1

I am reading several works of an author. I have identified in one of the works, what he meant by "Obviously $a \leq b$..." is the same as "Let $a \leq b$...", since (I think) it is absurd if the construction given implies $a \leq b$.

In another work of his, I have not managed to eliminate the possibility where due to the construction $c \leq d$, yet the sentence "Obviously $c \leq d$..." is present.

Thus I would like to know if this style of writing is common, or am I possibly missing something from the construction. The works of the author concerned is a famous (in the field) book, and a paper of his. This is the second paper I'm reading, that is why I'm still not used to the lingo.

I put the tag notation as I don't know what else suits this.

Edit:

The paper I am unsure about can be found here, proof of Thm 5, "Obviously, $x_0^T \leq w_0$."

The book where I thought the two phrases might be equivalent is Matching Theory, by Laszlo Lovasz and Michael Plummer. Can be found on page 470, proof of theorem 12.3.3, "Obviously $v \not\in T_1 \cup T_2$". I don't think $H'-E_i$ must not use $v$ as a vertex cover, that is why I thought "Obviously..." here means "Let...".

AyamGorengPedes
  • 964
  • Can you link the paper? It is hard to judge without context. – Andreas Tsevas Dec 16 '21 at 12:27
  • 1
    Please quote examples where you've seen obviously (or e.g. clearly) & let so we can understand why you think they're synonyms. They're not. See also here. – J.G. Dec 16 '21 at 12:28
  • 1
    Those phrases are certanly not equivalent. For the first, we must have defined both $a$ and $b$ and then we remark that the latter is not less than the former. For the second, we need not have defined either of them (perhaps we are just choosing two any two values which satisfy the desired inequality) or we have just defined one of the values and are choosing the other subject to that inequality. – lulu Dec 16 '21 at 12:28
  • @Andreas The paper can be found here http://ftp.cs.elte.hu/~lovasz/scans/perfect.pdf, proof of Thm 5, "Obviously, $x_0^T \leq w_0$." – AyamGorengPedes Dec 16 '21 at 12:28
  • @J.G. I will add it in the question – AyamGorengPedes Dec 16 '21 at 12:30
  • In that paper, what (partial?) order is the author using for vectors? What does it mean to write that $\vec v≤ \vec u$? Lexicographic? Something else? – lulu Dec 16 '21 at 12:32
  • @lulu the $a \leq b$ is just an example, but I edited the question and added the paper and the book. – AyamGorengPedes Dec 16 '21 at 12:37
  • @AyamGorengPedes In that example, "obviously" does not mean "let". The proof says: Let $w_0$ be any $(0,1)$-vector and use it as $w$ in equation (11). Then let $x_0$ be a solution of equation (11) with the greatest possible number of zeros. The statement "obviously, $x_0^T \leq w_0$" then follows from these definitions. – Andreas Tsevas Dec 16 '21 at 12:41
  • @AyamGorengPedes More generally, I have never seen "let" and "obviously" used synonymously. The former is a definition and the latter is a (supposedly) easy deduction. – Andreas Tsevas Dec 16 '21 at 12:42
  • @Andreas yes I just realized we can force each co-ordinate of $x_0$ to be 0 if the same co-ordinate for $w_0$= 0. The solution would still be feasible, and doesn't affect objective value. Thus the co-ordinates of $x_0$ are 1 only when the same co-ordinate for $w_0$ is not 0, and from here comes $x_0^T \leq w_0$. I should have realized this earlier, and not ask a silly question – AyamGorengPedes Dec 16 '21 at 12:46
  • 2
    @AyamGorengPedes Don't beat yourself up; asking questions is how you can learn. – Andreas Tsevas Dec 16 '21 at 12:47

0 Answers0