Symmetry of real and imaginary parts in FFT INPUT
x[0] = (0.00 + j 0.00)
x[1] = (1.00 + j 0.00)
x[2] = (2.00 + j 0.00)
x[3] = (3.00 + j 0.00)
x[4] = (4.00 + j 0.00)
x[5] = (5.00 + j 0.00)
x[6] = (6.00 + j 0.00)
x[7] = (7.00 + j 0.00)
x[8] = (8.00 + j 0.00)
x[9] = (9.00 + j 0.00)
x[10] = (0.00 + j 0.00)
x[11] = (0.00 + j 0.00)
x[12] = (0.00 + j 0.00)
x[13] = (0.00 + j 0.00)
x[14] = (0.00 + j 0.00)
x[15] = (0.00 + j 0.00)

FFT:
X[0] = (45.00 + j 0.00)
X[1] = (-25.45 + j 16.67)
X[2] = (10.36 + j -3.29)
X[3] = (-9.06 + j -2.33)
X[4] = (4.00 + j 5.00)
X[5] = (-1.28 + j -5.64)
X[6] = (-2.36 + j 4.71)
X[7] = (3.80 + j -2.65)
X[8] = (-5.00 + j 0.00)
X[9] = (3.80 + j 2.65)
X[10] = (-2.36 + j -4.71)
X[11] = (-1.28 + j 5.64)
X[12] = (4.00 + j -5.00)
X[13] = (-9.06 + j 2.33)
X[14] = (10.36 + j 3.29)
X[15] = (-25.45 + j -16.67)


From the above FFT out put I noticed the following: 


  Re(x[1])=Re(x[15]),  Im(x[1])=-Im(x[15])
  
  Re(x[2])=Re(x[14]),  Im(x[2])=-Im(x[14])
  
  Re(x[3])=Re(x[13]),  Im(x[3])=-Im(x[13])
  
  Re(x[4])=Re(x[12]),  Im(x[4])=-Im(x[12])


and so on

Is this a proven result that 

 Re(X[n])=Re(x[N-n]), and Im(X[n])=-Im(x[N-n])  for 0&lt;n&lt;N-1, where N is no. of DFT points?


If yes then is there any particular condition under which this is true ? What is the gerenralised result?   

If this is a general rule then I can save alot in memory and arithmatics as I am concerned only in magnitudes of the DFT output, and not in phase.
Yes, this is always true if the input to the DFT is real valued. It's called the "conjugate complex symmetry", because
$$
X_{N-n} = {X_n}^*
$$
where $X_n$ is the DFT output and $()^*$ denotes the conjugate. It can be proven by inserting the property into the transformation formula of time domain sequence $x_k$:
$$
X_n = \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k n}{N}}\\
X_{N-n} = \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k (N-n)}{N}}\\
=\sum_{k=0}^{N-1}x_k e^{-j 2\pi k}e^{j\frac{2\pi k n}{N}}
$$
Using $exp(-j2\pi k) = 1 \:\: \forall \: k$ we find
$$
X_{N-n} = \sum_{k=0}^{N-1}x_k e^{j\frac{2\pi k n}{N}}
$$
Now we exploit that $x_k$ is real for all $k$ to rewrite the above:
$$
X_{N-n} = \sum_{k=0}^{N-1}\left(x_k e^{-j\frac{2\pi k n}{N}}\right)^* = X_n^*
$$
It's now straightforward to show that the same is valid for the IDFT. (real spectrum -> complex conjugate symmetric time domain sequence). Moreover, the opposite statement is also true: if the input to the DFT (IDFT) is complex conjugate symmetric its output is real valued.