# Prove question

1. Nov 22, 2008

### sedaw

need to prove that if $$0\leq X<n$$

and also 0<n so x=0.

The attempt at a solution

assume that x>0

$$\exists n, x\ni(0,\infty) |n< x$$

for X<0 theres contradiction agin .

from here X have to be 0

is that right ?

2. Nov 22, 2008

### Pere Callahan

x=1, n=2>0 satisfies $0\leq x<n$, yet x does not equal zero.

3. Nov 22, 2008

### sedaw

that is right but the porve is for each X so if theres only one case its enough .

4. Nov 22, 2008

### Pere Callahan

What are you trying to prove? I was under the impression that it is the following:
Whenever 0<=x<n for some positive (integer?) n, then x must be zero.
I gave a counter example that this is not always the case. As you said, one example is enough to show that the statement above is wrong.

5. Nov 22, 2008

### sedaw

not integer all the real num.

so what u suggest ?

6. Nov 22, 2008

### Pere Callahan

My counterexample is still valid.
I suggest, you say what you are trying to prove and what you don't like about the counter example

7. Nov 22, 2008

### sedaw

what`s not clear i need to prove that if x a real positive num and $$0\leq X<n$$

for each real posiive uumber n so x=0 .

8. Nov 22, 2008

### HallsofIvy

Staff Emeritus
This makes no sense. Do you mean you want to prove:
"if $0\le x< n$ for all real numbers n> 0, then x= 0"?

[/quote]The attempt at a solution

assume that x>0

$$\exists n, x\ni(0,\infty) |n< x$$

for X<0 theres contradiction agin .

from here X have to be 0

is that right ?[/QUOTE]

9. Nov 22, 2008

### HallsofIvy

Staff Emeritus
I would suggest that you care enough about the problem to at least copy the problem correctly!

The proof you give is by contradiction:
Suppose x> 0 then there exist n> 0 such that x> n.

But you give no proof that the last statement "there exist n> 0 such that> n" is true itself!
You need to construct such a number. If x> 0 then what positive number is less than x?

10. Nov 22, 2008

### sedaw

0<x<n

x=0.01
n=0.02

n = all the real numbers so that n might be 0.00000000000000000001

so x have to be 0