/sci/ - Science & Math

[Anonymous No.7939172]

Prove me wrong.
Protip: You can't.

[Anonymous No.7939175]

The LHS is multivalued while the RHS is only positive.

[Anonymous No.7939174]

This is correct.

[Anonymous No.7939180]

That's a definition not a theorem

[Anonymous No.7939183]

>>7939175
>implying the input is what matters
The outcomes are both the same.

[Anonymous No.7939184]

>>7939175
LHS is not multivalued. It'd be if it had [math]\plusminus[/math]

[Anonymous No.7939185]

Why is x^1 not the same thing as x^(2/2)?

[Anonymous No.7939186]

>>7939172
an excellent post-satire meta-ironic deconstruction of the "i don't understand burden of proof" meme. 10/10.

>>7939175
No. The radical sign means principal square root

[Anonymous No.7939187]

[eqn] \sqrt{i^2} = \sqrt{-1} = i [/eqn]
[eqn] |i| = 1 [/eqn]

[Anonymous No.7939189]

>>7939185
but it is

[Anonymous No.7939190]

>>7939183
He's not implying that. He's saying this statement is true only when x is a real number.

[Anonymous No.7939192]

>>7939189
but x^(2/2) is abs(x), not x^1.

[Anonymous No.7939196]

>>7939187
>using imaginary numbers
Hard more: No imaginary numbers, let alone, complex numbers.

[Anonymous No.7939222]

let x = i

[Anonymous No.7939226]

>>7939222
>let me just disprove a real number statement with imaginary numbers
wew lad :^)

[Anonymous No.7939227]

>>7939222
trips checked

[Anonymous No.7939298]

>>7939227
im still yet to get them

[Anonymous No.7939332]

>>7939172
It fails if x is a complex number so no.

[Anonymous No.7939335]

>>7939226
Where was it stated that x is a real number? This is obviously true then.

[Anonymous No.7939361]

>>7939184

wait now you can

[\math] \color{red}>redtext [/math]

??

[Anonymous No.7939372]

>>7939184

>>7939184

wait now you can

[math] \color{red}>\redtext [/math]


??**

[Anonymous No.7939379]

>>7939196
let x be a polynomial in t
let x by a differential equation in x'(t)
let x be a vector in Rn, any fixed n > 1
let be any matrix such that it is square and has more than 1 row.

literally easy mode.

[Anonymous No.7939380]

>>7939372
[eqn]\color{cyan} { \mathfrak{Faggot} }[/eqn]

[Anonymous No.7939398]

Are you ready to be butthurt, OP?
You are so stupid you didn’t even specify that [math]x[/math] was a real number, so I guess you wouldn’t mind if I used matrices. I’ll even get the hard mode as described by that >>7939196 retard. :^)
[eqn]\sqrt{ \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}^2 } \,=\, \sqrt{ \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} } \,=\, \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}[/eqn]
[eqn]\left\| \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} \right\| \,=\, \mathrm{tr}\, \left[ \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} \,\times\, \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} \right] \,=\, \mathrm{tr}\, \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} \,=\, 1 \,+\, 1 \,=\, 2[/eqn]
Therefore, [math]\sqrt{ \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}^2 } \,\neq\, \left\| \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} \right\|[/math], Q.E.D. faggot.

[Anonymous No.7939412]

>>7939372
[math]\color{red}>implying anyone would not know how to redtext[\math]

[Anonymous No.7939429]

>>7939192
you're thinking of (x^2)^1/2

[Anonymous No.7939467]

i will help prove you right

first, take the completion of the rationals, call it N
elements of N are called Normans

we assume that the square root is the principal square root, making it a single-valued function of the positive Normans and the zero Norman
[math] x^2 = xx [/math] is a function of x whose image is the positive Normans, meaning you can restrict its codomain to the positive Normans

let the square root be defined as such
for any positive Norman x,
Sqrt(x) = x/Sqrt(x)
so for any x^(2n), we have
Sqrt(x^(2n)) = x^(2n)/Sqrt(x^(2n))
Sqrt(x^(2n))Sqrt(x^(2n)) = x^(2n)
Sqrt(x^(2n))^2 = x^(2n)
Sqrt(x^(2n)) = x^n
then for x^2 we have n = 1
then Sqrt(x^2) = x
because the codomain of Sqrt is the positive Normans, x is a positive Norman
so we know that x is at least always positive
now we prove that x is the absolute value of the original Norman
for any negative Norman y = -z, y^2 = (-z)(-z) = (-1)^2 z^2 = 1 z^2 = z^2 = (-y)^2
then Sqrt((-x)^2) = Sqrt(x^2) = x
we define the absolute value of a Norman to be -x when x is negative, and x when x is positive, and 0 when x is 0
so for our first case, -x, we have Sqrt((-x)^2) = x
then our next case, 0, we have Sqrt(0^2) = 0
and our final case, x, we have Sqrt(x^2) = x
this is equivalent to the absolute value function

there is a slight problem, however, and that is that the Normans do not exist

[Anonymous No.7939470]

>>7939467
>first, take the completion of the rationals, call it N
>elements of N are called Normans

[Anonymous No.7939482]

>>7939467
>>7939470

i don't get it

[Anonymous No.7939497]

>>7939398
wrong

>>7939379 is correct.

[Anonymous No.7939498]

>>7939398
If you did it consistently, then you would have applied the absolute value via spectral theorem, and OP was right again.

[Anonymous No.7939529]

>>7939380
can you get banned for [math]\mathbb{THIS}[/math] [math]\color{purple}{dumb}[/math] [math]\mathcal{SHIT} [/math] though?

[Anonymous No.7939560]

>>7939529
[eqn]\color{red} {\mathcal{A}} \color{green} {\mathcal{S}} \; \color{yellow} {\mathbf{F}} \color{purple} {\mathbf{A}} \color{grey} {\mathbb{R}} \; \color{cyan} {\mathcal{A}} \color{yellowgreen} {\mathcal{S}} \; \sqrt{ -1 } \; \color{magenta} {\mathfrak{K}} \color{cadetblue} {\mathfrak{N}} \color{white} {\mathfrak{O}} \color{gold} {\mathfrak{W}} ... \\ N^{O^{O^{OOOOOOOOOOOOOOOOOOOOOOOOOOO!}}}[/eqn]

[Anonymous No.7939574]

>>7939529
[eqn]\color{red}{ \mathcal{A} } \color{green}{ \mathcal{S} } \; \color{yellow}{ \mathbf{F} } \color{purple}{ \mathbf{A} } \color{grey}{ \mathbb{R} } \; \color{cyan}{ \mathcal{A} } \color{yellowgreen}{ \mathcal{S} } \; \sqrt{ -1 } \; \color{magenta}{ \mathfrak{K} } \color{cadetblue}{ \mathfrak{N} } \color{white}{ \mathfrak{O} } \color{gold}{ \mathfrak{W }} ... \\ NO![/eqn]
fug i hope it renders

[Anonymous No.7939606]

[math] \color{purple}{ \scriptstyle \text{>not purpleposting}} [/math]

[Anonymous No.7939618]

>>7939606

will people flock to this board now that you can

[math]\color{purple}{>}\mathcal{DA}\mathbb{NK} [/math] [math] \color{orange}{PO}\color{red}{ST}[/math]

?

[Anonymous No.7939621]

>>7939606
[eqn]\color{cyan}{ \scriptstyle \text{>not cyanposting}}[/eqn]

[Anonymous No.7939623]

>>7939618
>>7939606


[math]\color{purple}>\mathcal{DA}\mathbb{NK} [/math] [math] \color{orange}{PO}\color{red}{ST}[/math]

[Anonymous No.7939731]

>>7939172
>>7939196

let [math]x \in G, dom(G) = {1,g,g^2,g^3}[/math], and g does not equal one. [math]\sqrt(g^4) = sqrt(1) = g^2[/math], but this is not true. Therefore the operators cannot be applied in this manner, and OP has been proven wrong. In hardmode.

[Anonymous No.7939746]

>>7939172

This would be a legit question if counter-proofs didn't exist, yet they do.

Pro-tip, more than one counter works depending on how constrained your axioms are.

[Anonymous No.7939748]

>>7939731
this isn't math
this is sudo-cyanse

[Anonymous No.7939752]

>>7939429
Thats exactly the same thing dude,(x^y)^z=x^(yz)

[Anonymous No.7939754]

>>7939192
It's because the square root function always has two solutions for any non 0 number, but in order to be defined as a function, it discards the negative.

[Anonymous No.7939770]

>>7939172
X = -2
Rad (X^2) = abs (X)
Rad (-2*-2)=abs(-2)
Rad (-2)*rad (-2)=2
Rad(2)*i*rad(2)*i=2
-2=2

;^)

[Anonymous No.7939991]

>>7939172

Prove me wrong

[Anonymous No.7939998]

>>7939991
1 doesn't equal -1
QED

[Anonymous No.7940104]

>>7939752
>(x^y)^z=x^(yz)
For reals with x > 0, that holds. But it certainly doesn't hold for x=-1, y=2, z=1/2.

[Anonymous No.7940139]

>>7940104
But it does. 1^.5 is either 1 or -1.

[Anonymous No.7940166]

>>7939574
This is an amazing discovery, stupid admin
[math] \color{red} {\mathbf{(USER ~~WAS ~~BANNED ~~FOR ~~THIS ~~POST)} }[/math]

[Anonymous No.7940173]

>>7939529
At least on /g/ misusing code tags can lead to a ban

[Anonymous No.7940211]

>>7940166
any retard can do that m8
[eqn]\color{red}{\mathcal{USER WAS BANNED FOR THIS POST}}

[Anonymous No.7940212]

>>7940211
dumbass nigga
[eqn]\color{red}{\mathcal{USER WAS BANNED FOR THIS POST}}[\eqn]

[Anonymous No.7940213]

>>7940212
[eqn]\color{red}{\mathcal{why is this so hard}\[\eqn]

[Anonymous No.7940217]

>>7940213
[eqn]\color{red}{ \mathcal{top kek} } \[/eqn]

[Anonymous No.7940219]

>>7940211
>>7940212
>>7940213
>>7940217
why are you so shit at this?

[Anonymous No.7940249]

\color{red} {\mathbf{(USER ~~WAS ~~BANNED ~~FOR ~~THIS ~~POST)} }

[Anonymous No.7940251]

[eqn] \color{red} {\mathbf{(USER ~~WAS ~~BANNED ~~FOR ~~THIS ~~POST)} } [/eqn]

if this one doesnt work im literally killing myself

[Anonymous No.7940274]

>>7940251
why was he banned?

\color{red} {\mathbf{(USER ~~WAS ~~BANNED ~~FOR ~~THIS ~~POST)} }

[Anonymous No.7940517]

>>7939172
>>7939574
Take your pedophile cartoons back to >>>/a/.