Well, Veeky Forums?

y = y + 2
y - 2 = y

y is the ring of even integers

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

It is clear that y = infinity.

this

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

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

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]

I USE THIS IN MY PAPER, DELETE IT

OP, is the answer 0 = 2?

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

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

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

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

Those three lines are not equivalent.

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

Let y equal infinity

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

[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]

indeed

Stop using my formulas in your paper! Delete them!

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

this pretty much

[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].

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.

It depends on whether these numbers are cardinals or ordinals.

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

He meant dank number.

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

Fuck i actually laughed out loud

up your butt

wowzers

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

holy kek

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

the answer is my dick

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

nice

fucking idiot.

y=2.

because y/y=y

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

Stop using my paper in your formulas! Delete it!

Where did that 90% figure come from?

y + 2 = y

Therefore,

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

Wow, you're so fucking smart.

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

It's either 2 or -2

y=y-2

>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
>???

hehehe

The answer is: y + 2

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

There is nothing plus 2 that equals itself.

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

>cardinals

looks like the bird is perched on a dildo.

wtf

Do you know what math is?

I see someone isn't old enough to be on Veeky Forums.

no, i am an engineer major.

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

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

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

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

y = 0.99999999... = 1

I assumed as much. Civil engineer?

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

no retard, any number satisfies this relation mod 2

-1/6

What ring are we working in?

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.

Is the natural number 1 equal to the integer 1?

y is infinity

literally can be any number so, literally infinity

(y + 2 = y) = false

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

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]

y/y=1

[math]y=0/0[/math]

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

wtf are you doing

What

Lol

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

>hurr durr infinity is a number