Could someone help me with this problem please :( Let S be a non empty unbounded
ID: 3031022 • Letter: C
Question
Could someone help me with this problem please :( Let S be a non empty unbounded set. Show that for every >0 there is x in S such that 1/|x|<I know by the Archimedean property For every >0 there exists n in N such that 1/n<
And if we can prove 1/|x|<1/n< We can say that 1/|x|< right ?
But how can I prove 1/|x|<1/n by using the information of this question that S in a unbounded set. Could someone help me with this problem please :( Let S be a non empty unbounded set. Show that for every >0 there is x in S such that 1/|x|<
I know by the Archimedean property For every >0 there exists n in N such that 1/n<
And if we can prove 1/|x|<1/n< We can say that 1/|x|< right ?
But how can I prove 1/|x|<1/n by using the information of this question that S in a unbounded set. Let S be a non empty unbounded set. Show that for every >0 there is x in S such that 1/|x|<
I know by the Archimedean property For every >0 there exists n in N such that 1/n<
And if we can prove 1/|x|<1/n< We can say that 1/|x|< right ?
But how can I prove 1/|x|<1/n by using the information of this question that S in a unbounded set.
Explanation / Answer
Answer: Take > 0, some real number. Define x = 1/.
We know x exists because 0, and we know x is real because real numbers are closed for division
(by a nonzero number). We also know it is positive from the rules of division.
So x satisfies the Archimedean Property, and there exists some integer n greater than x. Therefore, we
can say about that integer n: n > 1/ since that is how we defined x.
All of these numbers are positive, so we can multiply both sides of the inequality by ; and divide both
sides by n; and we get: > 1/n and since we know n exists as a finite intger, 1/n > 0.