From: Giannis P. (el96045(@)central.ntua.gr)
Date: Τρι 30 Μαΐ 2000 - 21:16:02 EEST
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Kalhmera/spera/nyxta..
kanw forward thn apanthsh pou esteila sthn Pavlina gia ta apolyta.. To
problhma den einai katholou paradojo, apla prin bgalei kaneis ta
apolyta prepei na balei to arnhtiko proshmo se olh thn parastash
mesa..
Twra to eida..
..einai fanero oti isxyei: |1-3/2| = |2-3/2| htoi: |-0.5| = |0.5|
..twra, an esy thes na bgaleis ta apolyta, kane thn prajh prwta kai meta bgale ta... alliws prepei na peis:
1-3/2 = -(2-3/2) <=> 1-3/2 = -2+3/2 <=> ... <=> 3 = 3
..Giannis!
Kanena paradojo!
On Mon, 22 May 2000 16:10:41 +0300, you wrote:
>Ένα απλό μαθηματικό πρόβλημα....Μπορείτε να μου πείτε πώς και γιατί;
>Λοιπόν:
>-2 = -2 =>
>1-3 = 4-6 =>
>1-6/2 = 4-12/2 =>
>1-6/2 + 9/4 = 4 - 12/2 + 9/4
>(1-3/2)2 = ( 2 - 3/2 )2 (Είναι τετράγωνα...απλά δε μπορώ να τα
>γράψω... τέλεια τετραγωνα (α+β)2=α2+β2-2αβ οκ;)
>=> τετράγωνο με τετράγωνο φεύγουν όπως α2=β2 άρα α=β
>έτσι
>1-3/2 = 2- 3/2
edw einai to lathos^
1-3/2 = - (2-3/2)
h'
- (1-3/2) = 2-3/2
>ΕΤΣΙ
>1 = 2
>
>Πώς?????
>
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