What's a number greater than ||x|-|y|| but less than |x-y|?

• Eclair_de_XII
In summary, Triangle inequality requires that the sum of the squares of the differences between each {an} be less than the sum of the squares of the {an}. This can be proved using the absolute value inequality, and a counterexample is given where the {an} do not converge at all.
Eclair_de_XII

Homework Statement

"Prove that if ##\{a_n\}## converges to ##A##, then ##\{|a_n|\}## converges to ##|A|##. Is the converse true?"

Homework Equations

Triangle inequality: ##|x+y|≤|x|+|y|##

The Attempt at a Solution

I think I solved the first part...

"If ##\{a_n\}## converges to ##A##, then there must exist an ##N∈ℕ## such that for each ##\epsilon>0##, ##|a_n-A|<\epsilon## for all ##n≥N##. In turn, ##|a_n|=|a_n-A+A|≤|a_n-A|+|A|##, which implies that ##|a_n-A|≥|a_n|-|A|##. Taking the absolute values of both terms gives: ##|a_n-A|≥| |a_n|-|A| |##. Thus, ##\epsilon>|a_n-A|≥| |a_n|-|A| |## for all ##\epsilon>0## and for all ##n≥N##."

For the second part, I basically came up with a counterexample:

"Define ##\{a_n\}## by ##\{a_n:a_n=-1,∀n∈ℕ\}##. We note that ##\{a_n\}## does not converge to ##1##, but ##\{|a_n|:|a_n|=|-1|,∀n∈ℕ\}## does converge to ##1##."

But I don't think this will cut it... What I want to find now, is an ##\epsilon>0## such that ##\epsilon>| |a_n|-|A| |## but ##|a_n-A|>\epsilon##. Hence, I will need to find a number ##\epsilon## such that ##|a_n-A|>\epsilon>| |a_n|-|A| |##, to show that the converse is not true. I will need to find an appropriate choice of ##\epsilon## that satisfies these conditions, in other words.

Eclair_de_XII said:
For the second part, I basically came up with a counterexample:

"Define ##\{a_n\}## by ##\{a_n:a_n=-1,∀n∈ℕ\}##. We note that ##\{a_n\}## does not converge to ##1##, but ##\{|a_n|:|a_n|=|-1|,∀n∈ℕ\}## does converge to ##1##."

But I don't think this will cut it...
Why not? You may want to slightly modify and state the counterexample more carefully so that you can convince yourself if it is right or not.

You mean I should rearrange the order in which I state it, to make the implication--or lack thereof--clearer?

That is one thing that would help. Also, you haven't specifically stated what A is. Is it +1 or -1? Define your {an} and A first. Then show that is satisfies the absolute value limits. Finally show that it does not satisfy the non-absolute value limits. I don't think that you need to go into detail of the ε proof for such obvious limits, but I don't know what your class will want.

Eclair_de_XII said:
You might also be able to think of a closely related example where the {an} do not converge at all.

Oh, do you mean, ##\{a_n:a_n=(-1)^n,\forall n \in ℕ\}##?

1. What does the expression ||x|-|y|| mean?

The double absolute value bars in this expression indicate the magnitude or distance of the number within them from zero. So, ||x|-|y|| represents the absolute value of the difference between the absolute values of x and y.

2. How do I find a number that is greater than ||x|-|y||?

To find a number greater than ||x|-|y||, you need to determine the values of x and y first. Then, calculate the absolute value of the difference between the absolute values of x and y, and add any number greater than that result to ||x|-|y||.

3. Can the number be equal to ||x|-|y||?

No, the number must be greater than ||x|-|y||. This is because the expression ||x|-|y|| represents the minimum possible value that can be greater than or equal to |x-y|.

4. How do I know if a number is less than |x-y|?

To determine if a number is less than |x-y|, you need to calculate the absolute value of the difference between x and y. Then, check if the number in question is smaller than that result.

5. Is there a specific range of numbers that satisfy this condition?

Yes, there is a specific range of numbers that satisfy this condition. This range is between ||x|-|y|| and |x-y|, where the number must be greater than ||x|-|y|| but less than |x-y|.

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