/sci/ - Science & Math

y = (-2, 0, 2)

>>15177292
What does y+2 mean in that context then?

>>15177321
(-2, 0, 2) +2
Cant you read?

[eqn]\begin{align}
y+2 &= y \\
1 + \frac{2}{y} &= 1 \qquad (y\neq 0) \\
\frac{2}{y} &= 0 \\
y &= \infty
\end{align}[/eqn]

>>15177283
y is equal to infinite.
No need to thank me.

>>15177323
No

But, why does (-2, 0, 2) + 2 = (-2, 0, 2) ?

>>15177337
Because you take y to be 0
And +2 to that is ±2
Which is 0±2
Which is (-2, 0, 2)

>>15177283
2 = 0

>>15177346
Stupidest thing I've ever seen. Go think about what you've done.

>>15177351
+2
y +2 = y
But

0 +2 = 2
and
-2 +2 = 0

so y = (-2, 0, 2)

Is it really that hard?

>>15177361
You can't add a tuple of numbers, to a number. Sorry, but isn't defined under any standard notational conventions.

>>15177292
>>15177323
>>15177332
>>15177346
>>15177350
Dumbasses, The real solution is
[math] y + 2 = y [/math]
[math] 4y + 8 = 4y [/math]
[math] 8 = 0 [/math]
Then rub the equation vigorously to get
[math] 8 ============== 0 [/math]

Cup is LIV

>>15177366
And you are supposing it will equal 1 actual answer rather than array of answers ?
in this case y to y+2 is ±2
(y -2 is also y)
if you did not see

>>15177388
>array
Go back to programming and stop dirtying the field of mathematics.

>>15177361
Ok, ok, slow down Einstein. What do you mean when you write something like (x, y, z)? I'm genuinely curious now. And how do you define something like (x, y, z) + w for example?

>>15177414
He's being a dumbass and trying to convince himself that it is possible to give y 2 values at the same time like in his computer. Except unless we are talking quantum shit, if you can give y more than one value at the same time. You can have multiple solutions, but not shove them in y all at once, only one at once.

>>15177283
Let [math]y = \{ a : a = 2r, r\in \mathbb Z \}[/math].

[math]\begin{align*}
y+2 &= \{a : a = 2r, r\in \mathbb Z\} + 2\\
&= \{a + 2:a = 2r, r\in\mathbb Z\}\\
&= \{a : a = 2r, r\in \mathbb Z\}\\
&= y\\
\end{align*}
[/math]

QED.

>>15177414
The question is just retarded, dont really want to jump deeper into bullshit math that can cause you brain damage

There's nothing to solve. This is a tautology.
t. only works in finite fields of characteristic 2

>>15177283
nan

>>15177474
Lol you could have just used the integers themselves.

/int/ vs /vp/ is about to start, we're up next after them

>>15177520
You are assuming we are in a field

>>15177332
BASED

It's time.

We are about to engage in a game on which the fate of our team depends and it is important to remind all players that the moment has passed for looking to the rear; all our efforts must be directed to attacking and driving back the enemy. Under the circumstances which face us, no act of weakness can be tolerated.

>>15177332
Not the only solution, what about negative infinity?

>>15177283
What is there to solve?

>>15177768
valid.
\begin{align} y+2 &= y \\ 1 + \frac{2}{y} &= 1 \qquad (y\neq 0) \\ \frac{2}{y} &= 0 \\ |y| &= \infty \end{align}

That's a fortuitous goal, Shinichi

>that through ball to glitchart
fuck

HAWKING !

>we finally beat /gd/

Through to the knockouts with 9 points!

Points-wise, /sci/ is the best team right now.

y+2=y (mod 2)
Cope seethe &dilate

The answer is true.

>>15177283
Infinity. There. I said it.

>>15177332
>>15177624
blessed posts

>>15177939
We topped the group today, we're probably 9 point cursed, but that doesn't matter too much.

>90% can't solve it
and they're right

>>15177283
y is the string 2222...
concatenating a 2 to y leaves y unchanged

>>15178076
>concatenating
CS pleb spotted

Today's matchday
https://implying.fun/video/winter23/2023-02-04/?t=2:56:51.0

>>15177332
based engineer

>>15177283
Solution is {}.

>>15177627
I wouldn't be working on the problem at all if it wasn't in a field

>>15177283
You don't have to "solve" it, it can't be simplified by rules of algebra. The solution set is the entire complement of the finites, but you can't arrive at that conclusion by simplifying the equation, because numbers outside the finites don't follow those rules. If y isn't assumed finite, you can't subtract y from both sides and assume the equation that results is as true as the one you started with, i.e. the solution is not 2 = 0.