DFT Symmetry Property: Why Does the Answer Not Void This Property?

[Shaheers]

Homework Statement

As an example, if we find a DFT of x[n]={1,1,0,1}
the result will be X(m)={3,1,-1,1}

Homework Equations

My Question is that as we know DFT holds symmetry property, why this answer does not void for that property?

 

[arkajad]

Let's take the definition from Wikipedia:
$$X_k=\sum_{n=0}^{N-1}x_n\,e^{-i2\pi kn/N}$$
For real data you have then the symmetry:
$$X_k=\bar{X}_{N-k},\, (k=1,...,N).$$
In your case $N=4$, and you have $X_0=3,X_1,=1,X_2=-1,X_3=1,$ all real.

From the symmetry property you should have $X_4=X_0,X_3=X_1,X_2=X_2$.
And you have it ($X_4$ can be thought of as defined by the symmetry property).