2
$\begingroup$

I feel uncertain about what should be quite basic reasoning — my maths knowledge is elementary. I am reading the first chapter of Nikolai Piskunov's Differential and Integral Calculus.

A proof of the reverse triangle inequality is set out as follows:

The absolute value of a difference is no less than the difference of the absolute values of the minuend and subtrahend. $$|x-y|\ge |x|-|y| \text{, } |x|>|y|$$ Proof. Let $x-y=z$, then $x=y+z$ and from what has been proved $$|x|=|y+z|\leq |y|+|z|=|y|+|x-y|$$ whence $$|x|-|y|\le |x-y|$$ thus completing the proof.

("what has been proved" refers to $\forall \{x,y\}\in\mathbb{R}(|x+y|\leq |x|+|y|)$.)

I'm wondering why this proof would necessarily imply the condition $|x|>|y|$. Or have I perhaps misunderstood the formatting, and it does not signify a condition?

$\endgroup$

2 Answers 2

2
$\begingroup$

The condition $|x| > |y|$ is not at all necessary, and is not involved in the proof. It could have been left out without causing any problems. The only reason the author might have for mentioning it is that the inequality $|x - y| \ge |x| - |y|$ is trivial when $|x| \le |y|$, because we already know $|x - y| \ge 0$.

$\endgroup$
1
$\begingroup$

You are misreading the theorem , something like this :

$$\text{It is always true that } |x-y|\ge |x|-|y| \text{ , and } |x|>|y|$$

You should read it like this :

$$\text{It is always true that } |x-y|\ge |x|-|y| \text{ , when } |x|>|y|$$

Here, LHS is Positive , hence when RHS is Negative , it is tautologically true. Eg $|4-9|=+5$ , $|4|-|9|=-5$ , hence $|4-9| \ge 0 \ge |4|-|9|$ , via elementary considerations , not requiring reverse triangle inequality. Hence there is no necessity or requirement to state it or high-light it with that theorem.

What the theorem wants to focus on is when RHS too is Positive & comparing with LHS is necessary & non-trivial , that is , when $|x|-|y| \gt 0$ , which is actually when $|x| \gt |y|$

OP : wondering why this proof would necessarily imply the condition

It is not implying that condition , it is just taken from the theorem condition.
When we take opposite of that condition , it is automatically true , because Positive is always greater than Negative.

$\endgroup$
2
  • $\begingroup$ Yes, it's indeed me misinterpreting the layout. I gathered that from @robert 's response. Thanks for your response. $\endgroup$
    – Dongchen
    Commented Aug 9 at 14:59
  • $\begingroup$ No worries , I wanted to add a little more to it , might aid you or other later readers $\endgroup$
    – Prem
    Commented Aug 9 at 17:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .