If x is a real number then there exists a positive integer n such that -n < x <

Question

If x is a real number then there exists a positive integer n such that -n < x < n

Explanation / Answer

using the archimedean principle: If x > 0 and if y is an arbitrary real number, there is a positive integer n such that nx > y. Prove that if x is a real number then there exists a positive integer n such that -nx, we can make this assumption as the positive integers does not have a supremum. if x is a positive integer then x>0 and so y>x>0>-y or y>x>-y. If x were negative then we can start by multiplying it by -1, satisfy the above requirements ending with y>x>-y, multiplying by -1 we get -y

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.