0 + 0 + 0 + 0... = 0*infinity and zero times infinity does not equal zero.
If zero times infinity equaled zero, line segments would not exist. That is because a line segment contains an infinite amount of points of zero length, and 0*infinity must equal the length of the line.
If that doesn't convince you, look up what 0*infinity equals on other websites.
And one more thing. You say you will make a video soon explaining the flaw, but it's been almost two years and you didn't make one!
If the list was finite, shifting the parentheses would have resulted in a -1 all by itself at the end which compensates for the +1 at the beginning. But since there was no end to this set, this last -1 didn't exist. This is actually a mathematical proof that the associative property doesn't apply to an infinite expression. LOL
this is similar for if there was an odd number of 0's. so after showing that we know no matter how matter how many 0's you add to the equation you're either going to have an even number or an odd number of 0's at any given time so your result will ultimately be 0 no matter how big you make it
i looked at this in terms of an even or odd argument even though the guy saids it goes on infinitely. with 0=0+0+0+0..... if there were an even number of 0's then there would be an even number of ones and negative ones, so 0= (1-1) +(1-1)+(1-1).... an even number of times and of course using associativity and communitivity you can just say 0= 2k (number of ones, k is some integer) - 2k which is 0
As you've stated you have an infinite number of 1's and an infinite number of -1's, so no matter how you draw the parenthesize you get an infinite sum.
If we split the series in half we get:
1+1+1+1...= infinity
and
-1-1-1-1...= - infinity
Then if we add the two sub-series together we will obtain the original series, so:
-1+(1-1)+(1-1)+(1-1)+... = (-1+1)+(-1+1)+(-1+1)+... is wrong.
The problem comes because, in order to place the parentheses in the new order, you need to split the infinite series, but that's invalid when the pieces that are split don't converge.
But can't this regrouping be done? Yes, but the result will be different (not an error). Since the parts diverge, you must do it without manipulationing infinite series, and instead do it by looking at the partial sums, and then taking the limit.
And now, of course, FUCKING YOUTUBE won't let me type what comes next. I assume the next reordering step, and its consequences (0 = infinity), are obvious.
In the theory of infinite series it`s well known that, the addition of two divergent series is also a divergent series. So, 1+1+1... is divergent and so is, 1-1-1... Adding the two does NOT produce a series that converges to `0`.
you cannot apply the associativity in the addition law when you have the subtraction law in the equation so if you want to apply it then you will have :
0=1-(1+1)-(1+1)-(1+1)......
then 0 will be equal to a negative integer and this is absurd
i can prove 0=1 where 0 and 1 and algebraic entities and not integers so in the end we will have that
0 will always be equal to 0 and this IS PURE LOGIC
you dont even have to go into convergence/divergence to disprove this "proof" (btw, wtb QED pst). You write the series as 1 + (-1+1) + (-1 +1)...... forgetting that no matter how many iterations of the (-1+1), the series still has to end in a -1, which was mistakenly truncated during the bracket shift
No, the series doesn't end with a negative one. Part of the proof is that your list of 1 - 1 + 1 - 1 + 1... is infinitely long. So it doesn't *end* with anything. It's not truncated because both before and after the bracket shift, it's still infinitely long. The error is only that he is using the associative property for an infinite series, and it only applies to finite series. It's actually really clever.
There are not an even number of terms, nor are there an odd number of terms. There are an infinite number of terms. Infinity is neither even nor odd. This is why you can't apply the associative property. I mean, with the commutative property, you could group an infinite number of 1s and -1s anyway you want to get ANY sum you want. If I rearrange the list to go like this:
1 + 1 + 1 + 1 + 1 + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1), ... it's the same exact list of numbers but now it adds to 5.
Before you can insert and/or move around brackets in a series you must first show that it converges, and this one does not. The sequence of partial sums for the series on the right hand side is {1,0,1,0...} which does not converge as the term index goes to infinity. It oscilates, which is a form of divergence. Therefore the series diverges(never 'adds up' to anything) and you can't say it equals zero in the first place.
stupiiiiid u forgot one fucking -1 at the end this is why u have 0=1
nikosDmark 3 months ago
This has been flagged as spam show
thats not right
gurioper 1 year ago
0 + 0 + 0 + 0... = 0*infinity and zero times infinity does not equal zero.
If zero times infinity equaled zero, line segments would not exist. That is because a line segment contains an infinite amount of points of zero length, and 0*infinity must equal the length of the line.
If that doesn't convince you, look up what 0*infinity equals on other websites.
And one more thing. You say you will make a video soon explaining the flaw, but it's been almost two years and you didn't make one!
anticorncob6 1 year ago
You added a 1.
magicbeaver321 1 year ago
If the list was finite, shifting the parentheses would have resulted in a -1 all by itself at the end which compensates for the +1 at the beginning. But since there was no end to this set, this last -1 didn't exist. This is actually a mathematical proof that the associative property doesn't apply to an infinite expression. LOL
kdmq 1 year ago
derp
TheFlannelRevolution 1 year ago
1:13 Fail
GXTRON 1 year ago 6
Comment removed
GXTRON 1 year ago
this is similar for if there was an odd number of 0's. so after showing that we know no matter how matter how many 0's you add to the equation you're either going to have an even number or an odd number of 0's at any given time so your result will ultimately be 0 no matter how big you make it
XTC2525 1 year ago
i looked at this in terms of an even or odd argument even though the guy saids it goes on infinitely. with 0=0+0+0+0..... if there were an even number of 0's then there would be an even number of ones and negative ones, so 0= (1-1) +(1-1)+(1-1).... an even number of times and of course using associativity and communitivity you can just say 0= 2k (number of ones, k is some integer) - 2k which is 0
XTC2525 1 year ago
when u shifted the parenthesis u just changed (1-1) to (-1+1). u added 1 but only to the right side which is impossible
ZenGuy12 1 year ago 5
As you've stated you have an infinite number of 1's and an infinite number of -1's, so no matter how you draw the parenthesize you get an infinite sum.
If we split the series in half we get:
1+1+1+1...= infinity
and
-1-1-1-1...= - infinity
Then if we add the two sub-series together we will obtain the original series, so:
infinity + -infinity = 0 (not 1)
t007ray 1 year ago
@t007ray infinity - infinity = infinity. lmfao
TheFlannelRevolution 1 year ago
and of course If you repeat the infinite row you will get 1 - 1 + 1 - 1 ... Instead of 0 + 0 + 0 + 0
BLBLBLS 1 year ago
and of course If you repeat the infinite row you will get 1 - 1 + 1 - 1 ... Instead of 0 + 0 + 0 + 0
BLBLBLS 1 year ago
you start with 6 ones. You end with 7 ones. U may not add a one because then u'd have to add a minus 1 as well. What u didn't
BLBLBLS 1 year ago
-1+(1-1)+(1-1)+(1-1)+... = (-1+1)+(-1+1)+(-1+1)+... is wrong.
The problem comes because, in order to place the parentheses in the new order, you need to split the infinite series, but that's invalid when the pieces that are split don't converge.
But can't this regrouping be done? Yes, but the result will be different (not an error). Since the parts diverge, you must do it without manipulationing infinite series, and instead do it by looking at the partial sums, and then taking the limit.
VeryEvilPettingZoo 1 year ago
(cont)
And, since of course YouTube mysteriously rejected rejected everything I typed explaining this, I'll only state the result.
If you manipulate this using partial sums, the end result is:
-1+(1-1)+(1-1)+(1-1)+... = ( (-1+1)+(-1+1)+(-1+1)+... ) + (-1), which is correct.
VeryEvilPettingZoo 1 year ago
(cont)
Here's a more horrific example using the same idea:
0 = (1-1)+(2-2)+(3-3)+(4-4)+... Correct!
VeryEvilPettingZoo 1 year ago
(cont)
And now, of course, FUCKING YOUTUBE won't let me type what comes next. I assume the next reordering step, and its consequences (0 = infinity), are obvious.
VeryEvilPettingZoo 1 year ago
Correction: In my previous comment, replace `1-1-1...` with ` -1-1-1...`, sorry.
LeconsdAnalyse 1 year ago
In the theory of infinite series it`s well known that, the addition of two divergent series is also a divergent series. So, 1+1+1... is divergent and so is, 1-1-1... Adding the two does NOT produce a series that converges to `0`.
LeconsdAnalyse 1 year ago
0=(1-1)+(1-1)+(1-1)+....
you cannot apply the associativity in the addition law when you have the subtraction law in the equation so if you want to apply it then you will have :
0=1-(1+1)-(1+1)-(1+1)......
then 0 will be equal to a negative integer and this is absurd
i can prove 0=1 where 0 and 1 and algebraic entities and not integers so in the end we will have that
0 will always be equal to 0 and this IS PURE LOGIC
alfredoz87 2 years ago
What program did you use for this?
VampiricMidget 2 years ago
There are only 6 (ones) on your third equation. Once you rewrite it on fourth equation with the shifting of the brackets, there are 7 (ones).
0=1-1 is true, there should always be even number of (ones)
terryhtun 2 years ago
Woot System of a Down =D
DominoBrothers 2 years ago 2
Zajeba si se grdno brale. Triba si na kraju stavit -1, vako prčiš u prašinu, jebat ga...
luna500rr 2 years ago
The series is not absolutely convergent, so the movement of the parentheses is invalid.
incrediblub 2 years ago
WOW that's so smart !!!
you're so much better than my math teacher!!!
keep it up!!
...
...
kostikas2002 2 years ago
FAIL!
poon5000 2 years ago
you dont even have to go into convergence/divergence to disprove this "proof" (btw, wtb QED pst). You write the series as 1 + (-1+1) + (-1 +1)...... forgetting that no matter how many iterations of the (-1+1), the series still has to end in a -1, which was mistakenly truncated during the bracket shift
mnkyboi 2 years ago
"the series still has to end in a -1"
No, the series doesn't end with a negative one. Part of the proof is that your list of 1 - 1 + 1 - 1 + 1... is infinitely long. So it doesn't *end* with anything. It's not truncated because both before and after the bracket shift, it's still infinitely long. The error is only that he is using the associative property for an infinite series, and it only applies to finite series. It's actually really clever.
BailiffQuimby 2 years ago
@BailiffQuimby
"the series still has to end in a -1" is right isn't it? since 0 = 1-1, there has to be qual number of 1s and -1s...
Mifune00 2 years ago
There are not an even number of terms, nor are there an odd number of terms. There are an infinite number of terms. Infinity is neither even nor odd. This is why you can't apply the associative property. I mean, with the commutative property, you could group an infinite number of 1s and -1s anyway you want to get ANY sum you want. If I rearrange the list to go like this:
1 + 1 + 1 + 1 + 1 + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1), ... it's the same exact list of numbers but now it adds to 5.
BailiffQuimby 2 years ago
lol wtf
since when does (1-1)+(1-1) = 1 + (1-1)??
haha thats like saying
0=1
thus, i proved 0 = 1
lookwhoscomin 3 years ago
Cool to know some people actually enjoy math unlike me.
mobotix309 3 years ago
That was the fastest response in history...The page hadn't even finished loading to show my comment and I got an email saying you had responded.
ijpayne09 3 years ago
Comment removed
shambayati195 3 years ago
Before you can insert and/or move around brackets in a series you must first show that it converges, and this one does not. The sequence of partial sums for the series on the right hand side is {1,0,1,0...} which does not converge as the term index goes to infinity. It oscilates, which is a form of divergence. Therefore the series diverges(never 'adds up' to anything) and you can't say it equals zero in the first place.
ijpayne09 3 years ago
yep you got it
shambayati195 3 years ago