Exercise 1.7 (iii) [Section 1, BOMC2011]
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
Edit your post, why don't you?
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Nur Muhammad Shafiullah  Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi

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 Joined: Wed Sep 28, 2011 12:24 pm
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
1.7(i)
0 ≤ a ≤ b ≤ 1.
Does that imply,
a ≤ 1, b ≥ 0????
if so , we can't divide (b  a) by (1  ab) . if (ab = 1), (1 ab = 0) ??
b ≥ a
=> b  a ≥ 0
=> (b  a) / 1  ab ≥ 0 [ a≤b≤1 , ab≤1, 1  ab not equal to 0] ****
again,
so, b≤1
=> b  a ≤ 1 [ a≤b≤1 ]****
=> (b  a)/ 1  ab ≤ 1 [1ab>0]
so 0 ≤(b  a)/ 1  ab ≤ 1
help me. tell me if the **** noted assumptions r true or not.
0 ≤ a ≤ b ≤ 1.
Does that imply,
a ≤ 1, b ≥ 0????
if so , we can't divide (b  a) by (1  ab) . if (ab = 1), (1 ab = 0) ??
b ≥ a
=> b  a ≥ 0
=> (b  a) / 1  ab ≥ 0 [ a≤b≤1 , ab≤1, 1  ab not equal to 0] ****
again,
so, b≤1
=> b  a ≤ 1 [ a≤b≤1 ]****
=> (b  a)/ 1  ab ≤ 1 [1ab>0]
so 0 ≤(b  a)/ 1  ab ≤ 1
help me. tell me if the **** noted assumptions r true or not.
 nafistiham
 Posts: 829
 Joined: Mon Oct 17, 2011 3:56 pm
 Location: 24.758613,90.400161
 Contact:
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
it is said that $0\leq a\leq b\leq1$ so cant it be $a=b=0$?
at that case you cant divide anything by $1ab$
at that case you cant divide anything by $1ab$
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please cooperate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please cooperate.
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
$ab=0$ directly implies an equality, so we may as well exclude that from our proof.
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
To exclude triviality from your solution, just state and prove those cases first, and then complete the proof.
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi

 Posts: 8
 Joined: Wed Sep 28, 2011 12:24 pm
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
Prove which cases??? ****noted ones??
a≤b≤1
=> ba≤1
=> (ba)/1ab ≤1 [1≥b≥a so ab≤1 1ab ≥0]
does that satisfy??
a≤b≤1
=> ba≤1
=> (ba)/1ab ≤1 [1≥b≥a so ab≤1 1ab ≥0]
does that satisfy??
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
Just to explain this, you can divide the thing in two cases , where $ab=1$ and where $ab \leq 1$. Then you can divide something by $1ab$ when it is necessary.
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
Sad but true, আমি নিজেও সমাধান করতে পারিনি, উপরের ৪ টা সমধানের একটাও বুঝতে পারিনি!
Stay Hungry. Stay Foolish.
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
Please tell others in which part were you stuck. Then others can try and explain it to you.
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah  Mahi
Re: Exercise 1.7 (iii) [Section 1, BOMC2011]
$0 \leq a$ and $0 \leq b$. substracting we get $ba \geq 0$. again $a \leq 1$ and so is b. Similarly by substracting we get $ba \leq 0$ !!! walla! so a=b. and $\frac{1}{4ab}$ is abviously greater than 1; so our inequality is proven. and for the first one, we merely just have to see that $a \leq b$ $\rightarrow ba^2 \leq ab^2$.
@Sourav Das how did you write $4ab(1ab) \geq 4(ab^2  a^2 b)$ ?
@*mahi* how can you say $4a + \frac{1}{ab} \geq 4b$ if $1 \geq b^{\frac{3}{2}}$ ?
@MATHPRITOM what a solution! but how would I know I've to take $(2a \sqrt{b}  1)^2 \geq 0$ ? If we start from the backward it's not easy to say $\sqrt{b} \geq b \geq b^2$.
@*Mahi* "ab=0 directly implies an equality, so we may well exclude that from our proof"! why we are excluding that?the problem did asked ask to prove both inequality and equality. how can I show $\frac {ba}{1ab} \geq 1$ when ab=0 ? we can't divide with ab1 right?
rakeen
@Sourav Das how did you write $4ab(1ab) \geq 4(ab^2  a^2 b)$ ?
@*mahi* how can you say $4a + \frac{1}{ab} \geq 4b$ if $1 \geq b^{\frac{3}{2}}$ ?
@MATHPRITOM what a solution! but how would I know I've to take $(2a \sqrt{b}  1)^2 \geq 0$ ? If we start from the backward it's not easy to say $\sqrt{b} \geq b \geq b^2$.
@*Mahi* "ab=0 directly implies an equality, so we may well exclude that from our proof"! why we are excluding that?the problem did asked ask to prove both inequality and equality. how can I show $\frac {ba}{1ab} \geq 1$ when ab=0 ? we can't divide with ab1 right?
rakeen
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