Can you prove that |x|<y if -y<x<y?. In case x$\geq$, I can prove myself. But in case x$<$0, I can only infer x$<$0 and -y$<$0, I can not prove x$>$-y. Please prove that. Thank you.
$|x|<y$
#1
Đã gửi 30-05-2016 - 12:58
#2
Đã gửi 01-06-2016 - 19:12
Note that $|x|=|-x|$. So whether $x$ is positif or not, the result doesn't change a bit. Hence, you could assume withou loss of generality (WLOG) that $x\ge 0$. The rest is up to you.
- minhnghiem08 yêu thích
$$\text{LOVE}\left( x \right)|_{x = \alpha}^\Omega = + \infty $$
I'm still there everywhere.
#3
Đã gửi 03-06-2016 - 17:39
I do not still understand. What is WLOG? Can you prove it more clear? I just started to learn math myself 2 month ago. This begin step make me difficult. I really appreciate if you help me deal with this problem
#4
Đã gửi 04-06-2016 - 22:56
Don't be scared of the word "WLOG" (Without loss of generality, i've written it right before mentioning). It means that "the proof in a such case will still be available for the rest, so we may assume in this case".
If you wish a complete proof without using WLOG, here it is:
- minhnghiem08 yêu thích
$$\text{LOVE}\left( x \right)|_{x = \alpha}^\Omega = + \infty $$
I'm still there everywhere.
#5
Đã gửi 16-06-2016 - 13:07
I were known that conclusion of math is made from Anxioms. I am studying about real number system and I see that it has 9 anxioms. I know that if we want solve Corollary, we must base on Anxioms provided. In your answer, you say -y<x<0 => y>(-x)>0. I wonder anxiom talk this. Please explain me clearer! I am just beginner. Thank you
#6
Đã gửi 19-06-2016 - 21:12
I were known that conclusion of math is made from Anxioms. I am studying about real number system and I see that it has 9 anxioms. I know that if we want solve Corollary, we must base on Anxioms provided. In your answer, you say -y<x<0 => y>(-x)>0. I wonder anxiom talk this. Please explain me clearer! I am just beginner. Thank you
Of course, we can only have this implication "$-y<x<0 \Rightarrow y>(-x)>0$" in real number system. Furthermore, I have no idea about the "Axioms" you said. All I wrote above is the basis knowledge.
$$\text{LOVE}\left( x \right)|_{x = \alpha}^\Omega = + \infty $$
I'm still there everywhere.
1 người đang xem chủ đề
0 thành viên, 1 khách, 0 thành viên ẩn danh