Limits and Continuity - Absolute Value Technicality ....

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Discussion Overview

The discussion centers around proving an inequality involving absolute values as presented in an example from Manfred Stoll's "Introduction to Real Analysis." The focus is on the application of the triangle inequality in the context of limits and continuity.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Peter seeks help to rigorously prove the statement that if $$ |x - p| < 1 $$, then $$ |x| < |p| + 1 $$.
  • One participant explains that the proof relies on the triangle inequality, stating that with $$ a = x - p $$ and $$ b = p $$, the inequality can be expressed as $$ |x| = |(x - p) + p| \leqslant |x - p| + |p| < 1 + |p| $$.
  • Peter acknowledges the explanation and expresses appreciation for the clarity it provides.

Areas of Agreement / Disagreement

The discussion appears to be resolved in terms of understanding the application of the triangle inequality, as participants agree on the method used to prove the statement.

Contextual Notes

No limitations or unresolved issues are noted in the discussion.

Who May Find This Useful

Readers interested in mathematical proofs, particularly in the context of real analysis and the properties of absolute values.

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I am reading Manfred Stoll's Book: "Introduction to Real Analysis" ... and am currently focused on Chapteer 4: Limits and Continuity ...

I need some help with an inequality involving absolute values in Example 4.1.2 (a) ... Example 4.1.2 (a) ... reads as follows:View attachment 7247In the above text we read ...

"... If $$ \mid x - p \lvert \ \lt \ 1$$ then $$\mid x \mid \ \lt \ \mid p \mid \ + \ 1$$ ... "Can someone please show me how to rigorously prove the above statement ...

Peter
 
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Peter said:
"... If $$ \mid x - p \lvert \ \lt \ 1$$ then $$\mid x \mid \ \lt \ \mid p \mid \ + \ 1$$ ... "Can someone please show me how to rigorously prove the above statement ...
This comes from the triangle inequality $|a+b|\leqslant |a| + |b|$. With $a=x-p$ and $b=p$, that becomes $$|x| = |(x-p) + p| \leqslant |x-p| + |p| <1 + |p|.$$
 
Opalg said:
This comes from the triangle inequality $|a+b|\leqslant |a| + |b|$. With $a=x-p$ and $b=p$, that becomes $$|x| = |(x-p) + p| \leqslant |x-p| + |p| <1 + |p|.$$
Thanks Opalg ... obvious when you see it ... :( ...

Appreciate your help ...

Peter
 
Peter said:
Thanks Opalg ... obvious when you see it ... :( ...
One of those little tricks that eventually become second nature. :)
 

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