0=1, Prove me wrong \ stacker news ~science

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432 sats \ 31 comments \ @istealcheetos 21h science

So long stackers 😈 Identify the catch and get rewarded!

We know

0 = 0

and

1 - 1 = 0

(subtracting the number from itself) Thus

(1 - 1) - (1-1) - (1-1) = 0

[Grandi's Series]

Hence

(1-1)+(1-1)+(1-1)+\dots=0+0+0+\dots=0

Thus

1-1+1-1+1-1+.... = 0

1-(1-1+1-1+\dots) = 0

[Leave the first term and take the "-" common] Thus

1 = (1 -1 + 1 -1 + 1- 1....)

1 = 0

ha 😈

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6 replies \ @didiplaywell 20h

The catch is in the 6th step, where the extraction of the (-) is made so that it stealthy breaks the assumed parity of the series:

in 1 - (1 - 1 + 1 - 1 + ... ) there's a hidden (+ 1) at the end, so the series should actually look like:

1 - (1 - 1 + 1 - 1 + ... + 1 - 1 + 1)

0 sats \ 5 replies \ @istealcheetos OP 20h

so close! but the explanation is not quite right

1060 sats \ 4 replies \ @didiplaywell 20h

I could express it like so:

The proposition starts with ( 1 - 1 ) + ( 1 - 1 ) + ( 1 - 1 ) = 0

Then expands it, but it does so ignoring the last part. We can make the same infinite expansion keeping the representation of the last part, like so:

( 1 - 1 ) + ( 1 - 1 ) + ... + ( 1 - 1 ) = 0 , instead of ( 1 - 1 ) + ( 1 - 1 ) + ... = 0

If we take the (-1) from the new representation, we can see that in reality what we get is:

1 - (1 - 1 + 1 - 1 + ... + 1 - 1 + 1) = 0

while the 1 - (1 - 1 + 1 - 1 + ... ) = 0 notation makes it look like if the series ends, in the infinite, like

1 - (1 - 1 + 1 - 1 + ... + 1 - 1 ) = 0

Which is not the true form of the first proposed series.

20 sats \ 3 replies \ @istealcheetos OP 20h

nailed it! it's in the number of terms!

15 sats \ 2 replies \ @didiplaywell 20h

Awesome! Thank you for making this fun and for the bounty, much appreciated :)

20 sats \ 1 reply \ @j7hB75 17h

You played well.

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15 sats \ 9 replies \ @0xbitcoiner 21h

1-(1-1+1-1+\dots) = 1

[Leave the first term and take the "-" common] Thus

1 = 1+(1 -1 + 1 -1 + 1- 1....)

0 sats \ 8 replies \ @istealcheetos OP 21h

nope, no new term was introduced, the negative sign is just taken as common leaving the first term as is

0 sats \ 7 replies \ @0xbitcoiner 21h

1-(1-1+1-1+\dots) = 0

Isn't this equation wrong?

0 sats \ 6 replies \ @istealcheetos OP 21h

How? Just appearing on the equation at first glance will make that seem wrong if you don't follow the previous steps :) That's not the breaking point

0 sats \ 5 replies \ @0xbitcoiner 21h

1-(0) = 0

21 sats \ 4 replies \ @istealcheetos OP 21h

let me explain, it was 1 - 1 + 1 - 1 + 1 - 1..... So I kept the first 1, then took the negative sign common for the rest eg 8 - 3 - 5 = 0 or 8 - (3 + 5) = 0 It might look illusionary lol

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103 sats \ 1 reply \ @SpaceHodler 18h

(1-1)+(1-1)+(1-1)+... = 0 1-((1-1)+(1-1)+(1-1)+...+1) = 0 0 = 0

0 sats \ 0 replies \ @istealcheetos OP 17h

nice try, fed you're late but here's a consolation

15 sats \ 4 replies \ @rootmachine 20h

I looked through the comments to see if someone mentioned the order of operation. You could say it is this at a first glance, but... nope. False proofs always appear because they are generally presented with what seems to be valid math as in the example above. The mistake in this case lies in treating the infinite series 1-1+1-1+1-1+....= 0. This is the catch. Grandi’s series does not converge in the traditional sense - its partial sums just alternate between 1 and 0. You are basically manipulating the result of this and force it to 0 artificially.

This is my opinion.

0 sats \ 2 replies \ @istealcheetos OP 20h

force it to 0 artificially

that's the point of this trick!

0 sats \ 1 reply \ @rootmachine 20h

1−(1−1+1−1+…)=0 -> this is the moment of failure. You are implying Grandi's series is =1 when you said before it =0.

0 sats \ 0 replies \ @istealcheetos OP 18h

well, you are saying this #1008494

0 sats \ 0 replies \ @istealcheetos OP 20h

too close! too close! just one subtraction close!

15 sats \ 1 reply \ @fauxfoe 20h

Interesting. I've usually seen Grandi's series as 1-1+1-1+1-1+1-1...

That series doesn't converge to 0 by any accepted method I know. Infinite series are weird like that!

0 sats \ 0 replies \ @istealcheetos OP 20h

it is a divergent series and is still divergent! you can get 2 answers out of this :)

15 sats \ 1 reply \ @robbabuona 20h

hmmm...when you move first 1 left of equal sign you have one term left and so 5 terms right ( considering the example of 3 couples so 6 ones ). So in the end:

1 = 1-1+1-1+1

so 1 = 1

0 sats \ 0 replies \ @istealcheetos OP 20h

no! I'm not moving the ones anywhere :) I just took the minus common and the rest of the series to the other side but you're close, try again