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/sci/ - Science & Math

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8400563 No.8400563 [Reply] [Original]

.

>> No.8400569

>>8400563

x=x+1
x+1=x
(x+1)-1=x-1

x=x-1

>> No.8400573

>>8400563
[math]\{x \in \mathbb{C} \ | \ x = x + 1\} = \emptyset[/math]

>> No.8400585

x=x+1 (mod 1)

>> No.8400593

>>8400563
x =/= x+1
that was easy

>> No.8400599

>>8400563
x=x+1
x=(x+1)+1
x=((((x+1)+1)+1)+1)+1
x=x+n
alt:
x=(x+1)
x=x
alt:
x=x+1
(x^2)=(x^2)+2x+1
-1=2x

>> No.8400604
File: 85 KB, 1024x512, Cnt3qUSW8AE-lHS.jpg [View same] [iqdb] [saucenao] [google]
8400604

x=x+1
>x=0
>0=1
>So x=/=0
xº=(x+1)º
1=1
So x∈R if x=/=0

>> No.8400605

Y(Succ)

>> No.8400634

>>8400563
/g/ here.

x = x + 1;

x += 1;

x++;

Get on my level faggots.

>> No.8400642

>>8400634
>/g/
>simple python operation
As expected.

>> No.8400666

>>8400563
-1/12

>> No.8400679

x = Aleph Naught

>> No.8400682
File: 193 KB, 1500x1005, e_ph_0273_01.jpg [View same] [iqdb] [saucenao] [google]
8400682

x = + infinity or -infinity if were working over the extended reals

>> No.8400698

>>8400642
>thinks x++; is specificaly a python operation
>ignores that pretty much every language uses the same shorthand for increment
As expected

>> No.8400713

>>8400698
Thanks for the info.

Which p.language is more useful tho?

>> No.8400763
File: 194 KB, 542x441, 1475199905275.png [View same] [iqdb] [saucenao] [google]
8400763

>>8400563
x=x+1

x=(x+1)+1

x=((x+1)+1)+1

x=x+2+1

repeat three more times

x=x+3+2+1

continuing in this fashion,

x=...+3+2+1

x=-1/12

>> No.8400805

x=x+1
x^2=x(x+1)
x^2=x^2 + 2x + 1
0=2x+1
-1=2x
x=-1/2

>> No.8400828

>>8400805
>-1/2 = -1/2 + 1
your x is actually |x|

>> No.8400936

>>8400642
>python
>semicolon

>> No.8400941

>>8400763
kek'd

>> No.8400943

>>8400563
[eqn]x=\sum^\infty 1[/eqn]

>> No.8401011

>>8400805

Are you high?

>x^2=x(x+1)
>x^2=x^2 + 2x + 1

x^2=x(x+1)
x^2=x^2 + x

Think i just fell for bait

>> No.8401037

>>8400642
There's no semicolons in python

>> No.8401043

>>8400634
Pretty sure there's more
++x;
x=x++;
But not
x=++x;
Right?

>> No.8401057

>>8401011
No he just skipped a step

x(x+1)=(x+1)(x+1)=x^2+2x+1

>> No.8401213
File: 37 KB, 261x394, gendercucks.jpg [View same] [iqdb] [saucenao] [google]
8401213

>>8400573

Stop assuming [math]x[/math]'s domain, or the definition of [math]=[/math] and [math]+[/math].

TRIGGERED

>> No.8401393

>>8400682
We're not.

>> No.8401429

>>8401037
>>8400936
Aren't they optional? It's not standard and you really shouldn't but IIRC you can if you want to. Same with extra parentheses.

>> No.8401440
File: 1.40 MB, 1530x720, 1476007007.webm [View same] [iqdb] [saucenao] [google]
8401440

>>8400563
The answer is 2.

>> No.8401441
File: 47 KB, 641x696, maths.jpg [View same] [iqdb] [saucenao] [google]
8401441

>>8400713

Z3 is superior for shitposting.

>> No.8401447
File: 57 KB, 674x490, mathsinjava.jpg [View same] [iqdb] [saucenao] [google]
8401447

>>8401441

>not using superior Java

>> No.8401451

>>8400563
literally just no solutions

not very hard

>> No.8401457

>>8401451

There were "literally just no solutions" to sqrt(-1) unless maths evolved.

Give it time.

>> No.8401468

>>8400563
that's just how you increment a number in like any basic programming language

>> No.8401508

>>8401457
i know a good bunch of math that proves this iterative function does not converge

>> No.8401879

>>8400563
>8400563
[math]\frac{-1}{12}[/math]

>> No.8401885

>>8400563
[eqn]\infty = \infty + 1[/eqn]

;)

>> No.8402281
File: 42 KB, 679x522, Shitposting quota filled through algebra.jpg [View same] [iqdb] [saucenao] [google]
8402281

This kind of thing again? Pic related.
x=x+1
8x=8x+8
0=8
Now stroke the equation vigorously.
0======8

>> No.8402288
File: 313 KB, 499x618, do NOT save this.jpg [View same] [iqdb] [saucenao] [google]
8402288

>>8401393
says who?

>> No.8402309

>>8401885

Best solution

>> No.8402336

>>8401885
[math]-\infty = -\infty + 1[/math]
:6)

>> No.8402337

>>8400763
S

>> No.8402385

>>8400563
CHRISTINA! I DON'T HAVE TIME FOR YOUR SHENANIGANS! the organisation sleeper agent is trying to distract me with basic maths, i am resisting as best i can, but may need an assist, el psy congroo.

>> No.8402690

>>8402281
I remember that thread.

>> No.8402781
File: 371 KB, 1020x649, PANT PANT.png [View same] [iqdb] [saucenao] [google]
8402781

>>8402385
wait
hah hah

>> No.8402792

>>8401043
>x=++x;
This could work depending on the language. It's just a really retarded way of doing it.

>> No.8403026

Not an equation, but the notation for a progress.
A concept regarding infinity, if there has to be a "solution".

>> No.8403036

>>8400585
Literally the only correct one

>> No.8403068

>>8401885
same answer

>> No.8403222

the answer depends on how you define 1. /thread

>> No.8403230

>>8403222
It does, the same way my whole sentence depends on how I define "It", "does", "the", "same" and so on. Fuck off you retard. The set of solutions is the empty set.

>> No.8403246

>>8400563
shorter to type x++

>> No.8403255

>>8400585
but that's not the formula in the pic.

>>8400573
>>8400604
are correct.

>>8400642
>python
>post-increment operator

>> No.8403258

>>8400563
The trivial ring is the only ring satisfying such an equation for some x.

If you want something that's not a ring, look at the extended real numbers or something. If indeed you define \infty + 1=\infty.

>> No.8403259
File: 450 KB, 720x1154, Screenshot_2016-10-07-20-56-19-1.png [View same] [iqdb] [saucenao] [google]
8403259

>>8400763

>> No.8403263

what are we suppose to do?

>d/dx(x) = d/dx(x+1)
>1 = 1+0
>1 = 1

>> No.8403269 [DELETED] 

>>8403263
So the equations are only off by a constant. Brilliant. So every x is a solution to the equations modulo a constant. So $\forall x (x \equiv x + 1 \mod 1)$. Brilliant job Einstein.

>> No.8403272 [DELETED] 

>>8403269
[math]\forall x (x \equiv x + 1 (\mod 1))[/math]

>> No.8403277

>>8403263
So you've proved: [math]\exists y \forall x (x+y=x+1)[/math]

Nice job!

>> No.8403316
File: 217 KB, 1189x719, solve (2).png [View same] [iqdb] [saucenao] [google]
8403316

>>8400563
Let's get serious in here boys

>> No.8403375

>>8402385
I love you...

>> No.8403383

>>8403255
>but that's not the formula in the pic
>Literally no idea what he is talking
Back to /g/ with you.

>> No.8403388

x_{n+1} = x_n + 1

>> No.8403394

>>8400563
if cs, it's x++, or x += 1

if it's math, x = 0.

>> No.8403397

>>8403394
>x = 0
You didn't think this through, did you?

>> No.8403405

>>8400563
what means solving here ?

>> No.8403406

>>8400563
You can either understand this in the context of a programming loop, or in the context of the extended reals. In the later case, the solution is [math] x = \pm \infty [/math]

>> No.8403408

>>8402792
> not enjoying the insanity of preincrementation
I had it backwards anyway
x=++x;
works.
x=x++;
doesn't

>> No.8403409

>>8403406
Not only, the equation is true for any trivial group.

>> No.8403877
File: 38 KB, 413x395, 1467996062688.jpg [View same] [iqdb] [saucenao] [google]
8403877

>>8400763
Now THIS is the quality of shitposting that I'd expect from /sci/.

>> No.8403895

>>8400563
Only true when 1 represents the multiplicative identity. Solved.

>> No.8403897

>>8403895
go ahead and show us how that works then

>> No.8403904

>>8403895
1 always represents the multiplicative identity. You mean only true when the multiplicative identity is equal to the additive identity which is equivalent to when we're working in the trivial ring which is what I said here >>8403258

>> No.8403906

|x| = 0.5

>> No.8403909

>>8403906
|x| = -0.5

>> No.8403935

x = x+1
0 = 1

Equation doesn't work, not mathematically feasible.

>> No.8404038

>>8403935
I mean, yes, but technically the answer is the empty set.

>> No.8404174

>>8403935
>>8404038
The equation holds for every x in the trivial ring. Idiots.

>> No.8404203

>>8404174
no it doesnt you fucking retard. it holds SPECIFICALLY for the empty set and the empty set only.

>> No.8404216

>>8404203
Let [math]R = \{0\}[/math]. Define 0+0=0, 0*0=0. Now this is the trivial ring with 0=1 (where 1 is conventionally the additive identity).

Consider arbitrary [math]x\in R[/math]. Now we must have x=0 since there are no other choices. And by definition 0+1=0+0=0. QED

>> No.8404217

>>8400563
this is simply modular arithmetic. alternatively, prime field addition.

x congruent to zero modulo n. try harder OP

>> No.8404227

>>8404217
modulo 1, excuse me.

>> No.8405303

>>8400763

I've been coming to this board for like 5 years and this is probably the best post I've ever seen.

>> No.8405305 [DELETED] 
File: 35 KB, 732x234, give_up.png [View same] [iqdb] [saucenao] [google]
8405305

>>8400563
i give up

>> No.8405678

x=x+1
X=? In The first place?
Else x = 0
Meaning x = 1 or not mathematically plausable.
x=x+1/10 wasted time

X=0 0+1 =1

>> No.8405766

>>8404216
Okay yep, you're right. I apologize, I COMPLETELY forgot about that property of the trivial ring.

>> No.8405779

Y'all niggers. x=C*sqrt(-1)

>> No.8406045

X=infinity

>> No.8406127

>>8400943
Best answer, i think

>> No.8406156

1=1+1/x
0 = 1/x
x = infinity
:^)

>> No.8406723

>>8400763
Fucking hell I'm in Probability right now and I can't stop chuckling.

>> No.8406749

>>8400563
x is transfinite and I can't be arsed to write a proof

>> No.8406774

x = x +1
x - x = x +1 - x
0 = 1

>> No.8406793

Using boolean algebra the correct answer would be x=1

>> No.8406822

>>8400563
X=1+X
X=-1+X
X=1+1+1+1+1+1....
X=-1-1-1-1-1-1-1-1.....
Infinity < X < -infinity

>> No.8406999

>>8400642
python doesn't have x++ dumbass

>> No.8407508

>>8402336

Is the 6 supposed to be a like a cylindrical nose? If so this is now my favorite ironic emote face ever

>> No.8407549
File: 137 KB, 748x740, THSOvRA.jpg [View same] [iqdb] [saucenao] [google]
8407549

The only answer is a quantum number that simultaneously holds the value of 1 and 0.
> You too can construct a quantum number such as this one with the simple definition:
ϕ = a
ϕ = b
a =/= b

>> No.8407621

>>8400563
A math equation is a statement that has a true/false value. If the statement is true, algebra rules will get you to another equation that is also a true statement. If the original statement is false, you will end up with a final statement (the "solution") which is also false. Any other fun I can ruin for you?

>> No.8407628

>>8400563
Should be fine for any ordinal number that is not a natural number

>> No.8407691

>>8400763
fuck you nice shitpost

>> No.8409103

x=x+1
x^2=x (x+1)
x^2=x^2+x
x^2-x^2=x
x=0

>> No.8409118

>>8400563
x=/=0 since 0=1
Therefore, {x|x ∈ R, x=/=0}.

>> No.8409140

>>8400563
(x)=(x+1)
1=(x+1)/(x)
1 = 1 + 1/x
0 = 1/x
This is what you started with. x = infinity.

>> No.8410024

>>8400563
x=x+1
subtract x from both sides
=1
the answer is 1.

>> No.8410028

>>8400563
I am assuming that the elements are coming from some ring R, and that 1 is the multiplicative identity.

then

x=x+1
0=1

So we are working in the trivial ring (0), so x=0=1 for all x.

>> No.8410835

[math]x=x+1[/math]
Given

[math]\frac{x}{x} = \frac{x+1}{x}[/math]
Divide both sides by x

[math] 1=1+\frac{1}{x}[/math]
x/x = 1

[math] 0 = 1/x [/math]
subtract both sides by 1

Therefore,

[math]\lim_{x\rightarrow \infty}=0 [/math]

There is no definite outcome, however the answer can be expressed as a limit of 1/x as x approaches infinity.

>> No.8410843

>>8400763
I fucking lol'd

>> No.8410850

>>8403255
>>8400573
>>8400604
Waaaaay overthinking it.

>>8410835
Correct.

>> No.8410853
File: 13 KB, 166x196, philosophicalyellowbobcat.jpg [View same] [iqdb] [saucenao] [google]
8410853

>>8400763
Can someone please explain this meme to me?

>> No.8410888

>>8410853
it's nonsense but -1/12 is the value of the infinite series of the natural numbers which is an epic meme

>> No.8410962

>>8410835

how do you know x isn't 0?

>> No.8410966

>>8410835
>subtract both sides by 1

subtract 1 from both sides*

are you some kind of foreigner?

>> No.8410971

>>8400763

I GET IT I GET IT

BEST JOKE EVER BECAUSE I GET IT

>> No.8410976

>>8400763
>>8405303
>>8410888

Why aren't you allowed to substitute like that?

>> No.8411146

>>8410888
>>8410976
I just watched the proof video and I can't figure it out. How can it be wrong and still work in physics?

>> No.8411175

>>8410962
Because 0+1 =/= 0

>> No.8411179

>>8403909
That's not right If
|x|=-c, there is no solution.

>> No.8411180

>>8411146
By traditional summation, it's incorrect (and rightfully is divergent), but when plugging it into the Riemann-zeta function, you get -1/12. It's the Riemann-zeta answer that's important in physics.

>> No.8411183

>>8411175
1 = 0 in the trivial ring

>> No.8411254

>>8400563
Hmm nothing is given about what the addition operation is. So I'll just conclude that for whatever it is, 1 is the identity.

>> No.8411341

>>8400763
You are a fascinating creature.

>> No.8411437

>>8405303
I concur

>> No.8412571

>>8410024
It's simple: "=1".
Underrated post.

>> No.8412574
File: 14 KB, 326x262, 5962636+_745be62738178521a40cba17b0bc0283.jpg [View same] [iqdb] [saucenao] [google]
8412574

>>8400763

>> No.8412753

X is unaffected by change in quantity.

>> No.8412901

>>8400563
So I think everyone else is on a higher realm of thinking than me. I'm just trying to think through this by using simple algebra.

Am I solving for x?

It's false if this is the case then. x-x = 0
0 =/= 1

>> No.8412944

>>8400763
We need more people like you
Stay on 4chan for ever please

>> No.8412947

>>8412901
This is exactly right. Everyone else in this thread is not thinking on a higher level, they are just trolling. As you say, some simple algebra yields 0 = 1, which is a contradiction, which indicates that there are no values of x that satisfy the equation.

>> No.8412962

>>8412901
>higher realm of thinking
Is this what kids call all mathematics outside of grade school addition these days?

>> No.8412969

>>8412947
>0 = 1, which is a contradiction
It is only a contradiction on most sets, as noted many times in this thread (including on the 3rd post), you can have algebraic structures that have 0 = 1, so that this leads to no contradiction and leads to the solution {1} or {0} since 1 = 0.
Otherwise the solution is indeed the empty set.

>> No.8413023

let F be a field of characteristic 1. then 0=1 and for all x in F, x+1=x.

>> No.8413114

>>8410966
Are you fucking retarded?

>> No.8414280

>>8400563
x = 0 = 1 over the trivial ring
it has no solutions somewhere else homeboy

>> No.8414375

>>8410962
Because you can't divide by zero

>> No.8414438

>>8413023
>field of characteristic 1

Good job

>> No.8414459

>>8400563
I'm pretty sure 0 =/= 1

>> No.8414563

>>8414459
you're not using the right ring then

>> No.8414619
File: 190 KB, 443x1347, 1472324801159.jpg [View same] [iqdb] [saucenao] [google]
8414619

>>8400563

>> No.8414814

>>8410976
Assigning a value to a divergent series is essentially trolling.

>> No.8414820

>>8400563
x = x +1
x - x = x + 1 - x
0 = 1
>x is not real

x = x + 1
1 = (x+1)/x
x = +-inf

>> No.8414968
File: 251 KB, 983x1013, 1407050522436.jpg [View same] [iqdb] [saucenao] [google]
8414968

>>8400763

>> No.8415032

>>8414820
>assuming that 0 = 1 can't be true
you must be at least 18 to post on 4chan

>> No.8415062

>>8400563
[eqn]\mathbb{F}_1[/eqn]

>> No.8415074

>>8415032
>not understanding what is meant by "real"
You must be at least 18 to bait on 4chins.

>> No.8415168

>>8415074
>implying that that statement was explicit and not ambiguous

>> No.8415173

>>8415168
Maybe once you attain the age of 18, you will be able to interpret the meaning of statements based on context rather than complaining that everything is ambiguous.

>> No.8415175

>>8415173
you realize that the trivial ring is a real ring where 0=1 right?

>> No.8415189

>>8414619
what is this monstrosity

>> No.8415195

>>8400563

easy

x = x-1

so now

x-1 = x-1 +1 now add 1 to each side.

x = x-1

Done.

>> No.8415222

>>8415175
The trivial ring is a ring, yes. The elements of the trivial ring are not "real" in the sense of being real numbers, i.e. elements of the unique uniformly complete archimedean field.

>> No.8415269

>>8415189
Standard model langragian.

Below it is the Einstein's Field Equations.

>> No.8415422

x = infintity

>> No.8416697

x = +/- 1/2

>> No.8416708

>>8400569
x=x + 1
-x
0=1
There is no real answer.

>> No.8416736

>>8400763
this is why I love sci

>> No.8416763

Lim x ==> infinity

X = x + 1

The relationship is only true when x is equal to infinity