Can sci construct the real numbers from scratch?

Can sci construct the real numbers from scratch?

suppose there exists a number. it is real

...

this dude rules over nerds like sci. without him, itd be noather chernobyl

Dadekind cut ur dick

For any real number you can construct a sequence of rationals that converges to it,

N is trivial to construct :
{}, {{}}, {{{}}}, {{{{}}}}, ...
Q is just NxN*
R is just the limit of every convergent sequence in Q

sqrt(2)

yes

the limit... of every convergent... sequence?
in Q?
*breathes in*
aahaahhHAHAHAHAHAHAHAHAHAHAHAAAAA

>R is just the limit of every convergent sequence in Q

Is this what passes for a rigorous proof on Veeky Forums nowadays? Simply throwing words together does not create a new concept in itself, as N. Wildberger explained.

Would
>R is just the teapot of some water donkey in Q
be a proof to you as well?

see

>redditfrog

Post discarded.

>reddittext

Post discarded.

Riddet can't into continued fractions,
poor babby.

>rigorous proof on Veeky Forums nowadays
no rigor on Veeky Forums now-a-days, sorry Grandpa

Let's assume we have N, then Z = N U N- U {0}
N x Z -> Q
Q c R, and for R axiom about if A

You should specify the metric, when talking about convergence. Otherwise you could end up with something different then IR.

I'm gonna start off with Q cause anything below that is tedious and boring.

Define a dedekind cut as (A,B) where A and B are in P(Q)

Reqs to be a dedekind cut:
>if creverse is true for D
>A and B are disjoint
>for all c in A, all d in B, c < d
>A has no max, B has no min
>|Q/(A U B)| = 1 or 0 (something like this, I forget the exact rule to prevent an interval of rational numbers inside the cut)

Take the set of all valid rational dedekind cuts and you have the real number line.

Addition
>(A1,B1)+(A2,B2) = (A3,B3)
>A3 = {a1+a2 | a1 € A1, a2 € A2}
>same for b3

Multiplication is similar but much more gross to define since negative numbers wrap around back to postitive

You can prove this satisfies the requirements to be a field but that'd take all day.

This definition proves that the reals are complete - that every bounded sequence has a least upper bound.

The least upper bound of a bounded set S, which contains (A1,B1), (A2,B2),...
Is the following real number:
(A1 U A2 U A3 .... , B1 ∩ B2 ∩ B3 ....)

Rate my construction of the reals, go easy bc I have never took an analysis class

The absolute state of Veeky Forums.

>that pic
engineer detected, lol.

>reddit

Post retarded.

someone get a doctor there are people in this board who haven't taken their bergerpill yet

I'm sorry, you can't just throw some symbols and words around and expect anyone to take you seriously.

Im just happy no one is attacking my use of the "€" symbol

I always preferred constructing them as the equivalence classes of cauchy sequences in Q. Was more intuitive to me.

"convergent sequences in Q" converge to an element of Q, friendo.

You meant equivalence classes over Cauchy sequences, not "limit of convergent sequences".

>constuction of real numbers
>uses concepts whose defenitions rely on real numbers

It's not a proof, it's a conjecture

What are you talking about? Dedekind cuts are contained in the powerset of Q

How about we start with the naturals as the *smallest" inductive set, add the concept of zero if you're not French and then multiply by -1 after you've proven basic operations
After whole numbers move to all condensed fractions again proving the group is closed and adheres to your basic axiomatic operations and so on, this is first week calc stuff, y'all are dumb nignogs

dedekind cuts masterrace

...

Haven't you been bergerpilled yet? Everyone knows that squares with integer side lengths literally don't have diagonals.

What? Bergerpill me.

Consider a set of real numbers constructed from scratch...

how can the length of something be an irrational number?

That's fucking retarded. Think about it for a second, if the length of something is irrational then it doesn't have a definite size.

Hmm, this really makes me think. So you had a 2 and then drew some funny looking lines around it and now you claim that is a number? Well, can't argue with that. See my picture:

Behold! I've brought you a number.

>start with naturals
>multiply by -1

>calling other people dumb

It clearly does have a definite size. We just can't state it explicitly.

Just consider it as an axiom

just the result of multiplying complex numbers with their respective conjugates, branlets

No.