MHB Limits and Continuity - Absolute Value Technicality ....

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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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