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 Community Discussion Question:
PROVE
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 12:23:42 IST
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QUESTION 3
The positive integers a,b,c,d,e,f are such that (a/b) >(c/d) > (e/f )and af-be= 1 prove that
0< (cf-de)/df < 1/bf
NOTE:Please show working.ThankU
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let a=bx,c=dy,e=fz
x>y>z
also from 2nd eqn v ge bf(x-z)=1 or bfx=1+bfz
to prove 1st inequality
v have to prove cf>de or dyf>dfz or y>z which is aldready true
now for 2nd inequality
v have to prove: cf-de<d/b
or cfb<d+bde
or bfy<bfz+1
but v know bfx=1+bfz and bfx>bfy(as x>y)
so bfy<bfz+1
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Nitwit Blubber Odment Tweak
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23 Mar 2008 01:18:46 IST
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Thanx for the working bro.
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23 Mar 2008 01:21:06 IST
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cheers :)
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