0
$\begingroup$

I don't really understand why is statement 1 ≥ statement 2 in the attached picture. From what I understand the negative term in statement 2 must be greater than or equal the negative term in statement 1 if statement 1 ≥ statement 2 but I don't really know how. Any help on this matter will be really appreciated. Thanks!

Statement 1 and Statement 2 refers to the red highlighted boxes, from the attached picture, tagged as 1 and 2 respectively.

Original source: http://oucsace.cs.ohiou.edu/~razvan/courses/cs4040/lecture15.pdf

enter image description here

$\endgroup$
  • $\begingroup$ I don't understand what you mean by statement 1 ≥ statement 2. What is statement 1? What does it mean for one statement to be greater than or equal to another? $\endgroup$ – D.W. Jun 7 '18 at 0:00
  • $\begingroup$ @D.W. I edited the post and the attached image. $\endgroup$ – D-PUNK-R Jun 7 '18 at 6:35
1
$\begingroup$

Slides are not a substitute for a textbook or careful exposition. This is especially true when reading a proof. Slides are intended for presentation in real-time, and as a result often leave out some details. If there is some aspect you don't understand when reading a set of slides, usually the best thing to do is to find a proper written exposition of the subject -- usually a textbook (but sometimes written lecture notes can be adequate, if written in enough detail).

In this case, on the previous slide, the statement of the theorem mentions "nonincreasing order of $p_i/w_i$", which means that $p_1/w_1 \ge p_2/w_2 \ge \cdots$. In particular, $p_k/w_k \ge p_i/w_i$ when $k<i$. The inequality you are referencing then follows.

$\endgroup$
  • $\begingroup$ Ok, I get it now. Thanks a lot. Those slides are not from my school. I just found the slides on the internet. I usually rely on the book but the book I am using in my university doesn't even include the proof. So, I started looking for it on the internet and I found a lot of proofs that were confusing to me, but I was able to grasp this one a lot better than the other ones so I decided to stick to this one. $\endgroup$ – D-PUNK-R Jun 7 '18 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.