Soldato
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- 17 Oct 2002
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True. If 0.9r < 1, then there exists a number between them, (0.9r + 1)/2 say. But if 0.9r is the "closest number to 1" then (0.9r+1)/2 is 0.9r isn't it?Originally posted by VDO
So tell us what you need to add. What number is between 0.9r and 1?
(0.9r+1)/2 = 0.9r
0.9r + 1 = 2*0.9r
Take 0.9r from both sides
1 = 0.9r
Hence if you asume 0.9r is the closest number to 1, you still prove it is infact 1
