My Math Forum 3 Problems I am really stuck on

 Calculus Calculus Math Forum

 January 1st, 2017, 02:10 PM #11 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 Note -- I have no issue with your proof of the converse (i.e., that the product of two elements of $R$ can also be expressed as a sum of two elements of $R$).
January 1st, 2017, 02:13 PM   #12
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 Originally Posted by romsek nope, just that there exist an $a, b$ and they sum to $A$ I see the problem. the assertion is false for probably most $A$, but it's true for all $A$ made up of the sum of two members of $R$
That's what you would need to prove.

As of now, you haven't proved it.

 January 1st, 2017, 03:01 PM #13 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 @Rachel1234: These aren't normal HW problems. They are more like competition problems. My guess is that these problems were offered as challenge problems for extra credit, and that you are not required (and probably not expected) to submit solutions. Right?
January 1st, 2017, 03:09 PM   #14
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 Originally Posted by quasi That's what you would need to prove. As of now, you haven't proved it.
yes, I see the issue.

I'm beginning to wonder if the statement is true.

It seems to rely on rational square roots existing

January 1st, 2017, 03:20 PM   #15
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 Originally Posted by romsek yes, I see the issue. I'm beginning to wonder if the statement is true. It seems to rely on rational square roots existing
given $A = a + \dfrac 1 a + b + \dfrac 1 b$

the required product is

$A = \left(\sqrt{a b}+\dfrac{1}{\sqrt{a b}}\right)\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b }{a}}\right)$

and there is no guarantee that

$\sqrt{a b},~\sqrt{\dfrac{a}{b}} \in \mathbb{Q}^+$

addendum: we have to factor in the info that $A \in \mathbb{Z}^+$, I'll continue to look at it

Last edited by romsek; January 1st, 2017 at 03:23 PM.

 January 1st, 2017, 03:24 PM #16 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 The statement is true -- I can prove it.
January 1st, 2017, 03:56 PM   #17
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 Originally Posted by quasi The statement is true -- I can prove it.
some reason you haven't already done so?

January 1st, 2017, 04:45 PM   #18
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 Originally Posted by romsek some reason you haven't already done so?
The proof is elementary, but not a quick post -- no time right now.

Besides, my guess is that the problem is an extra credit problem for Rachel's class, so I'm not sure posting a full solution is warranted.

I also have an elementary proof of part (B):

"(B) Show that there are infinitely many positive integers $A$ that cannot be written as the sum of two elements of $R$."

But I don't have a proof of part (C):

"(C) Show that there are infinitely many positive integers $A$ that can be written as the sum of two elements of $R$."

In fact, unless I'm missing something, part (C) is quite hard, and will probably require advanced methods.

 January 1st, 2017, 05:40 PM #19 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 Ok, Rachel made it clear that the problems were from her math team class, assigned for fun, not for credit, so when I get a chance (sometime this week), I'll post my proofs for parts (A) and (B) (unless someone posts the same proof or a better one before me). Thanks from Joppy
January 2nd, 2017, 08:49 AM   #20
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 Originally Posted by rachel1234 Problem 1) A k-city is a k by k grid of positive integers so that every integer greater than 1 is the sum of one integer from the same row and one integer from the same column. Below is an example of a 3-city: 1 1 2 2 3 1 6 4 1 Part 1: Make a 5-city that has an integer that is greater than or equal to 150. Part 2: Show that for all k ≥ 1, the largest entry in a k-city is, at most, 3^ (k choose 2)
This is the way I would set it up (think about it):

Let $\displaystyle a_{ij}$ be a typical grid element.
Then $\displaystyle a_{ij}=a_{ik}+a_{mj}, \ k\neq i, m\neq j$

EDIT: Whoops. Just saw this as a separate problem. Replied to that too.

Last edited by zylo; January 2nd, 2017 at 08:54 AM.

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