Well, /sci/?

>>8831727
>subtract y from both sides
>2=0
impossible

>>8831727
Z^2

assuming modulo arithmetic 2

y is an even number

OP here, you can actually get a value for why if you jump through some hoops for it

>>8831731
>>8831742

forgot to add post

Wait what the fuck

>>8831742
>for why
i'm actually retarded

>>8831727
y = [0] and [1] in z_2

>implicit differentiation

>>8831727
y = y + 2
y - 2 = y

y is the ring of even integers

>>8831727

y + 2 = y
y(1 + 2/y) = y
1 + 2/y = 1
2/y = 0

It is clear that y = infinity.

>>8831785
this

>>8831785
>not using a limit
[math]\lim_{y\to\infty}y+2=\infty[/math]

>>8831727
y=N where N is defined as a number where N = N+2 is true

>>8831727
>>8831727
solution exists in groups isomorphic to the additive integral group of order 2

in particular, 2 is congruent to zero modulo 2.

We do not consider the case that this is modulo 1, since the trivial group is boring.

Introduce the dark number (designated ע) that is always equal to itself plus any natural number.

[math]
\begin{align*}
\sqrt{1} + 2 &= \sqrt{1} \\
-1 + 2 &= 1 \\
1 &= 1
\end{align*}
[/math]

>>8832067
I USE THIS IN MY PAPER, DELETE IT

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>>8831727
OP, is the answer 0 = 2?

Because I thought 0 had to equal 0, how does this work?

>>8831772
>y is the ring of even integers
Where's the 'ring'? I don't see any rings.

>>8832067
{-1,1} + 2 = {1, 3}
They share an element. They are not equal.

>>8832097
The 2 is a dark number when you start taking more complex physics you'll understand.

>>8832067
Those three lines are not equivalent.

>>8832158
yes they are, when you squareroot 1, you collapse it's wave function

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>>8831727
Let y equal infinity

>>8832147
how do dark numbers differ from light numbers?
quantum stuff?

>>8832121
>>8832158

[math]
\begin{align*}
\sqrt{x} &= 2 \\
x &= 4 \\
\{-2,2\} &= \{2\}
\end{align*}
[/math]

And you should define equivalence, and what "set plus number" means.
And if the 3 lines were not equivalent, then you could for example not solve:

[math]
ax^2+bx+c = 0
[/math]

>>8832163
indeed

>>8832084
Stop using my formulas in your paper! Delete them!

>>8832067
>>8832084
>>8832208

you guys should just peer review and cite each other. ez publishing guaranteed.

let y = x where x is defined as the answer to the question

>>8831785
this pretty much

>>8831785
>>8831823
>>8832232
>>8831785
>>8831823
>>8832232
[math] 2 + \omega = \omega [/math], but [math] \omega + 2 \neq \omega [/math].
2 plus "infinity" = "infinity" but "infinity" plus 2 not equal "infinity".
And [math] \infty [/math] is not a number. So, not solvable in [math]\mathbb{R}[/math].

>>8831785
it doesn't work like that m8

A number is a number. There are an infinite amount of them. A number that's an infinite distance away is still a unique number.

You're thinking of when you have 1/y and 1/(y+2) and you send y to infinity, these will both go to 0 and therefore be the same.

>>8832299
It depends on whether these numbers are cardinals or ordinals.

>>8831727
>y + 2 = y
> 2 =0
substitute back in...
>y + 0 = y
>y = y
>1 = 1
the equation is solved by any y

>>8832169
He meant dank number.

>>8831727
divide through by y
1 + 2/y = 1
2/y = 0
divide both sides by 2
1/y = 0
y => infinity
qed
/sci/ btfo

>>8831751
Fuck i actually laughed out loud

>>8832101
up your butt

>>8832375
wowzers

>>8831727
y+2=y
dy/dx=1+c=1
c=0

y+2=y
y+1.99999...9=y
y+1.99999...8=y
continue this 2*infinity times
y+0=y
divide by zero
y/0 +0/0 = y/0
multiply by zero
0=y
here's the tricky part
if 0=y, then y=-1/12

>>8831742
>>8831751
holy kek

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come on guys are you fucking dumb this shit is easy just add 4 to both sides nigga

the answer is my dick

>>8832588
ebin

[eqn]9/11w.a.s.a.n.i.n.s.i.d.e.j.o.b[/eqn]

y is any real number
any real number + 2 = any real number

>>8832067
nice

>>8832602
fucking idiot.

y=2.

because y/y=y

y+2=y
(y+2)/2=y/y
2=y

>>8832208
Stop using my paper in your formulas! Delete it!

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>>8831727
Where did that 90% figure come from?

>>8831727

y + 2 = y

Therefore,

y = 2 + (2 + (2 + ... ))

>>8831731
Wow, you're so fucking smart.

>>8831727
Infinity.

Infinity + 2 = infinity.

there is an answer in characteristic zero if we choose to work in the additive group of integers

let Y = 2Z

>>8831727
It's either 2 or -2

>>8831727
y=y-2

>>8831727
>y+2=y
>(y+2)/y=1
>1+2/y=1
>substitute 1, 2
>x+z/y=x
>z=0
>substitute OP with quality content
>???

>>8833248
hehehe

The answer is: y + 2

>>8831727
Define "2" as being "0". Problem solved

There is nothing plus 2 that equals itself.

>>8832067
radical 1 does not equal -1. negative radical 1 equals -1.

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>>8832304
>cardinals

>>8835166
looks like the bird is perched on a dildo.

>>8831727
wtf

>>8835090
Do you know what math is?

>>8834933
I see someone isn't old enough to be on 4chan.

>>8835178
no, i am an engineer major.

y+2=y
(y+2)/y = 1
1+2/y = 1
2/y = 0
y=2

>>8835210
oh fuck nvm
2/y =0 is not true unless y = infinity

>>8835210
how does the y in (y+2) get turned to 1 after dividng it by y?

>>8835218
(y+2)/y = y/y + 2/y = 1 + 2/y

>>8831727
y = 0.99999999... = 1

>>8835197
I assumed as much. Civil engineer?

only possible answer is plus or minus infinity, no finite number satisfies this

>>8831734
no retard, any number satisfies this relation mod 2

>>8833120
-1/6

>>8831727
What ring are we working in?

>>8831955
The dark number would prevent the number line from satisfying the conditions to be a field.

Assuming the infinity you're talking about is aleph0, aleph0+2 and aleph0 have the same cardinality but are different ordinals. So this solution really depends on what youre talking about.

>>8835864
Is the natural number 1 equal to the integer 1?

y is infinity

literally can be any number so, literally infinity

>>8831727
(y + 2 = y) = false

>>8831727
Pfft, easy.
y=y+2
y=y(1+2/y)
1+2/y=1
2/y=0
y=2/0
y=infinity
Infinity=-1/12
y=-1/12

>>8835878
In the context of this problem yes. But if you're talking about infinities as solutions you should be clear on whether what you're doing is ordinal arithmetic or cardinal arithmetic

1. Define A to be a finite set
2. Define the following relation: [math]A + b = A \cup \{b\}[/math]
Then:[eqn]Y = \{y_1,y_2,y_3,...y_n\}[/eqn][eqn]Y + 2 = \{y_1,y_2,y_3,...y_n, 2\}[/eqn][eqn]\{y_1,y_2,y_3,...y_n\}=\{y_1,y_2,y_3,...y_n, 2\}[/eqn][eqn] 2\in Y[/eqn]Therefore, any set [math]Y[/math] such that [math]2\in Y[/math] is a solution to [math]Y+2=Y[/math].[/math]

>>8832639
y/y=1

>>8831727
[math]y=0/0[/math]

>>8831727
ln(y+e^ipi / y-e^ipi)=0 has no solution. Qed

>>8837347
wtf are you doing

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>>8837347
What

>>8831751
Lol

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you can do this easily by solving for the eigenvalues of a hamiltonian operator

>>8835604
>hurr durr infinity is a number