- #1

- 62

- 0

and also 0<n so x=0.

The attempt at a solution

assume that x>0

[tex]\exists n, x\ni(0,\infty) |n< x [/tex]

contradiction !

for X<0 theres contradiction agin .

from here X have to be 0

is that right ?

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- Thread starter sedaw
- Start date

- #1

- 62

- 0

and also 0<n so x=0.

The attempt at a solution

assume that x>0

[tex]\exists n, x\ni(0,\infty) |n< x [/tex]

contradiction !

for X<0 theres contradiction agin .

from here X have to be 0

is that right ?

- #2

- 586

- 1

x=1, n=2>0 satisfies [itex]0\leq x<n[/itex], yet x does not equal zero.

- #3

- 62

- 0

x=1, n=2>0 satisfies [itex]0\leq x<n[/itex], yet x does not equal zero.

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

- #4

- 586

- 1

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

What are you trying to prove? I was under the impression that it is the following:

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

- 62

- 0

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 ?

- #6

- 586

- 1

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

- #7

- 62

- 0

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 [tex]0\leq X<n[/tex]

for each real posiive uumber n so x=0 .

- #8

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

This makes no sense. Do you mean you want to prove:need to prove that if [tex]0\leq X<n[/tex]

and also 0<n so x=0.

"if [itex]0\le x< n[/itex] for

[/quote]The attempt at a solution

assume that x>0

[tex]\exists n, x\ni(0,\infty) |n< x [/tex]

contradiction !

for X<0 theres contradiction agin .

from here X have to be 0

is that right ?[/QUOTE]

- #9

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

I would suggest that younot integer all the real num.

so what u suggest ?

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

- #10

- 62

- 0

I would suggest that youcareenough 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 thenwhatpositive 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

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