Proof involving the Archimedean Property

[pissedoffdude]

Homework Statement

If x is a real number, show that there is an integer m such that:
m≤x<m+1
Show that m is unique

Homework Equations

Archimedean Property: The set of natural numbers has no upper bound

The Attempt at a Solution

I'm having trouble with showing that m is unique. If x is a real number, I can find integers that are smaller and bigger than it. If m=x, then m≤x. By the Archimedean property, m+1>x and m+1>m, so m≤x<m+1

 

[HallsofIvy]

Yes, that's fine if x happens to be an integer (that's what you imply when you say "if m= x". What if it isn't?

Whatever x is, you are correct when you say that the Archimedean property says that there exist integers larger than or equal to x. What can you say about the set of integers larger than or equal to x?

 

[pissedoffdude]

Yes, that's fine if x happens to be an integer (that's what you imply when you say "if m= x". What if it isn't?

Whatever x is, you are correct when you say that the Archimedean property says that there exist integers larger than or equal to x. What can you say about the set of integers larger than or equal to x?

Thanks for the advice.

So if I understand correctly, we have three cases:
1) x is an integer. Then, we can say m=x
2) x is rational. Then by the Archimedean property, we an find integers that are strictly greater and less than x, so we can let m be an integer such that m=x+1/2, then m<x<m+1 and x is halfway point here
3) x is irrational, then suppose we have the interval (m, m+1) where m is an integer, we let x be an irrational number somewhere in that interval