Introduction to Analysis by Bilodeau. Problem 1.3.3

[Phylosopher]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >[/color]

I couldn't find more informative title!

I find difficulties with proofs. So my solution might be weird

The problem says
"Suppose that x is a fixed non-negative real number such that, for all positive real numbers ε, 0≤ε<x. Show that x=0."

My attempt was:
Assume 0≤x and x <ε
Case 1:
0 <x
For evey ε, 0<x and x <ε is false

Case 2:
0=x
For evey ε, 0=x and x <ε is trueMy professor was not happy with making cases to proof the statement, and he said this is not a real proof. He suggested starting with the assumption 0 <x and from that I proof 0=x is the wanted fixed non negative real number that satisfies the statement.

I am confused. How am I supposed to proof it?

 

[BvU]

Can you check the problem statement ?

all positive real numbers ε, 0≤ε<x

1 is a positive real number, but 0≤ε<0 is definitely not right for ε = 1

 

[PeroK]

The problem says
"Suppose that x is a fixed non-negative real number such that, for all positive real numbers ε, 0≤ε<x. Show that x=0."

You've got this the wrong way round. That would force $x$ to be infinite. Instead, you should have:

$\forall \epsilon > 0, \ x < \epsilon$

For the solution, did you consider a proof by contradiction?

 

[Phylosopher]

Can you check the problem statement ?
1 is a positive real number, but 0≤ε<0 is definitely not right for ε = 1

Oh sorry, you are right.

The question should be the following:
"Suppose that x is a fixed non-negative real number such that, for all positive real numbers ε, 0≤x<ε. Show that x=0."

You've got this the wrong way round. That would force $x$ to be infinite. Instead, you should have:

$\forall \epsilon > 0, \ x < \epsilon$

For the solution, did you consider a proof by contradiction?

Sorry, I wrote the wrong way.

The question should be the following:
"Suppose that x is a fixed non-negative real number such that, for all positive real numbers ε, 0≤x<ε. Show that x=0."

 

[PeroK]

Oh sorry, you are right.

The question should be the following:
"Suppose that x is a fixed non-negative real number such that, for all positive real numbers ε, 0≤x<ε. Show that x=0."
Sorry, I wrote the wrong way.

The question should be the following:
"Suppose that x is a fixed non-negative real number such that, for all positive real numbers ε, 0≤x<ε. Show that x=0."

Any ideas about a proof? First, perhaps, can you see and explain why it's true?

 

[Phylosopher]

Any ideas about a proof? First, perhaps, can you see and explain why it's true?

Discussion:
x fixed non-negative real number
ε is a positive real number

There exist no fixed non-negative real number x such that it's less than every positive number ε, except 0.

My Proof:

Assume 0≤x and x <ε
Case 1:
0 <x
For every ε, 0<x and x <ε is false (Since there is always a positive number that is less than x "i.e x >ε" when 0<x. Another way to state it would be that since x and ε here are both positive real numbers, there will always be x >ε an x <ε which is a contradiction with one of the axioms)

Case 2:
0=x
For every ε, 0=x and x <ε is true (Since x <ε, ε≠0 "Axiom". Thus 0<ε is always true since by definition a positive real number is ε>0 )

 

[PeroK]

My Proof:

Assume 0≤x and x <ε
Case 1:
0 <x
For every ε, 0<x and x <ε is false (Since there is always a positive number that is less than x "i.e x >ε" when 0<x. Another way to state it would be that since x and ε here are both positive real numbers, there will always be x >ε an x <ε which is a contradiction with one of the axioms)

I think you've got the idea, but this proof is not really right. You need to think about a specific $\epsilon$ if $x > 0$. Any ideas?

 

[Phylosopher]

I think you've got the idea, but this proof is not really right. You need to think about a specific $\epsilon$ if $x > 0$. Any ideas?

Well. This is confusing for me! I am supposed to find a fixed x, why should I search for a fixed ε?

Since x>0 is assumed, 0<x<ε. There is no such fixed ε that satisfy 0<x<ε for all x.

 

[PeroK]

Well. This is confusing for me! I am supposed to find a fixed x, why should I search for a fixed ε?

Since x>0 is assumed, 0<x<ε. There is no such fixed ε that satisfy 0<x<ε for all x.

Okay, I claim that $10^{-6}$ is a number that is less than every positive number. How would you disprove that?

 

[Phylosopher]

Okay, I claim that $10^{-6}$ is a number that is less than every positive number. How would you disprove that?

I would say that 10^-7 is less than 10^-6, thus 10^-6 is not less than every positive number

 

[PeroK]

I would say that 10^-7 is less than 10^-6, thus 10^-6 is not less than every positive number

Okay, so what about $10^{-8}$?

 

[Phylosopher]

Okay, so what about $10^{-8}$?

A number that is less than 10^-7, but yet not less than 10^-9, which mean it is not less than every positive number

 

[PeroK]

A number that is less than 10^-7, but yet not less than 10^-9, which mean it is not less than every positive number

So, using what you've done for these numbers, what about any number $x > 0$? Why isn't that less than every positive number?

 

[Phylosopher]

So, using what you've done for these numbers, what about any number $x > 0$? Why isn't that less than every positive number?

Because there is always a positive real number that is less than x. It also can be shown by number line: Since x>0, x/2>0 and thus x/2<x. So there is always a positive number that is less than another positive number.

 

[PeroK]

Because there is always a positive real number that is less than x. It also can be shown by number line: Since x>0, x/2>0 and thus x/2<x. So there is always a positive number that is less than another positive number.

Yes. Note you also have $x \not < x$. In other words, any positive $x$ is not less than itself.

Can you do the proof now?

 

[Phylosopher]

Yes. Note you also have $x \not < x$. In other words, any positive $x$ is not less than itself.

Can you do the proof now?

Proof:
let x>0
there exit ε for any x, 0<x<ε | this is false since there exit x>ε, which contradicts with an Axiom.
Thus x=0 is the only non-negative real number x that satisfies 0≤x<ε (0=0<ε).

 

[PeroK]

Proof:
let x>0
there exit ε for any x, 0<x<ε | this is false since there exit x>ε, which contradicts with an Axiom.
Thus x=0 is the only non-negative real number x that satisfies 0≤x<ε (0=0<ε).

What happened to $x/2$?

 

[Phylosopher]

What happened to $x/2$?

for x>0, x/2>0 which is still not less than all positive numbers ε
for x=0, x/2=0 which is less than all positive numbers

 

[PeroK]

for x>0, x/2>0 which is still not less than all positive numbers ε
for x=0, x/2=0 which is less than all positive numbers

No, you haven't understood the idea at all. It's difficult now to give you any more hints. Let me give you a written answer and you can try to translate that into maths:

Suppose $x$ is greater than zero, then $x$ is not less than $x/2$, which is itself a positive number, so $x$ is not less than all positive numbers ...