Let d f and n be integers with nd1 also let AkZn fk0 Prove
Let d, f and n be integers with n>d>1 also let A={[k]Zn | [f][k]=[0]}. Prove if d=gcd(f, n), then the set A contains more than one element and fewer than n elements. [Note that Zn contains exactly n elements: [0], [1], …, [n-1].]
Solution
Given:Let d, f and n be integers with n>d>1
and if d=gcd(f, n)
From the relation: n>d>1 or 1<d<n
that means d is between 1 and n.
if d=gretest common division
it may be 1 to n.
and set A={[k]Zn | [f][k]=[0]}.
it is depend on the values in k, but
this only contain these in Zn.
so, A contains more than one element and fewer than n elements.
![Let d, f and n be integers with n>d>1 also let A={[k]Zn | [f][k]=[0]}. Prove if d=gcd(f, n), then the set A contains more than one element and fewer than Let d, f and n be integers with n>d>1 also let A={[k]Zn | [f][k]=[0]}. Prove if d=gcd(f, n), then the set A contains more than one element and fewer than](/WebImages/25/let-d-f-and-n-be-integers-with-nd1-also-let-akzn-fk0-prove-1065777-1761557464-0.webp)