$X$ represents sample in frequency domain and $x$ represents samples in time domain.

**NOTATION 1**

$X(k) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $
   
$
x\left[ n \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } 
$



**NOTATION 2**     

$
X\left[ k \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } 
$

$
x\left[ n \right] = \sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } 
$

**NOTATION 3**

$
X\left[ k \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } 
$

$
x\left[ n \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } 
$

**NOTATION 4**

$
X\left[ K \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{j\frac{{2\pi }}{N}nk} } 
$

$
x\left[ n \right] = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {X\left[ K \right]e^{ - j\frac{{2\pi }}{N}nk} } 
$

Please observe the scaling factor $\frac{1}{N}$ and change of a negative sign $-$ over the exponent term ${e^{\pm j\frac{{2\pi }}{N}nk} }$. Why is every notation of DFT valid?