Proof That If 0 ≤ X < n, X Must Equal 0

[sedaw]

need to prove that if 0\leq X&lt;n

and also 0<n so x=0.

The attempt at a solution

assume that x>0

\exists n, x\ni(0,\infty) |n&lt; x

contradiction !

for X<0 there's contradiction agin .

from here X have to be 0

is that right ?

 

[Pere Callahan]

x=1, n=2>0 satisfies 0\leq x&lt;n, yet x does not equal zero.

 

[sedaw]

x=1, n=2>0 satisfies 0\leq x&lt;n, yet x does not equal zero.

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

 

[Pere Callahan]

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

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.

 

[sedaw]

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.

not integer all the real num.

so what u suggest ?

 

[Pere Callahan]

not integer all the real num.

so what u suggest ?

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

 

[sedaw]

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

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

for each real posiive uumber n so x=0 .

 

[HallsofIvy]

need to prove that if 0\leq X&lt;n

and also 0<n so x=0.

This makes no sense. Do you mean you want to prove:
"if 0\le x&lt; 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&lt; x

contradiction !

for X<0 there's contradiction agin .

from here X have to be 0

is that right ?[/QUOTE]

 

[HallsofIvy]

not integer all the real num.

so what u suggest ?

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?

 

[sedaw]

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?

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