Proving |x| < |p| + 1 from |x - p| < 1

[Math Amateur]

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

 

[Opalg]

"... 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 $$|x| = |(x-p) + p| \leqslant |x-p| + |p| <1 + |p|.$$

 

[Math Amateur]

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

 

[Opalg]

Thanks Opalg ... obvious when you see it ... :( ...

One of those little tricks that eventually become second nature. :)