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# Impossible math 2

### #11

Posted 16 July 2007 - 03:08 AM

### #12

Posted 17 July 2007 - 04:24 PM

Sorry mate but you are wrong on this one.

You can't say that the sqrt of -1 is -1 if you do not include the imanginary i in the formula... You just can't!

However, I have seen this:

0 = 0+0+0+0...

= (1-1)+(1-1)+(1-1)+(1-1)+... [to infinity]

= 1+(-1+1)+(-1+1)+(-1+1)+... [there will always be another +/- to pair with]

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

= 1

You can therefore get m=0=n, where n and m are any real or imaginary number.

### #13

Posted 17 July 2007 - 10:48 PM

nick, that was a good use of the aleph-null paradox!

### #14

Posted 22 July 2007 - 02:45 PM

### #15

Posted 23 July 2007 - 09:43 AM

Comperr,

Sorry mate but you are wrong on this one.

You can't say that the sqrt of -1 is -1 if you do not include the imanginary i in the formula... You just can't!

However, I have seen this:

0 = 0+0+0+0...

= (1-1)+(1-1)+(1-1)+(1-1)+... [to infinity]

= 1+(-1+1)+(-1+1)+(-1+1)+... [there will always be another +/- to pair with]

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

= 1

You can therefore get m=0=n, where n and m are any real or imaginary number.

Cool! If anyone didn't know about aleph-null, this would sure baffle 'em!

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