associativity of addition and multiplication) For all x and y in R, x + y = y + x and x * y = y * x. (commutativity of addition and multiplication) For all x, y, and z in R, x * (y + z) = (x * y) + (x * z). (distributivity of multiplication over addition) For all x in R, x + 0 = x. (existence of additive identity) 0 is not equal to 1, and for all x in R, x * 1 = x. (existence of multiplicative identity) For every x in R, there exists an element −x in R, such that x + (−x) = 0. (existence of additive inverses) For every x ≠ 0 in R, there exists an element x−1 in R, such that x * x−1 = 1. (existence of multiplicative inverses) 2. (R, ≤) forms a totally ordered set. In other words, For all x in R, x ≤ x. (reflexivity) For all x and y in R, if x ≤ y and y ≤ x, then x = y. (antisymmetry) For all x, y, and z in R, if x ≤ y and y ≤ z, then x ≤ z. (transitivity) For all x and y in R, x ≤ y or y ≤ x. (totalness) 3. The field operations + and * on R are compatible with the order ≤. In other words, For all x, y and z in R, if x ≤ y, then x + z ≤ y + z. (preservation of order under addition) For all x and y in R, if 0 ≤ x and 0 ≤ y, then 0 ≤ x * y (preservation of order under multiplication) 4. The order ≤ is complete in the following sense: every non-empty subset of R bounded above has a least upper bound. In other words, If A is a non-empty subset of R, and if A has an upper bound, then A has a least upper bound u, such that for every upper bound v of A, u ≤ v. The final axiom, defining the order as Dedekind-complete, is most crucial. Without this axiom, we simply have the axioms which define a totally ordered field, and there are many non-isomorphic models which satisfy these axioms. This axiom implies that the Archimedean property applies for this field. Therefore, when the completeness axiom is added, it can be proved that any two models must be isomorphic, and so in this sense, there is only one complete ordered Archimedean field. When we say that any two models of the above axioms are isomorphic, we mean that for any two models (R, 0R, 1R, +R, *R, ≤R) and (S, 0S, 1S, +S, *S, ≤S), there is a bijection f : R → S preserving both the field operations and the order. Explicitly, f is both injective and surjective. f(0R) = 0S and f(1R) = 1S. For all x and y in R, f(x +R y) = f(x) +S f(y) and f(x *R y) = f(x) *S f(y). For all x and y in R, x ≤R y if and only if f(x) ≤S f(y).

Once we have shown that we have axiomatically defined the Reals to be both a group under addition and multiplication, (proof left as an exercise to the reader), we start by defining 2 (two). examining the natural numbers, ℕ⊆ℝ, our first element in the set is 1. We define 2 to be the following: 1+1 = 2. Therefore, by transitive property, we find 2 + 2 = 1+1+1+1, which by the successor function we define to be the fourth element in the natural numbers, which is 4.

GET NUMBERTHEORIED

Answer 9

GET COPY PASTED!

Answer 10

I feel like proving all that stuff doesn't prove it a field but a vector space?