Recent content by mathstudent1

they said in question x<y , also i can found m and n but i do not know who i sketch the proof

yes i can take m=3 and n=2

you are right , then how i prove it ?

by archimedean property right ? there is n,m which are integers s.t n>1/y and 1/m>-1/x then 1/n<y and 1/m<-x , add them we will get 1/m+1/n <y-x.

i think my assumption is false i don ot know how i complete

i try but i do not know how to complete and i think my assumption is false

Let x, y ∈ R be such that x < y. Prove that there exist natural numbers m and n such that x +1/m < y −1/n?

do you know any books, videos, or notes that can help me to understand these topics: 1-primitive roots for primes 2-the existence of primitive roots I am using right now elementary number theory and its application by Kenneth H. Rosen to understand these topics, do you know another source that...

because 1 is not congruent to 0 mod 2. so, this prove holds only for odd primes? the reason that I ask is because my teacher wrote for any prime proof that

thank you

a+1 terms and since a must be even we will have even+1= odd and since sigma is multiplicative, we can distribute the sigma so we have this odd*odd*.....= odd that is why sigma is odd right?

so we can use this form? 1+p+P^2+.......+p^a?

Do you mean we can use this form p^a+1 -1/p-1 ? and since sigma is a multiplicative function, we can distribute the sigma?

n^2=p1^2r1*p2^2r2*..........*p^2rk?