Limits and Continuity - Absolute Value Technicality ....

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SUMMARY

The discussion centers on proving the inequality involving absolute values from Manfred Stoll's "Introduction to Real Analysis," specifically Example 4.1.2 (a). The statement asserts that if $$ |x - p| < 1 $$, then $$ |x| < |p| + 1 $$. The proof utilizes the triangle inequality, $$ |a + b| \leq |a| + |b| $$, where $$ a = x - p $$ and $$ b = p $$, leading to the conclusion that $$ |x| \leq |x - p| + |p| < 1 + |p| $$.

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  • Understanding of absolute value inequalities
  • Familiarity with the triangle inequality
  • Basic knowledge of limits and continuity in real analysis
  • Experience with mathematical proofs and rigor
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  • Study the triangle inequality in depth
  • Explore additional examples in "Introduction to Real Analysis" by Manfred Stoll
  • Learn about rigorous proof techniques in real analysis
  • Investigate applications of absolute value inequalities in calculus
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Students of real analysis, mathematics educators, and anyone seeking to deepen their understanding of limits and continuity through rigorous proofs involving 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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