is this solveable with the amount of current mathematics we currently know?

[Anonymous No.14769816]

[Anonymous No.14769822]

Only if 0 = 2, which is true in groups with integers mod 2.

[Anonymous No.14769825]

>>14769816
This is simple to solve

Y+2 = Y

Multiply both sides by 4, giving us 4Y + 8= 4Y

Resulting in

8=0

Now rub the equation vigorously

8=======0
shit posting quota complete

[Anonymous No.14769830]

>>14769822
I fucking love algebra!

[Anonymous No.14769853]

Infinity

[Anonymous No.14769864]

>>14769816
When the key conditions are defined as such

Y = -2 when you are adding 2 to Y

But Y = 0 when Y is presented as a lone figure

[Anonymous No.14769881]

>>14769853
[eqn]\omega + 2 \neq \omega[/eqn]
in ordinal arithmetic.

[Anonymous No.14770029]

>>14769816
Sure, just make an equivalence relation on formal expressions that identifies every expression with itself, except that y + 2 and y are identified with each other.

Pretty satisfying, right?

[Anonymous No.14770035]

Suppose we're working in binary, y is a 1-bit signal, plus is the OR operation.

Y + 2 would be Y OR 2 but keeping only the least significant bit.

In that case Y + 2 = Y.

[Anonymous No.14770045]

>>14769822
Mom always told me to keep away from rings of characteristic 2.

[Anonymous No.14770670]

>>14769816
y+2=y
2=y-y
2=0
ez, nowhere it says "solve for y"

[Anonymous No.14770685]

(y + 2) = y |^2
y^2 + 4y + 4 = y^2
4y + 4 = 0
y = -1.

[Anonymous No.14770731]

>>14770685
>-1+2=-1

[Anonymous No.14771186]

>>14769816
>DUDE SOLVE THIS UNSOLVABLE EQUATION
>IF YOU CANT SOLVE IT YOURE A DUMBASS LMAOOOOOOOOO

[Anonymous No.14771300]

>>14769816
>infinity minus 1 = infinity
y = infinity

[Anonymous No.14771592]

a = y+2 = y
2a = 2y+4 =2y
2a=
y+2= y
0+y+2= y
y+2+y+2 = y+y
(y+2)+(y+2) = (y+y)
2(y+2) = 2y
[2(y+2)]/2 = 2y/2
y+2=y
subtract y from both sides
0+2=0
subtract 2 from both sides
0= 0-2
divide both sides by -2
0/-2 = -2/-2
solve
[math]\frac{0}{-2}=1[/math]
-2/-2 = 1
-1.5/-2 = 0.75
-1/-2 = 0.5
-0.5/-2 = 0.25
0/-2 = 0

0=1

add 3 to both sides
3=4
divided by 3
1=1.333333333
divided by 3
0.333... = 0.444...

times 3
[math]0.999... = 1[/math]

[Anonymous No.14771615]

y=infinite
y+2=infinite

Easy

[Anonymous No.14771616]

>>14771592
kek.
0.999... ≤ 1

[Anonymous No.14773846]

>>14769816
[math]y = \sqrt{1}+1[/math]
[math]y = 0[/math] and [math]y = 2[/math]
[math]0+2 = 2[/math]
ez

[Anonymous No.14775334]

>>14773846
*sigh*

[Anonymous No.14775473]

>>14773846
Sqrt is defined to always be positive.

[Anonymous No.14776794]

>>14775473
https://en.wikipedia.org/wiki/Square_root

[Anonymous No.14776820]

>>14775473
>Imaginary numbers

[Anonymous No.14776821]

>>14776794
That's objectively wrong. Sqrt(x^2) = |x|

The equation x^2 = y has 2 solutions, +-x but the square root of a number is always positive.

[Anonymous No.14776825]

>>14776821
So you're smarter than Wikipedia huh

[Anonymous No.14776957]

>>14776825
Anyone can edit Wikipedia and if that is actually in the wiki, it is wrong.

[Anonymous No.14777247]

>>14769816
Maybe with complex numbers.
I mean:

xi + 2 = xi +2


Or I am retarded...

[Anonymous No.14777297]

>>14769816
Just use group theory m8

[Anonymous No.14777339]

>>14769881
hes talking about [math]\infty+2=\infty[/math]

[Anonymous No.14777347]

>>14776820
>being to much of a brainlet to let nomenclature created by dudes that didn't like "imaginary" numbers being... well just that was enough to send them in a tizzy, them being
just call them lateral numbers if you are to mentally impaired to see past their name

[Anonymous No.14778370]

>>14776957
It's correct under the definition they've provided for it. The square root function is defined as always being positive, but the square root as a solution to a polynomial is not. Nowhere in that definition do they actually invoke the square root function.

[Anonymous No.14778405]

y+2=y
2=0y
y=2/0
seems like a new type of numbers with some interesting properties
2/0+2=2/0
2/0-2/0=-2
2/0-2/0=2

[Anonymous No.14778483]

>>14769816
the solution is "false"

[Anonymous No.14778493]

>>14769816
>doesn't know that parallel lines can intersect in non-euclidean geometry

[Jealous No.14778497]

>>14777339
it's implicit that you're working in R, or maybe C.

[Anonymous No.14778500]

>>14778370
>4 and -4 are the square roots of 16.
They implicitly stated it here. The square root of 16 is always 4.

If instead they said the solutions to x^2 = 16 is +-4, then they would be correct as you and I have already stated.

[Anonymous No.14778515]

So how does x^1/2 work? Is ^1/2 equivalent to the square root symbol, giving the positive principal root of x? Or can x^1/2 be both positive and negative?

[Anonymous No.14778670]

>>14778515
One of the main motivations for modern algebra was to ensure that every polynomial of degree n has n solutions. yconsider. the 2nd degree polynomial ax^2 + bx + c = y,

If you look at just x^2 = y, it should have 2 solutions, +-x. The square root function was defined such that sqrt(x^2) = |x|. So we have y = |x| which is just +-x.

This is a common point of confusion even for grad students.

[Anonymous No.14779618]

>>14769822
It's solvable, for some values of plus

[Anonymous No.14780407]

>>14769816
y=7

[Anonymous No.14780498]

17 = 23
Solve.
Show your working.

[Anonymous No.14780577]

>>14769816
it increment's y by 2

[Anonymous No.14780591]

>>14769816
>y = 0 * -
>0 * -(+2) = 0
easy

[Anonymous No.14780611]

>>14771615
I like you.

[Anonymous No.14780901]

>>14769816
y = y - 2
Solved

[Anonymous No.14780934]

>>14769825
kek

[Anonymous No.14781596]

Solve this, /sci/.

[Anonymous No.14781783]

>>14769816
>10% of math graduates solve this
bad schools

[Anonymous No.14781821]

>>14769816
The statement "y+2 = y" is always false for any y \in R. There isn't any y for which this expression is true, this is your answer.

[Anonymous No.14781985]

>>14769816
y+2=y
y+2=y
2=y-y
2=0 -> False
Therefore there is no (known) solution.
OR
We could take y=y+2 as saying y is y but the value is always y+2

[Anonymous No.14782433]

>>14769816
y + 2 = y <==> FALSE

For a solution that isn't math but still relevant for a change.