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robarth
Rank #8119 on Comments
Level 227 Comments: Mind Blower
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Gender:
male
Date Signed Up:
3/26/2011
Stats
Comment Ranking:
#8119
Highest Content Rank:
#18798
Highest Comment Rank:
#3934
Content Thumbs:
41
total, 65
, 24
Comment Thumbs:
3110
total, 3451
, 341
Content Level Progress:
76.27% (45/59)
Level 0 Content: Untouched account → Level 1 Content: New Here
Comment Level Progress:
70% (70/100)
Level 227 Comments: Mind Blower → Level 228 Comments: Mind Blower
Subscribers:
0
Content Views:
3263
Total Comments Made:
555
FJ Points:
2748
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FUTURE
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What Do Your...
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latest user's comments
#13
 ( ͡° ͜ʖ ͡°)
[+]
(3 replies)
12/03/2017 on
Not trying anymore
+5
#34

anotheroneonearth
(12/03/2017)
[]
I've visited Laos once. 10/10 would visit again.
even though it's one of the poorest countries in asia, people were all nice and kind.
they never tried to scam or rip me off, which felt almost contrary to Bangkok.
I was saved by local people many times. my faith in humanity was restored.
plus the country has a lot of stunning landscapes. possibly one of the chillest place in the world.
#37

therealhimmler
(12/03/2017)
[]
>Saved
Ummmm....what the fuck happened?
#49

anotheroneonearth
(12/03/2017)
[]
there is this backpacker destination called Vang Vieng, which is known for laid back landscapes and 'tubing'.
basically you use a big tire tube as a swim ring and go down the river.
I tried it just like any other traveler did.
while I was being mesmerized by majestic nature, I noticed the sun was setting.
yes I wasn't sober
looked like I missed the point where I was supposed to get ashore at.
so I had to wander a jungle and rice fields for a while.
no money, no phone, no food, no clue where to go. the only possession was a big fucking tire tube.
after a while I finally found some local people, but I could speak little to no their language.
so using hand gesture I desperately asked them where was the town I stayed at.
after walking for a while as they suggested, I finally found a paved road.
then some local man on a scooter asked me where I was going.
when I answered the name of the town, he said he was going there too and took me to the town, even though he came from the opposite direction.
arriving the town, I tried to pay him some money but he said "No Problem!" and left.
tl;dr
they help people simply out of kindness and never ask for money. this happened many times in my trip.
#39
 It's also true that a number is multiple of nine if and only i…
11/25/2017 on
Nines complements *head...
+1
#7
 The final solution to your cities slipping problems
11/19/2017 on
Roses are red, butter is...
+3
#13
 Comment deleted
11/04/2017 on
we wuz wolves n shiet
0
#86
 Russian?
10/30/2017 on
Stuff #152
+14
#6
 Isn't drawing also a kind of a project?
[+]
(4 replies)
10/25/2017 on
life of a project
+69
#7

funkyiestmonk
(10/25/2017)
[]
#8

blueflags
(10/25/2017)
[]
#9

funkyiestmonk
(10/25/2017)
[]
OH SHIT
i've been bamboozled
#39

dragunovx
(10/26/2017)
[]
#43
 Nonono, never would I say that x/0 is infinity. But you did it…
[+]
(1 reply)
09/29/2017 on
Trivial memes
0
#44

tariv
(09/29/2017)
[]
Okay, let's assume k=infinity with your infinities
c/infinity+f=2
f=2
I mean, it's not wrong, but only because the value was 2. Of course if it wasn't then we wouldn't have this problem, but infinity just got injected in there like a disease. c is just dead. I don't like your infinities.
#37
 I only wanted to say, that if you accept carrot to be infinity…
[+]
(5 replies)
09/29/2017 on
Trivial memes
0
#79

thekeeperofthumbs
(09/29/2017)
[]
bruh I got that too
>>
#78
#42

blargchikahonkhonk
(09/29/2017)
[]
#39

tariv
(09/29/2017)
[]
I realized another problem with carrot=infinity
The original equation without the other variable's values, c/k+f=2
Solve for k:
f + c/k = 2
Bring f + c/k together using the common denominator k:
(c + f k)/k = 2
Multiply both sides by k:
c + f k = 2 k
Subtract c + 2 k from both sides:
k (f  2) = c
Divide both sides by f  2:
Answer: k = c/(f  2)
Here, if c=0 then k=0 and you'd be dividing by 0 in the original equation, but c=1 so we're fine. Also, if f=2 you have outright division by 0, and f=2 so k is undefined.
And if you say that x/0=infinity I will blow a gasket.
#43

robarth
(09/29/2017)
[]
Nonono, never would I say that x/0 is infinity. But you did it again. You are trying to "solve for k" using process that conserves the truth value of the expression if f,c and k were all real numbers. But if you consider k to be infinity then those equivalent changes will cease to be equivalent. Therefore you conclusion is still false.
#44

tariv
(09/29/2017)
[]
Okay, let's assume k=infinity with your infinities
c/infinity+f=2
f=2
I mean, it's not wrong, but only because the value was 2. Of course if it wasn't then we wouldn't have this problem, but infinity just got injected in there like a disease. c is just dead. I don't like your infinities.
#35
 Well, the reals are a field. In a field you have two operation…
[+]
(8 replies)
09/29/2017 on
Trivial memes
+1
#49

godisbert
(09/29/2017)
[]
You and this other guy are operating on different levels of math, its seems
#36

tariv
(09/29/2017)
[]
Apparently you and I subscribe to different schools of infinity. Also you are a philistine and a heathen.
My alternative was going to be just saying it's 1/(k+2)=2, and the man with the pretty variables is an idiot who thinks adding happens first or something, which doesn't have to deal with questionable mathematics.
#37

robarth
(09/29/2017)
[]
I only wanted to say, that if you accept carrot to be infinity, then the solution for donut is this:
www.wolframalpha.com/input/?i=1%2F(2*...
#79

thekeeperofthumbs
(09/29/2017)
[]
bruh I got that too
>>
#78
#42

blargchikahonkhonk
(09/29/2017)
[]
#39

tariv
(09/29/2017)
[]
I realized another problem with carrot=infinity
The original equation without the other variable's values, c/k+f=2
Solve for k:
f + c/k = 2
Bring f + c/k together using the common denominator k:
(c + f k)/k = 2
Multiply both sides by k:
c + f k = 2 k
Subtract c + 2 k from both sides:
k (f  2) = c
Divide both sides by f  2:
Answer: k = c/(f  2)
Here, if c=0 then k=0 and you'd be dividing by 0 in the original equation, but c=1 so we're fine. Also, if f=2 you have outright division by 0, and f=2 so k is undefined.
And if you say that x/0=infinity I will blow a gasket.
#43

robarth
(09/29/2017)
[]
Nonono, never would I say that x/0 is infinity. But you did it again. You are trying to "solve for k" using process that conserves the truth value of the expression if f,c and k were all real numbers. But if you consider k to be infinity then those equivalent changes will cease to be equivalent. Therefore you conclusion is still false.
#44

tariv
(09/29/2017)
[]
Okay, let's assume k=infinity with your infinities
c/infinity+f=2
f=2
I mean, it's not wrong, but only because the value was 2. Of course if it wasn't then we wouldn't have this problem, but infinity just got injected in there like a disease. c is just dead. I don't like your infinities.
#31
 Well, infinity can be considered a number, if you take proper …
[+]
(10 replies)
09/29/2017 on
Trivial memes
+1
#33

tariv
(09/29/2017)
[]
How so? Any value of the form x/y that is then multiplied by y/1 will equal x, and that is what was done. If k=infinity then obviously infinity=k and k can remain in the equation to represent the value and it can be manipulated as usual. 0=1
Besides, take the infinity that equals k, multiply by two and you have an infinity twice the infinity of k. Add one to that infinity and it is a distinct, new infinity slightly larger than just the twice k infinity. Now divide this twice k plus one infinity by the infinity of k and you have a value infinitesimally greater than 2, which does not equal 2.
2=2.00...01
Subtract 2
0=0.00...01
Multiply by 10^infinity
0=1
Don't divide by 0
#35

robarth
(09/29/2017)
[]
Well, the reals are a field. In a field you have two operations, + and *. These operations are defined in certain way. One is that all numbers (except 0) have multiplicative inverse. Meaning given nonzero a, real, then there exist real number b such that a*b=1. But when you add infinity into the mix then the properties of these operations are only defined for the reals and not the infinities. You need to define how these operations act on these infinities. The standart way is to define it as such:
infinity * x = infinity, whenever x>0.
infinity * infinity = infinity.
1/infinity = 0.
(there are some formalities with order and such but are, for brevity, left out)
But there are still some operation left undefined and those are:
infinity/infinity and 0*infinity.
When you did (2 k+1)/k and multiplied it by k then you had (2 k + 1)* (k/k), but guess what, if you consider k=infinity and k/k is undefined, then you used an undefined term, and your conclusion is therefore false.
#49

godisbert
(09/29/2017)
[]
You and this other guy are operating on different levels of math, its seems
#36

tariv
(09/29/2017)
[]
Apparently you and I subscribe to different schools of infinity. Also you are a philistine and a heathen.
My alternative was going to be just saying it's 1/(k+2)=2, and the man with the pretty variables is an idiot who thinks adding happens first or something, which doesn't have to deal with questionable mathematics.
#37

robarth
(09/29/2017)
[]
I only wanted to say, that if you accept carrot to be infinity, then the solution for donut is this:
www.wolframalpha.com/input/?i=1%2F(2*...
#79

thekeeperofthumbs
(09/29/2017)
[]
bruh I got that too
>>
#78
#42

blargchikahonkhonk
(09/29/2017)
[]
#39

tariv
(09/29/2017)
[]
I realized another problem with carrot=infinity
The original equation without the other variable's values, c/k+f=2
Solve for k:
f + c/k = 2
Bring f + c/k together using the common denominator k:
(c + f k)/k = 2
Multiply both sides by k:
c + f k = 2 k
Subtract c + 2 k from both sides:
k (f  2) = c
Divide both sides by f  2:
Answer: k = c/(f  2)
Here, if c=0 then k=0 and you'd be dividing by 0 in the original equation, but c=1 so we're fine. Also, if f=2 you have outright division by 0, and f=2 so k is undefined.
And if you say that x/0=infinity I will blow a gasket.
#43

robarth
(09/29/2017)
[]
Nonono, never would I say that x/0 is infinity. But you did it again. You are trying to "solve for k" using process that conserves the truth value of the expression if f,c and k were all real numbers. But if you consider k to be infinity then those equivalent changes will cease to be equivalent. Therefore you conclusion is still false.
#44

tariv
(09/29/2017)
[]
Okay, let's assume k=infinity with your infinities
c/infinity+f=2
f=2
I mean, it's not wrong, but only because the value was 2. Of course if it wasn't then we wouldn't have this problem, but infinity just got injected in there like a disease. c is just dead. I don't like your infinities.
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