My Math Forum 3 Problems I am really stuck on

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 December 31st, 2016, 08:47 AM #1 Newbie   Joined: Dec 2016 From: New Jersey Posts: 7 Thanks: 1 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) Problem 2) Let there be the set R = {x + 1/x , where x ranges over all positive rational numbers} (A) Let A be a positive integer. Show that A is the sum of two elements of R if and only if A is the product of two elements of R. (B) Show that there are infinitely many positive integers A that cannot be written as the sum of two elements of R. (C) Show that there are infinitely many positive integers A that can be written as the sum of two elements of R. Problem 3) George, Harry, and Ethan are perfect logicians that always tell the truth. George decides to pose a puzzle to his friends: he tells them that the day of his birthday is at most the number of the month of his birthday. Then George announces that he will whisper the day of his birthday to Harry and the month of his birthday to Ethan, and he does exactly that. After George whispers to both of them, Harry thinks a bit, then says “Ethan cannot know what George's birthday is.” After that, Ethan thinks a bit, then says “Harry also cannot know what George's birthday is.” This exchange repeats, with Harry and Ethan speaking alternately and each saying the other can’t know George's birthday. However, at one point, Ethan instead announces “Harry and I can now know what George's birthday is. Interestingly, that was the longest conversation like that we could have possibly had before both figuring out George's birthday.” Find George's birthday. Last edited by skipjack; January 2nd, 2017 at 07:05 PM.
December 31st, 2016, 09:35 AM   #2
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 Originally Posted by rachel1234 Problem 2) Let there be the set R = {x + 1/x , where x ranges over all positive rational numbers} (A) Let A be a positive integer. Show that A is the sum of two elements of R if and only if A is the product of two elements of R.
a) assume $A = a + b,~a,b \in R$

let $a = x y + \dfrac {1}{x y}$

let $b = \dfrac x y + \dfrac y x$

$A = x y + \dfrac {1}{x y} + \dfrac x y + \dfrac y x$

$A = \left(x + \dfrac 1 x\right)\left(y + \dfrac 1 y\right)$

clearly $x + \dfrac 1 x \in R,~y + \dfrac 1 y \in R$

so $A$ is the product of two elements of $R$

now assume $A = x y,~x,y \in R$

$A = (a + 1/a)(b + 1/b) = ab + \dfrac a b + \dfrac b a + \dfrac{1}{ab}$

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

$x = ab + \dfrac {1}{ab}$

$y =\dfrac a b + \dfrac {1}{\frac b a}$

$x, y \in R$

$A = x + y$

Last edited by skipjack; January 1st, 2017 at 01:28 PM.

January 1st, 2017, 01:22 PM   #3
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 Originally Posted by romsek a) assume $A = a + b,~a,b \in R$ let $a = x y + \dfrac {1}{x y}$ let $b = \dfrac x y + \dfrac y x$
I don't see how you can justify the above "let"s.

Last edited by skipjack; January 1st, 2017 at 01:27 PM.

 January 1st, 2017, 01:26 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,954 Thanks: 1601 They are definitions.
January 1st, 2017, 01:27 PM   #5
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 Originally Posted by quasi I don't see how you can justify the above "let"s.
definition of the set $R$

January 1st, 2017, 01:45 PM   #6
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 Originally Posted by skipjack They are definitions.
In this part of the proof, the hypothesis is:

"A is a positive integer such that $A = a + b,~a,b \in R$."

So you can't define $a,b$.

If you assert "let $a$ = ..." and "let $b$ = ...", it must be the case that
the claimed forms for $a,b$ follow from the hypothesis.

That's what I don't see.

January 1st, 2017, 01:54 PM   #7
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 Originally Posted by romsek definition of the set $R$
$a,b$ are arbitrary elements of R.

So you can claim $a = u + 1/u$ and $b = v + 1/v$ for some positive rational numbers $u,v$, but I don't see how you can assert, without justification, that there exist positive rational numbers x,y such that $u = xy$ and $v =x/y$.

January 1st, 2017, 01:58 PM   #8
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 Originally Posted by quasi $a,b$ are arbitrary elements of R. So you can claim $a = u + 1/u$ and $b = v + 1/v$ for some positive rational numbers $u,v$, but I don't see how you can assert, without justification, that there exist positive rational numbers x,y such that $u = xy$ and $v =x/y$.
I'm saying pick $x,y \in \mathbb{Z}^+$

and form $a,b$

I'm not specifying the value of $A$, just that it's made up of the sum of two members of $R$

January 1st, 2017, 02:05 PM   #9
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 Originally Posted by romsek I'm saying pick $x,y \in \mathbb{Z}^+$ and form $a,b$ I'm not specifying the value of $A$, just that it's made up of the sum of two members of $R$
You asserted a particular form for $a,b$. I agree that if $a,b$ have that form, then $a,b \in R$. Are you claiming that all pairs $a,b \in R$ can be put in that form? I don't see that.

January 1st, 2017, 02:07 PM   #10
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 Originally Posted by quasi You asserted a particular form for $a,b$. I agree that if $a,b$ have that form, then $a,b \in R$. Are you claiming that all pairs $a,b \in R$ can be put in that form? I don't see that.
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 $R$

Last edited by romsek; January 1st, 2017 at 02:09 PM.

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