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Algebra

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17 Feb 2010 12:31:52 IST

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{x} , [x] & x are in GP find common ratio. and x is a +ve real no.

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{x} , [x] & x are in GP find common ratio. and x is a +ve real no.

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gaurav jindal

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Joined: 17 Feb 2010

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17 Feb 2010 14:31:00 IST

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(1 + root of 5)/2

Hari Shankar

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17 Feb 2010 20:12:42 IST

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Let x = I + f so that [x] = I>=0 and {x} = f.

So the equation can be written as I^2 = (I+f)f

Now $f = \frac{I^2}{ I+f} > \frac{I^2}{I+I} = \frac{I}{2}$

$\because 0\le f<1, \frac{I}{2}<1 \Rightarrow I<2 \Rightarrow I = 0,1$

I = 0 gives f=0

I =1 gives the quadratic f^2-f-1 = 0 which gives the permissible root $f = \frac{\sqrt 5-1}{2}$

Hence, there is a unique value of x among positive reals which is

$x = I+f = 1+ \frac{\sqrt 5-1}{2} = \frac{\sqrt 5+1}{2}$

akl

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Joined: 6 May 2008

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17 Feb 2010 21:26:40 IST

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sir how I/2<1 ?

Hari Shankar

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18 Feb 2010 09:24:51 IST

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We first proved that f > I/2

We also know that 1>f

Combining the two, we get 1 > I/2.

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