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1+2+3+4+ ... equals?

1+2+3+4+ ... equals?

Posers and Puzzles

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1+2+3+4+ ... equals?

ZERO

Obviously 1+2+3+4+ ... = (2-1)+(3-1)+(4-1)+(5-1)+...

=2+3+4+5+... - (1+1+1+1+...)

= inf - inf

= 0

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FAIL

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Repeat after me "one-to-one correspondence"... "one-to-one correspondence"... "one-to-one correspondence"...

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😕

Of course!

I'm trying to point out the flaw in our opponents' argument in the "many balls" thread

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Originally posted by wolfgang59
😕

Of course!

I'm trying to point out the flaw in our opponents' argument in the "many balls" thread
Are there more even numbers than natural numbers?

Yet if I start with this sequence

{2} and {1,2}
{2,4} and {1,2,3,4}
...
{2,...,2n} and {1,2,...,n,...,2n}
...
n goes to infinity...
...

The difference in cardinality is always increasing in n!

Yet...the cardinality of the countably infinite sets is the same.

Now repeat after me "one-to-one correspondence"...

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Originally posted by Palynka
Are there more even numbers than natural numbers?

Yet if I start with this sequence

{2} and {1,2}
{2,4} and {1,2,3,4}
...
{2,...,2n} and {1,2,...,n,...,2n}
...
n goes to infinity...
...

The difference in cardinality is always increasing in n!

Yet...the cardinality of the countably infinite sets is the same.

Now repeat after me "one-to-one correspondence"...
yes but did you know that there are more evens than odds?

for the odd nmber 1 there is 1+1=2

for the odd nmber 3 there is 3+1=4

for the odd nmber 3 there is 5+1=6

for the odd nmber Nthere is N+1

So for every odd there is an even ... but then we include ZERO

QED There are more evens than odds







😉

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Originally posted by wolfgang59
1+2+3+4+ ... equals?

ZERO

Obviously 1+2+3+4+ ... = (2-1)+(3-1)+(4-1)+(5-1)+...

=2+3+4+5+... [b]-
(1+1+1+1+...)

= inf - inf

= 0[/b]
inf - inf is not zero. It's not even defined. Therefore your calculation is flawed.

If we define inf - inf, inf / inf, and such, as a number then we get funny results, as the one above.

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Did you know that 0 = 1?

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

QED

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Originally posted by FabianFnas
inf - inf is not zero. It's not even defined. Therefore your calculation is flawed.

If we define inf - inf, inf / inf, and such, as a number then we get funny results, as the one above.
I think I knew that!
😕

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Originally posted by Thomaster
Did you know that 0 = 1?

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

QED
So which standard operation breaks down when you reach an infinite sum? Could you get that result by extrapolating from the union of finite sequences of sums?

1 edit
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If I remember correctly, it's all to do with whether the series is absolutely convergent or not.

SUM(x_i) is absolutely convergent <=> SUM(|x_i|) is convergent.

Absolutely convergent series behave when you rearrange them. Series that are just convergent (e.g. 1 - 1/2 + 1/3 - 1/4 + ...) can converge to different values (or not at all) depending on the order.

For example:
1 - 1/2 + 1/3 - 1/4 + 1/5 +... = ln 2
1 + 1/3 + 1/5 + ..... - 1/2 - 1/4 - 1/6 - ... does not converge

1 - 1 + 1 - 1 + ... is very obviously not absolutely convergent!

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Originally posted by wolfgang59
I think I knew that!
😕
I know you know. And I know also that you know that I know that you know it.

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Originally posted by FabianFnas
I know you know. And I know also that you know that I know that you know it.
I didnt know that!

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