Troll thread

[Anonymous No.2033085]

<span class="math">-1 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{(-1) \cdot (-1)} = \sqrt{1} = 1[/spoiler]

[Anonymous No.2033093]

aw sheet.
I guess we're all fucked then.

[Anonymous No.2033106]

sqrt(ab)=abs(sqrt(a)sqrt(b))
fucking fags don't know basic algebra

[Anonymous No.2033118]

All numbers equal all other numbers

The following example uses division by zero to prove that 2 = 1, but can be modified to prove that any number equals any other number.

1. Let ''a'' and ''b'' be equal non-zero quantities
<span class="math">a = b [/spoiler]
2. Multiply through by ''a''
<span class="math">a^2 = ab [/spoiler]
3. Subtract <span class="math">b^2 [/spoiler]
<span class="math">a^2 - b^2 = ab - b^2 [/spoiler]
4. Factor both sides
<span class="math">(a - b)(a + b) = b(a - b) [/spoiler]
5. Divide out <span class="math">(a - b) [/spoiler]
<span class="math">a + b = b [/spoiler]
6. Observing that <span class="math">a = b [/spoiler]
<span class="math">b + b = b [/spoiler]
7. Combine like terms on the left
<span class="math">2b = b [/spoiler]
8. Divide by the non-zero ''b''
<span class="math">2 = 1 [/spoiler]

[Anonymous No.2033129]

>>2033085
>>2033085
>>2033085
I FUCKING BROUGHT THIS TO /SCI/ AGES AGO AND YOU GO AND STEAL MY FUCKING THUNDER!!!!ONE!!!ELEVEN!!!FORTY-TWO!!!!

[Anonymous No.2033132]

A proof that all positive integers are equal to zero follows.

Suppose we have the following system of linear equations:
<span class="math">\left\{
\begin{matrix}
c_1x_1 + c_1x_2 + \cdots + c_1x_n = c_1\\
c_2x_1 + c_2x_2 + \cdots + c_2x_n = c_2\\
\vdots \\
c_nx_1 + c_nx_2 + \cdots + c_nx_n = c_n
\end{matrix}\right.
[/spoiler]

Dividing the first equation by c1, we get x1 + x2 + · · · + xn = 1. Let us now try to solve the system via Cramer's rule:

<span class="math">\begin{bmatrix}
c_1 & c_1 & c_1 & \dots\\
c_2 & c_2 & c_2 & \dots\\
\vdots&\vdots&\vdots&\\
c_n & c_n & c_n & \dots
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
\vdots\\
x_n
\end{bmatrix} =
\begin{bmatrix}
c_1\\
c_2\\
\vdots\\
c_n
\end{bmatrix}<span class="math">

Since each column of the coefficient matrix is equal to the resultant column vector, we have
A_i = A \implies |A_i| = |A| \implies {|A_i| \over |A|} = 1 = x_i

for all i. Substituting this back into x1 + x2 + · · · + xn = 1, we get

\sum_{i=1}^n 1 = 1[/spoiler][/spoiler]

[Anonymous No.2033134]

-1 = sqrt(1·i)

you first statement is false idiot.

[Anonymous No.2033135]

1 + 1 = 11

[Anonymous No.2033137]

>>2033132
>I can't tex

[Anonymous No.2033140]

A proof that all positive integers are equal to zero follows.

Suppose we have the following system of linear equations:
<span class="math">\left\{
\begin{matrix}
c_1x_1 + c_1x_2 + \cdots + c_1x_n = c_1\
c_2x_1 + c_2x_2 + \cdots + c_2x_n = c_2\
\vdots \
c_nx_1 + c_nx_2 + \cdots + c_nx_n = c_n
\end{matrix}\right.
[/spoiler]

Dividing the first equation by c1, we get x1 + x2 + · · · + xn = 1. Let us now try to solve the system via Cramer's rule:

<span class="math">\begin{bmatrix}
c_1 & c_1 & c_1 & \dots\
c_2 & c_2 & c_2 & \dots\
\vdots&\vdots&\vdots&\\
c_n & c_n & c_n & \dots
\end{bmatrix}
\begin{bmatrix}
x_1\
x_2\
\vdots
x_n
\end{bmatrix} =
\begin{bmatrix}
c_1\
c_2\
\vdots\
c_n
\end{bmatrix}<span class="math">

Since each column of the coefficient matrix is equal to the resultant column vector, we have
A_i = A \implies |A_i| = |A| \implies {|A_i| \over |A|} = 1 = x_i

for all i. Substituting this back into x1 + x2 + · · · + xn = 1, we get

\sum_{i=1}^n 1 = 1[/spoiler][/spoiler]

[Anonymous No.2033142]

no come back! solve you're fail tex

[dolan !Uq56cE.45s No.2033147]

[Anonymous No.2033150]

A proof that all positive integers are equal to zero follows.

Suppose we have the following system of linear equations:
<span class="math">\left\{
\begin{matrix}
c_1x_1 + c_1x_2 + \cdots + c_1x_n = c_1\
c_2x_1 + c_2x_2 + \cdots + c_2x_n = c_2\
\vdots \
c_nx_1 + c_nx_2 + \cdots + c_nx_n = c_n
\end{matrix}\right.
[/spoiler]

Dividing the first equation by c1, we get x1 + x2 + · · · + xn = 1. Let us now try to solve the system via Cramer's rule:

<span class="math">\begin{bmatrix}
c_1 & c_1 & c_1 & \dots\
c_2 & c_2 & c_2 & \dots\
\vdots&\vdots&\vdots&\
c_n & c_n & c_n & \dots
\end{bmatrix}
\begin{bmatrix}
x_1\
x_2\
\vdots
x_n
\end{bmatrix} =
\begin{bmatrix}
c_1\
c_2\
\vdots\
c_n
\end{bmatrix}<span class="math">

Since each column of the coefficient matrix is equal to the resultant column vector, we have
A_i = A \implies |A_i| = |A| \implies {|A_i| \over |A|} = 1 = x_i

for all i. Substituting this back into x1 + x2 + · · · + xn = 1, we get

\sum_{i=1}^n 1 = 1

for fucks sake where is the invalid control[/spoiler][/spoiler]

[Anonymous No.2033163]

yall mad

[Anonymous No.2033185]

We are going to prove that 1=0. Take the statement

<span class="math">x = 1[/spoiler]

Taking the derivative of each side,

<span class="math"> \frac{d}{dx}x = \frac{ d}{dx}1 [/spoiler]

The derivative of x is 1, and the derivative of 1 is 0. Therefore,

<span class="math"> 1 = 0 [/spoiler]

<span class="math">Q.E.D.[/spoiler]

[Anonymous No.2033200]

>sqrt(-1*-1)
and that's where you failed

[Anonymous No.2033207]

We are going to prove that 1=0. Start with the addition of an infinite succession of zeros
<span class="math">0 = 0 + 0 + 0 + \cdots[/spoiler]
Then recognize that <span class="math">0 = 1 - 1[/spoiler]
<span class="math">0 = (1 - 1) + (1 - 1) + (1 - 1) + \cdots[/spoiler]
Applying the associative law of addition results in
<span class="math">0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + \cdots[/spoiler]
Of course <span class="math">-1 + 1 = 0[/spoiler]
<span class="math">0 = 1 + 0 + 0 + 0 + \cdots[/spoiler]
And the addition of an infinite string of zeros can be discarded leaving
<span class="math">0 = 1 \,[/spoiler]

<span class="math">Q.E.D.[/spoiler]

[Anonymous No.2033212]

[Anonymous No.2033283]

oh lawd

[Anonymous No.2033293]

Proof d/dx x = 1

d/dx x

Divide the d's

1/x x

Simplify

x/x

d/dx x = 1

Problem?

[Anonymous No.2033309]

>>2033207
Never seen this one before. Good one. How do you rigorously show this formulation to be invalid?

[Anonymous No.2033335]

>>2033309

The error is that the associative law cannot be applied freely to an infinite sum unless the sum is absolutely convergent (see also conditionally convergent). Here that sum is 1 − 1 + 1 − 1 + · · ·, a classic divergent series. In this particular argument, the second line gives the sequence of partial sums 0, 0, 0, ... (which converges to 0) while the third line gives the sequence of partial sums 1, 1, 1, ... (which converges to 1), so these expressions need not be equal.

[Anonymous No.2033346]

>>2033309
He's missing a -1

xn = (1 - 1)

So the equation should have an equal amount of positive and negative 1s, but he gets rid of one.

So it should be

0=1+(−1+1)+(−1+1)+(−1+1)-1

[Anonymous No.2033348]

>>2033346
trooooooooooooooololololol!!!!

[Anonymous No.2033352]

>>2033346
In an infinite set of (-1 + 1) there is no -1 at the end. For every -1 you put at the end, I could argue that there is a + 1 behind it (as per the pattern).

[Anonymous No.2033362]

>>2033085
this is a true statement

[Anonymous No.2033372]

We are going to prove that 2π = 0


x = 2π
sin(x) = 0
x = arcsin(0)
x = 0
2π = 0

Q.E.D.

[Anonymous No.2033378]

[Anonymous No.2033387]

We are going to prove that 1 = 3.

From Euler's formula we see that

<span class="math">e^{\pi i} = \cos(\pi) + i \sin(\pi) = -1 + 0i = -1 [/spoiler]
and
<span class="math">e^{3\pi i} = \cos(3\pi) + i \sin(3\pi) = -1 + 0i = -1 [/spoiler]
so we have
<span class="math">e^{\pi i} = e^{3\pi i} [/spoiler]
Taking logarithms gives
<span class="math">\ln(e^{\pi i}) = \ln(e^{3\pi i}) [/spoiler]
and hence
<span class="math">\pi i = {3\pi i} [/spoiler]
Dividing by <span class="math">\pi i[/spoiler] gives
<span class="math"> 1 = 3 [/spoiler]

<span class="math">Q.E.D.[/spoiler]

[Anonymous No.2033406]

:D

[Anonymous No.2033418]

We will now prove that ∞ equals -∞.

Let x = ∞
Let x = -∞
Since x = x, ∞ = -∞.
QED

[Anonymous No.2033436]

WHAT THE FUCK DOES Q.E.D. MEAN?????'''

[Anonymous No.2033448]

>>2033436
>QED
>Qan't Even Deny

[Anonymous No.2033449]

>>2033436
quod erat demonstrandum
something along the lines of hence what was needed was shown

[Anonymous No.2033451]

why not just:

-1 = sqrt(1) = 1

[Anonymous No.2033460]

>>2033449
what was to be shown

[Anonymous No.2033529]

>>2033436

Quantum Electrodynamics

L2science