I had a question regarding a formula derivation from a cryptography class I am taking. Not really a crypto question.. Just more wondering how does one go from LHS to RHS in above equality ?
$d<=(N^.25)/3 => (\frac{1}{2d^2} ) - (\frac{1}{N^.5} ) >= (\frac{3}{N^.5} ) $ [Implication-1]
I have attached a screenshot from where the lecturer mentions the above equivalence. He seems to be skipping some steps.. so to complete my understanding, I filled them in...like this ->>>
square both sides, then invert (flipping inequality direction):
$d<=(N^.25)/3 \implies (\frac{1}{d ^2} ) >= (\frac{9}{N^.5} )$
multiply both sides by 1/3:
$ (\frac{1}{3} ) * (\frac{1}{d ^2} ) >= (\frac{1}{3} ) * (\frac{9}{N^.5} ) $
and we obtain this ->
$ (\frac{1}{3*d ^2} ) >= (\frac{3}{N^.5} ) $
Note that the RHS is same as RHS in [Implication-1] But the LHS -- which in the above is $(\frac{1}{3*d ^2} ) $ -- does not equal the original in [Implication-1] -- which is $(\frac{1}{2d} )^2 - (\frac{1}{N^.5} ) $
Thanks in advance for any guidance/tips/etc !
