MHB Solving Inequality: Get Help with (Uw-UL)<a-B

  • Thread starter Thread starter martyfarty
  • Start date Start date
  • Tags Tags
    Inequality
martyfarty
Messages
4
Reaction score
0
Hi, I am reading a paper and in this equation is given.

View attachment 6477

I don't quite follow how they end up with the last (Uw-UL)<a-B. If I do it myself I get the inequality sign wrong.

Any help?

Thx
 

Attachments

  • help.jpg
    help.jpg
    4.9 KB · Views: 84
Mathematics news on Phys.org
Welcome martyfarty,

The inequality holds if and only if $\alpha > \beta$. If you had the inequality reversed, then perhaps you had $\alpha < \beta$.
 
Indeed \alpha > \beta is true. How would you step by step come to the last equation given that \alpha > \beta?
 
Let's first subtract $$\frac{1}{2}$$ from both sides:

$$\frac{\left(U^W-U^L\right)}{2(\alpha-\beta)}<\frac{1}{2}$$

If $\alpha-\beta>0$, then we may multiply through by $2(\alpha-\beta)$ to get:

$$U^W-U^L<\alpha-\beta$$
 
MarkFL said:
Let's first subtract $$\frac{1}{2}$$ from both sides:

$$\frac{\left(U^W-U^L\right)}{2(\alpha-\beta)}<\frac{1}{2}$$

If $\alpha-\beta>0$, then we may multiply through by $2(\alpha-\beta)$ to get:

$$U^W-U^L<\alpha-\beta$$

However, $$\alpha$$ and $$\beta$$ are reversed right? Like this

$$\frac{\left(U^W-U^L\right)}{2(\beta-\alpha)}<\frac{1}{2}$$
 
martyfarty said:
However, $$\alpha$$ and $$\beta$$ are reversed right? Like this

$$\frac{\left(U^W-U^L\right)}{2(\beta-\alpha)}<\frac{1}{2}$$

No, I reversed them to get rid of the leading negative for that term. :D
 
MarkFL said:
No, I reversed them to get rid of the leading negative for that term. :D

Okay wow, that option did not even cross my mind haha. Thanks for your help!
 
Back
Top