# Discrete Math Question

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1. Oct 23, 2014

### bensoa1

1. The problem statement, all variables and given/known data
Show that,
n=2[0,1 - 1/n] = [0,1)

2. Relevant equations

3. The attempt at a solution

2. Oct 23, 2014

### Dick

Well, explain why the union of all of those intervals contains everything in [0,1] except for 1.

3. Oct 23, 2014

### bensoa1

This was the plan of action I wanted to take, however, I'm not sure of the appropriate way to do that

4. Oct 23, 2014

### Dick

Start by explaining why it doesn't contain 1. Then continue by explaining why it does contain 99/100. Then extrapolate from there. Think about it.

5. Oct 23, 2014

### bensoa1

My method was to first show that when n = 2 the answer was [0,1). Then I had an arbitrary integer greater than 0, such that ∪n=2[0,1-1/(k+2)) = [0,1). Would this be sufficient?

6. Oct 23, 2014

### Dick

n goes from 2 to infinity in the union. You can't pick it to be 2. Start by explaining why 1 is not in the union.

7. Oct 23, 2014

### bensoa1

What is the mathematical proof to use in order to show that it isn't?

8. Oct 23, 2014

### bensoa1

Okay so I did this, ∪n=2∞[0,1 - 1/n] = [0,1/2), [0,2/3), [0,3/4),...,[0,n-1/n). Since n-1/n < 1, by union properties ∪n=2∞[0,1 - 1/n] = [0,1). Would this suffice?

9. Oct 23, 2014

### Dick

The union is [0,1/2]U[0,2/3]U[0,3/4]U[0,4/5]... Don't you see why 1 isn't in it and any number 0<=x<1 is? Just explain in words.

10. Oct 25, 2014

### HallsofIvy

Staff Emeritus
What do you mean "when n= 2 the answer was [0, 1)"? When n= 2. [0, 1- 1/n]= [0, 1- 1/2]= [0, 1/2]. That is NOT "[0, 1)"!

You need to focus on 1- 1/n. Do you see that there is NO n such that $1\le 1- 1/n$? Why not?

On the other hand, if x is any positive number there exist n such that 1/n< x. Why? Then what can you say about 1- 1/n?

Do you understand that the right hand side is
$$[0, 1/2]\cup[0, 2/3]\cup[0, 3/4]\cup[0, 4/5]\cdot\cdot\cdot$$?

Last edited: Oct 25, 2014