Richard Williamson

x 2x^{2}

Newer test

yx 5z yx^{5}z
(1)x 2=5 x^{2} = 5

And then

(2)y x=z y^{x} = z
By (1), we have that …, whilst by (2) we have that …

a) Hello

b) Hello

n2\lfloor \frac{n}{2} \rfloor

i=3,5,7,1122i i=15,2122i\sum_{i=3, 5, 7, 11} \lfloor \frac{22}{i} \rfloor - \sum_{i=15, 21} \lceil \frac{22}{i} \rceil.

We have that the number of primes in RR is less than or equal to

i=3,5,7,1122i i=15,2122i, \sum_{i=3, 5, 7, 11} \lfloor \frac{22}{i} \rfloor - \sum_{i=15, 21} \lceil \frac{22}{i} \rceil,

which is equal to

Last revised on October 4, 2018 at 16:20:36. See the history of this page for a list of all contributions to it.