From: DERREN <dez.cartledge@btinternet.com>
To: Teacher2Teacher Public Discussion
Date: 2009041517:13:36
Subject: SQRT Too ( (1^2) & (1^2)/3 ).. ?

	If the series (0+)1+3+5+7... All Odd Numbers ( logiclly 0 Being Even,
it can be equally divided by two, with both having nothing. Even
though 0^2=0 Good old logic ???) added together, ignoring the
functions we know make up the square's. 
Why then is there no mention of these more logical Square roots. 
If 1(^2) is to 4 as 1 is to 2^2 (sqrt1<sqrt4) and 2^2 is to 9 as 4 is
to 3^2 (sqrt4<sqrt9). Surly however illogical on paper, in our maths
scribble. There has to be an exact whole part. Decimals though
beautiful, have made lazy guess work of most of maths problems.

So (X by X) 
eg. 2 by 2 = a square made of 4. To make a square made of 9 from this,
then two rectangles(1 has three squares added) made of X by Y (Y =
X/(Z)or (X/X)/z = 1/z) and a square made of Y^2. 
ie. 4 (2 by 2) + 5 = 9 == 2*2 (X*X) + 2 by 1 *2 (=2(XY)) + 1 by 1 (=Y
by Y)). 
9 TO 16 (3 by 3 TO 3*3 + (1^2 by 3)*2 + 1 by 1). No simplifyng, yet. 
sqrt 2 = 1 by 1 + 1/3 by 1 + 1/3 by 1/3 = 1 & 1/3. 

1, whole + 3, 3rds (individually) 5, 5ths... etc 

1/1 + 1/3 + 1/3 + 1/3 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/7 + 1/7 + 1/7
+ 1/7 + 1/7 + 1/7 + 1/7.. etc 

1, 1&1/3, 1&2/3, 2, 2&1/5, 2&2/5, 2&3/5...4&2/9 (4&3/9=(4&1/3))...etc 

So 4/6 by 4/6 or the square of two thirds would have a diaganal of
ONE. 

Can someone tell me the logical expression that disprove's this,
because i keep going round in circles (And that remind's me, don't get
me started on pi) and if not the sqrt series what the hell have I
found?.
At least the easiest estimation for students???
DE2.

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