Prove |A^2 - B^2| \leq \frac{1}{2} \{2|B|+\frac{1}{2}\}

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Discussion Overview

The discussion revolves around proving the inequality |A^2 - B^2| ≤ 1/2 {2|B| + 1/2} for real numbers A and B, given the condition |A - B| < 1/2. The scope includes mathematical reasoning and problem-solving strategies related to inequalities and algebraic identities.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant seeks hints for proving the inequality, expressing frustration over their inability to progress.
  • Another participant suggests rewriting |A^2 - B^2| as a product of two expressions, indicating it is a well-known identity.
  • A participant attempts to apply the identity and expresses uncertainty about the next steps, questioning the use of the given inequality |A - B| < 1/2.
  • Another participant emphasizes the need to prove the inequality rather than assume it, restating the equality |A^2 - B^2| = |A - B||A + B| and suggesting to derive an estimate for |A + B| using the given inequality.
  • A later reply mentions the necessity of using the triangle inequality in the proof process.

Areas of Agreement / Disagreement

Participants generally agree on the need to prove the inequality and the approach of using algebraic identities. However, there is no consensus on the specific steps to take next, and uncertainty remains about how to apply the given conditions effectively.

Contextual Notes

Participants have not fully resolved the mathematical steps needed to connect the given inequality with the desired proof, and there are dependencies on the definitions and interpretations of the expressions involved.

bullpup
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There are many questions i can't get. This is just one. Can anyone give me a hint on what to do? It's probably really simple :(

If A and B are real numbers such that [tex]|A - B| < \frac{1}{2}[/tex], show that:

[tex]|A^2 - B^2| \leq \frac{1}{2} \{2|B|+<br /> \frac{1}{2} \}[/tex]
 
Last edited:
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Welcome to PF!
Try to rewrite [tex]|A^{2}-B^{2}|[/tex] as a product of two expressions..
(It's a well known identity)
 
Is this what you meant?

[tex]|(A+B) (A-B)| \leq |B| + \frac{1}{4}[/tex]

If so, i already had that but didn't know where to go from there...already wasted a lot of paper going no where

Am i supposed to use [tex]-\frac{1}{2} < A - B < \frac{1}{2}[/tex] somewhere?

thanks
 
Last edited:
First of all:
You are to PROVE that inequality, not ASSUME it!
Hence, what you start with, is the following EQUALITY:
[tex]|A^{2}-B^{2}|=|A-B||A+B|[/tex]
(Ok so far?)
Now, use the inequality you've been given to derive:
[tex]|A^{2}-B^{2}|=|A-B||A+B|\leq\frac{1}{2}|A+B|[/tex]
Agreed?
Now you'll need to use your given inequality to provide an estimate of |A+B|!
 
PS:
You'll need to use the triangle inequality as well..
 

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