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To: donh

>> Tell ya what, here's a simple set of arithmetic identities, all valid in finite math, and an equivalent set could easily be part of a useful program. tell me what c resolves to and provide the proof of your answer.

a = b + 1
b = a - 1
c = b + 1
<<

Unlees I'm missing something This appears to just be a linear relationship (i.e. a straight line) a=b+1 where a and c are interchangable. Am I missing something?

The line would be 45 degrees with intercepts of (0,1) and (-1,0)

The system could be reduced to 2 lines a=b+1 and A=c

I guess I've got to be missing something.


478 posted on 05/10/2005 7:03:21 PM PDT by paul_fromatlanta
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To: paul_fromatlanta; AmishDude
Unlees I'm missing something This appears to just be a linear relationship (i.e. a straight line) a=b+1 where a and c are interchangable. Am I missing something?

I asked you to resolve c, not to solve the equations. I will recast the problem in a more classic form so as to avoid this interpretation:


b     = 1

a     = f1(b) - 1
f1(b) = f1(a) + 1


solve for a, providing a proof of your answer

Please note that I can write a useful program that looks exactly like this--so there's no traction in claiming that it doesn't really exist.
592 posted on 05/11/2005 1:23:22 PM PDT by donh
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