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