Consider x a rational number. Let epsilon >=0 be the minimal value such that x+epsilon is irrational, and let also gamma>0 be the minimal value such that x+gamma is rational. How to compute gamma/epsilon?

Owen Mathis

Owen Mathis

Answered question

2022-11-18

Consider x a rational number. Let ϵ 0 be the minimal value such that x+ϵ is irrational, and let also γ > 0 be the minimal value such that x + γ is rational. How to compute γ / ϵ?

Answer & Explanation

Kaeden Lara

Kaeden Lara

Beginner2022-11-19Added 23 answers

There are no such minimal values, in either of the two cases. If some positive number x + ε is irrational, then x + ε is not the minimal value for which that holds, because x + ε 2 is also irrational. To prove this, suppose it were rational. Then x + ε 2 = m n for some integers m and n. It follows that ε 2 = m n x and so ε 2 , being a difference between two rational numbers, is rational. And therefore so is ε, so x + ε must be rational, and we have a contradiction.
It's even simpler if x + γ is rational. Since x is rational, and x + γ is rational, we have γ = ( x + γ ) x, a difference of two rational numbers, which is therefore rational. Since γ is thus proved to be rational, so is γ / 2, and then so is x + γ / 2.
It's even simpler if x + γ is rational. Since x is rational, and x + γ is rational, we have γ = ( x + γ ) x, a difference of two rational numbers, which is therefore rational. Since γ is thus proved to be rational, so is γ / 2, and then so is x + γ / 2.

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