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

  • Thread starter bullpup
  • Start date
  • Tags
    Proof
In summary, the conversation discusses how to prove an inequality involving real numbers A and B, given the condition |A-B| < \frac{1}{2}. It is suggested to rewrite the expression as a product of two expressions and use the triangle inequality to provide an estimate for one of the expressions. The conversation also addresses a misunderstanding about assuming vs proving the inequality.
  • #1
bullpup
2
0
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|+
\frac{1}{2} \}[/tex]
 
Last edited:
Physics news on Phys.org
  • #2
Welcome to PF!
Try to rewrite [tex]|A^{2}-B^{2}|[/tex] as a product of two expressions..
(It's a well known identity)
 
  • #3
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 :grumpy:

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

thanks
 
Last edited:
  • #4
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|!
 
  • #5
PS:
You'll need to use the triangle inequality as well..
 

1. What does the equation "Prove |A^2 - B^2| \leq \frac{1}{2} \{2|B|+\frac{1}{2}\}" mean?

The equation is an inequality that is asking for proof that the absolute value of the difference between A squared and B squared is less than or equal to half of the sum of twice the absolute value of B and one-half.

2. How do you prove the given equation?

To prove the equation, you can start by assuming that A and B are any real numbers. Then, using algebraic manipulations and properties of absolute values, you can show that the left side of the inequality is always less than or equal to the right side.

3. What is the significance of the absolute value in the equation?

The absolute value ensures that the expression inside the brackets is always positive. This is important because the inequality is asking for a proof that the left side is less than or equal to the right side, regardless of the values of A and B.

4. Can you provide an example to illustrate the equation?

Sure, let's say A = 3 and B = 2. Plugging these values into the equation, we get |3^2 - 2^2| \leq \frac{1}{2} \{2|2|+\frac{1}{2}\} which simplifies to 5 \leq 5. This is true, so the equation holds for these values.

5. What are some real-world applications of this equation?

One possible application could be in physics, where this inequality could be used to prove the stability of a system. It could also be used in optimization problems, where the equation could represent a constraint that needs to be satisfied in order to find the optimal solution.

Similar threads

Replies
5
Views
273
Replies
3
Views
939
Replies
4
Views
979
Replies
4
Views
193
Replies
1
Views
1K
  • Calculus
Replies
24
Views
1K
Replies
3
Views
1K
Replies
16
Views
2K
Replies
3
Views
219
Replies
3
Views
1K
Back
Top