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Q&A #3880 |
From: DERREN
To: Teacher2Teacher Public Discussion
Date: 2009041516:13:36
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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