A proof with epsilon

[dunkelheit]

Prove that if [MATH]|a| \leq \varepsilon[/MATH] for all [MATH]\varepsilon>0[/MATH] then [MATH]a=0[/MATH].
Proof: Suppose that [MATH]a \neq 0[/MATH], so is [MATH]a>0[/MATH] or [MATH]a<0[/MATH].
By the arbitrary [MATH]\varepsilon>0[/MATH], we can choose [MATH]\varepsilon=\frac{|a|}{2}[/MATH]; but by hypothesis it is [MATH]|a| \leq \varepsilon =\frac{|a|}{2}[/MATH] and this is a contradiction.
Is this correct? Thanks for your time.

 

[Steven G]

What is the contradiction you arrived at? It seems that since you chose \(\displaystyle \epsilon=\dfrac{|a|}{2}\) then that is the contraction, that is that \(\displaystyle \epsilon\neq\dfrac{|a|}{2}\)

I would try using the squeeze theorem. If you need further help then post back and if you can use the squeeze theorem then please back with your work so others can see it.

 

[Steven G]

The contradiction you want is that a>0 or a<0 is not possible and yes that is what you showed. Good job.

Something has been bothering about your proof. For what it is worth I hated to grade proofs as they are so hard to judge if they are correct or not when written by beginning students.

What your proof left out is stating what was given and that was that \(\displaystyle \epsilon \leq |a|\).

 

[pka]

Prove that if [MATH]|a| \leq \varepsilon[/MATH] for all [MATH]\varepsilon>0[/MATH] then [MATH]a=0[/MATH].
Proof: Suppose that [MATH]a \neq 0[/MATH], so is [MATH]a>0[/MATH] or [MATH]a<0[/MATH].
By the arbitrary [MATH]\varepsilon>0[/MATH], we can choose [MATH]\varepsilon=\frac{|a|}{2}[/MATH]; but by hypothesis it is [MATH]|a| \leq \varepsilon =\frac{|a|}{2}[/MATH] and this is a contradiction.

No is is not correct, but lets make it work.
If \(a\ne 0\) then \(|a|>0\). We know that \(1>\frac{1}{2}\). Multiply by \(|a|\) giving \(|a|>\frac{|a|}{2}\).
If \(\varepsilon =\frac{|a|}{2}\). Then we have \(\varepsilon>|a|>\frac{|a|}{2}=\varepsilon.\) WHY?
Now do we see the contradiction?

 

[dunkelheit]

@pka: Thanks for your answer, but I don't understand where is the mistake. I've negated the thesis and get a contradiction with the hypothesis, I get [MATH]|a| \leq \varepsilon = \frac{|a|}{2} \Rightarrow |a| \leq \frac{|a|}{2} \Rightarrow 1 \leq \frac{1}{2}[/MATH] which is absurd; so isn't this proven?

 

[pka]

@pka: Thanks for your answer, but I don't understand where is the mistake. I've negated the thesis and get a contradiction with the hypothesis, I get [MATH]|a| \leq \varepsilon = \frac{|a|}{2} \Rightarrow |a| \leq \frac{|a|}{2} \Rightarrow 1 \leq \frac{1}{2}[/MATH] which is absurd; so isn't this proven?

Your proof may be clear to you but not to a grader.
Do you see how the one I posted is easy to follow? It may not be for you!
That is a good example why face-to-face mathematics instruction is important.