Proof Help: Converting Inequality from sn ≤ M < A to |sn - A| ≥ A - M

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SUMMARY

The discussion focuses on converting the inequality from \( s_n \leq M < A \) to \( |s_n - A| \geq A - M \). The user identifies the need to manipulate the inequalities involving three variables. Key steps include recognizing that \( s_n - A \leq M - A \) and \( A - s_n \geq A - M \) can be utilized to derive the desired absolute value inequality. The conclusion emphasizes the importance of understanding the properties of inequalities in multi-variable contexts.

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  • Understanding of inequalities involving multiple variables
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  • Basic knowledge of limits and sequences
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  • Study the properties of absolute values in inequalities
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Students in advanced mathematics courses, particularly those studying real analysis or calculus, as well as educators looking for methods to teach inequality manipulation.

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Homework Statement



I reached the point of

sn ≤ M < A

and I need to get to

|sn - A| ≥ A - M

before I can move on and reach my final conclusion.

(Don't worry about what I'm trying to prove; just help me on this little step)

Homework Equations



Not sure because I've never worked with a 3-thing'd inequality a ≤ b < c.

The Attempt at a Solution



I mean, it's obvious looking at the problem on a number line. The steps just aren't coming to me ...
 
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You know that:
[tex] s_{n}-A\leqslant M-A[/tex]
Also
[tex] A-s_{n}\geqslant A-M[/tex]
What can we say from this?
 
hunt_mat said:
You know that:
[tex] s_{n}-A\leqslant M-A[/tex]
Also
[tex] A-s_{n}\geqslant A-M[/tex]
What can we say from this?

We can't say |A - sn| ≥ A - M
 

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