The discussion revolves around two distinct proofs in real analysis. The first proof concerns the expression \(2^n + 3^n\) and its divisibility by 5 for all odd natural numbers \(n\). The second proof addresses the existence of a rational number \(r\) such that the absolute difference between a real number \(a\) and \(r\) is less than \(1/n\) for any natural number \(n\).
Some participants have provided insights and alternative approaches to the problems, while others have expressed uncertainty about the classification of the first problem as a real analysis question. The discussion appears to be ongoing, with multiple interpretations being explored.
There is a mention of the need for clarity regarding the classification of the first problem and the assumptions underlying the proofs being discussed.
Yes, saying that |a- r|= |r- a|< 1/n is the same as saying that -1/n< r- a< 1/n or that a- 1/n< r< a+ 1/n. Since there exist a rational number in any interval, there exist a rational number in the interval (a- 1/n, a+ 1/n).vyro said:Also, another proof:
Homework Statement
Prove that for each a that exists in real numbers and n that exists in natural numbers there exists a rational r such that abs(a-r) < 1/n
Homework Equations
The Attempt at a Solution
I believe it has something to do with the theorem on density of rationals.
HallsofIvy said:Yes, saying that |a- r|= |r- a|< 1/n is the same as saying that -1/n< r- a< 1/n or that a- 1/n< r< a+ 1/n. Since there exist a rational number in any interval, there exist a rational number in the interval (a- 1/n, a+ 1/n).