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

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432 sats \ 31 comments \ @noknees 17 Jun 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 17 Jun

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 \ @noknees OP 17 Jun

so close! but the explanation is not quite right

1060 sats \ 4 replies \ @didiplaywell 17 Jun

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 \ @noknees OP 17 Jun

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

15 sats \ 2 replies \ @didiplaywell 17 Jun

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

20 sats \ 1 reply \ @j7hB75 17 Jun

You played well.

21 sats \ 0 replies \ @didiplaywell 17 Jun

Finally :')

103 sats \ 1 reply \ @SpaceHodler 17 Jun

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

0 sats \ 0 replies \ @noknees OP 17 Jun

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

15 sats \ 4 replies \ @rootmachine 17 Jun

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 \ @noknees OP 17 Jun

force it to 0 artificially

that's the point of this trick!

0 sats \ 1 reply \ @rootmachine 17 Jun

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 \ @noknees OP 17 Jun

well, you are saying this #1008494

0 sats \ 0 replies \ @noknees OP 17 Jun

too close! too close! just one subtraction close!

15 sats \ 1 reply \ @fauxfoe 17 Jun

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 \ @noknees OP 17 Jun

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

15 sats \ 1 reply \ @robbabuona 17 Jun

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 \ @noknees OP 17 Jun

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

15 sats \ 9 replies \ @0xbitcoiner 17 Jun

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 \ @noknees OP 17 Jun

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 17 Jun

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

Isn't this equation wrong?

0 sats \ 6 replies \ @noknees OP 17 Jun

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 17 Jun

1-(0) = 0

21 sats \ 4 replies \ @noknees OP 17 Jun

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