Why the Nyquist frequency is 0.5 of Fs, why not 0.55 or 0.65?, brief explanation This my elaboration of the aliasing issue:
a continuous signal can be represented by factors of :
$e^{(i2{\pi}ft)}$ if we sample this signal then I will get:
$e^{(i2{\pi}fk/N)}$ where $k=0,1,2.., N-1$
this point can be represented by the below factor as well:
$e^{(i2{\pi}f_pt)}$
if we put all together:
$e^{(i2{\pi}ft)} = e^{(i2{\pi}f_pt)}$
$e^{(i2{\pi}(f_p-f)t)} = 1$
this is satisfied only if :
$(f_p-f) 2{\pi} /N = 2{\pi}m,$ this lead to
$f_p-f = N*m$
now here it comes the analysis
$N$ is actually the sampling frequency, let say $N$ = 100
so if a say my signal is a single frequency $f$ = 100 then my aliasing frequency (or the other signal frequency) would be:
$f_p = f + m*N = 100 + m*100 = 100(1+m),$
$f_p(m=-1) = 0,$ issue,
$f_p(m=0) = 100,$ no issue,
$f_p(m=1)= 200,$..issue,
now if my $f = 50$
$f_p = f + m*N = 50 + m*100 = 100(0.5+m),$
$f_p(m=-1) = -50,$ issue
$f_p(m=0) = 50,$ no issue
$f_p(m=1)= 150,$..issue
now if my $f = 25$
$f_p = f + m*N = 25 + m*100 = 100(0.25+m),$
$f_p(m=-1) = -75,$ no issue
$f_p(m=0) = 25,$ no issue
$f_p(m=1)= 125,$.no .issue
now if my $f = 75$
$ff = f + m*N = 75 + m*100 = 100(0.75+m),$
$ff(m=-1) = -25,$ issue
$ff(m=0) = 75,$ no issue
$ff(m=1)= 175,$.no .issue
so then after tabulating all possible frequencies I can say that the best frequencies where $f_p$ doesn't exist is when $f &lt; N / 2$
so What do you think about this reasoning. is there other way to explain the 1/2 term that multiplies $N$ to give us the right signal frequency ( that can be correctly sampled with $N$  samples)?
Appreciate your opinion
From the definition of sampling, and the definition of periodicity follows that any signal is identical to signal at a frequency higher by an integer amount of sampling frequencies.
Since for real signals, negative frequencies aren't distinguishable from positive ones, it directly follows that for every $0.5f_\text{sample}\le f \le f_\text{sample}$, there is a negative frequency $-0.5f_\text{sample}\ge \tilde f=1- f \ge f_\text{sample}$ that looks the same as a positive frequency.
Hence, only frequencies lower than $|f| &lt; 0.5f_\text{sample}$ can be uniquely represented.