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9654163 No.9654163 [Reply] [Original] [archived.moe]

prove sqrt2 exists. i'll wait

>> No.9654165

define "exists"

>> No.9654182

[eqn]\left\{x\,\in\,\mathbf Q \mid x^2\ \leqslant\ 2\right\}[/eqn]
[eqn]\begin{cases} u_0\ =\ 1 \\ \forall n\,\in\,\mathbf N,\, u_{n\,+\,1}\ =\ \frac12\,\left( u_n\,+\,\frac2{u_n} \right) \end{cases}[/eqn]

>> No.9654208


The true redpill is when you realize that in order to prove that (exists x:T,P x) -> Q, where Q is a sentence (not containing x as a free variable) and T is any type, it is never necessary to prove that there is actually an x in T such that P x.

In fact the sentence "(exists x:T,P x) -> Q" is logically equivalent to "forall x:T, (P x -> Q).

Discarding the math derived as a consequence of the existence of square roots (i.e. simply using them from the definition) under the pretext that "maybe there are no square roots" shows a lack of understanding of the hypothetical reasoning.

>> No.9654219

prove Q exists

>> No.9654233

Q:= Z x (N\{0})/~ (quotient set) where (a,b)~(c,d) iff ad=bd and Z = NxN / ~~ where (a,b) ~~ (c d) iff a+d = b+c.
operations on Z,Q on demand (you just have to check some compatibility conditions with the equivalence relationship above but calculations are what you think they are).

>> No.9654729

It doesn't.

>> No.9654963
File: 17 KB, 581x538, 1518904147146.png [View same] [iqdb] [saucenao] [google] [report]

hey mathfags
prove [math]i[/math] exists. i'll wait

>> No.9655008


>> No.9655015

Look at the alphabet more carefully.. after h

>> No.9655027

In [math]\mathbf R\left[ X \right] / \left(X^2\,+\,1\right)[/math],
[eqn]{\overline X}^2\ =\ \overline{X^2}\ =\ \overline{X^2\,+\,1\,-\,1}\ =\ \overline{X^2\,+\,1}\,-\,\overline 1\ =\ \overline 0\,-\,\overline 1\ =\ -\overline 1.[/eqn]
You're welcome.

>> No.9655035

f(x) = g(x) * g(x) = x * x is a continuous function, because g(x) = x is continuous. (property that comes from limit of sequences)

f(0) = 0, f(2) = 4

there must then exist a c for which 0 < f(c) = 2 < 4, i.e. c * c = 2 (intermediate value theorem)

>> No.9655045

Let (K,+,x) be a commutative ring with unit 1 and L:= K^2 (the cartesian product).
Is (a,b) and (c,d) are in K, define (a,b)+(c,d):= (a+c,b+d) and (a,b)*(c,d):= (ac-bd, ad+bc). Define also i:= (0,1). It turns out that (L,+,*) is also a commutative ring with unit = (1,0), the map f:x->(x,0) from K to L is an injective ring homomorphism, and for every x,y in K we have (x,y)= f(x)+ i * f(y) (since f i s injective, if we identify t and f(t) for every t, we get the more familiar notation (x,y) = x+i*y). Wed also have i*i = -1 from the definition.

In addition if K is a field and for every x in K, x^2 is different from -1, then L is also a field (for every a,b in K, we cant have a^2+b^2=0 otherwise we'd get a^2=-b^2 then (a/b)^2=-1 contradicting the above: that being said, you can check that (a/(a^2+b^2), -b/(a^2+b^2) ) is the inverse of (a,b) for every a,b in K.

>> No.9655255
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Prove me that numbers exist.

>> No.9655777
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read plato

>> No.9655795

define "sqrt2"

>> No.9655807

its actually sqrt(2) kek

>> No.9656287


>> No.9656315

Checked for truth
Anyone who isn't a Platonist is a brainlet

>> No.9657569

Platonism is bullshit.

>> No.9659187
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>> No.9659204

Not all of it, no. Besides, the dualistic properties of nature immensely influenced Aristotle's Metaphysics.

>> No.9659237

>implying sqrt (2) can't exist because it's irrational in R.

There are some groups where sqrt (2) is rational. Therefore it exists even by your shitty definition.

>> No.9659267
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>tips fedora

>> No.9659605

What book is this excerpt taken from?

>> No.9659611

The Foundations of Mathematics by Kunen

>> No.9659633

No, it's the truth. If you actually care about knowledge and truth you will be a platonism.

>> No.9659859

which edition?

>> No.9660429

It's the sidelength of a square of area 2. So 1.414^2=2. If you don't like it change your scale.

>> No.9660456

There's only one...

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