December 15th, 2016, 05:07 PM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 155 Thanks: 5  Some Fun and Simple Questions About the Rationals
This is meant to be a bit of a fun brain teaser. It sure has me stumped. 1 Let w, x be any two rational numbers in $\mathbb{Q} \cap$(0, 1). 2 Where r is any irrational number > 0, let y, z be any two rational numbers in $\mathbb{Q} \cap$(r, 1+r). Question: Is it possible for there to exist k = w  x such that there does not exist a y  z = k? Question: Is it possible for there to exist k = y  z such that there does not exist a w  x = k? I believe the answer is "yes" to both, but in assuming I am wrong (I've made a few mistakes with this stuff recently), then perhaps consider two sets: 3 Let A = { w  x : x $\in \mathbb{Q} \cap$(0, 1) }, where w is a rational in $\mathbb{Q} \cap$(0, 1). 4 Let B = { y  z : z $\in \mathbb{Q} \cap$(r, r+1) }, where y is a rational in $\mathbb{Q} \cap$(r, 1+r). Question: Is it possible that A = B? If you answered “no” to the above two questions, then presumably your answer to this question must be yes. I believe the answer is no, but again assuming I am wrong and that A = B, then: 5 Let w $\sim$ y $\Rightarrow$ A = B. 6 Let m = y  w. The difference between any two rationals is also a rational, so m is rational. 7 Note that where w  x = y  z and y = m + w, in substituting for y we get w  x = m + w  z $\Rightarrow$ z = x + m $\land$ x = z  m. Each element of A takes the form w  x and each element of B takes the form y – z, so where w $\sim$ y, we can assert that there is a one to one relationship between each possible x and z where w  x = y  z such that $f$(x) = m + x is a bijection between them. 8 Let X = m + ( $\mathbb{Q} \cap$(0, 1) ) = { m + x : x $\in \mathbb{Q} \cap$(0, 1) } = $\mathbb{Q} \cap$(m, 1+m). 9 Note that where m + w = y, y is in X. Also, where the function $f$(x) = m + x from step 7 is a bijection between all possible x and z where w  x = y  z, we know that each possible value for z is included in X. 10 Notice that X contains all rational numbers on the interval (m, 1+m), but by definition it should also contain all rationals on the interval (r, 1+r) at this point. Clearly, something is wrong now because m is rational while r is irrational, and there are an infinite number of rationals between any rational and an irrational. Question: If you answered “no” to the first two questions, then do you think your answers may have been wrong? Question: Is this paradoxical and can anyone explain this? Question: Does this mean there exists a set C = { x : $\nexists$z where y  z = w  x } for any given y and w? Likewise, a set D = { z : $\nexists$x where y  z = w  x } for that same y and w? Last edited by AplanisTophet; December 15th, 2016 at 05:19 PM. 
December 15th, 2016, 06:02 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,343 Thanks: 2082 Math Focus: Mainly analysis and algebra 
I think the answer is "no" to both, and I'd prove it by selecting any rational$ $a$ and $b$ within the given interval and considering possible values of $xa$, $aw$, $zb$ and $by$.

December 15th, 2016, 06:42 PM  #3  
Senior Member Joined: Jun 2014 From: USA Posts: 155 Thanks: 5  Quote:
What about the remaining questions? This isn't supposed to be like a proof where you hit something you don't agree with and stop. Naturally, people should assert no to both of the initial questions. It's a brain teaser but I am in fact looking for legitimate answers to the questions.  
December 15th, 2016, 08:15 PM  #4 
Senior Member Joined: Jun 2014 From: USA Posts: 155 Thanks: 5 
I'm going to make a remarkable claim now, but I realize that such claims are seldom correct. This will be a very easy one to verify though so just take this with a crankery warning (I'm just experimenting). Here it is. While the cardinalities of the sets are of course the same, there is one and only one 'more' rational in (r, 1+r) than there is in (0, 1). To show it, I actually have to give a picture, so it's arts and crafts time and I'll upload a .pdf file in a moment. Grab your popcorn and get ready to potentially laugh at me for being an idiot out on a limb. 
December 15th, 2016, 08:32 PM  #5  
Senior Member Joined: Jun 2014 From: USA Posts: 155 Thanks: 5  Quote:
 
December 15th, 2016, 08:53 PM  #6  
Senior Member Joined: Feb 2016 From: Australia Posts: 725 Thanks: 271 Math Focus: Yet to find out.  Quote:
 
December 15th, 2016, 09:28 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,343 Thanks: 2082 Math Focus: Mainly analysis and algebra 
There was never any claim that $r \lt 1$. You might not have a bijection by adding $m$, but that doesn't mean that there is no bijection (which there is because the cardinalities are equal).

December 15th, 2016, 09:30 PM  #8  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,343 Thanks: 2082 Math Focus: Mainly analysis and algebra  Quote:
 
December 15th, 2016, 09:44 PM  #9  
Senior Member Joined: Aug 2012 Posts: 824 Thanks: 140  Quote:
Each element in $\mathbb Q \cap (r, 1 + r)$ is simply some element in $\mathbb Q \cap (0, 1)$ shifted $r$ units to the right. If $r = \pi$ then $(0,1)$ shifts to $(\pi, \pi + 1)$. Any two points $w, x \in (0,1)$ get shifted to $w + \pi, \ x + \pi \in (\pi, \pi + 1)$ respectively. Now $(w + \pi)  (x + \pi) = w  x = k$. So in fact $y = w + \pi$ and $z = x + \pi$ are two elements of $\mathbb Q \cap (r, 1 + r)$ that satisfy your requirement. You are just shifting an interval to the right (although r can be negative too and you can shift left). It's a principle of measure theory that shifting an interval (or any measurable set) does not alter its measure. Interval relationships between pairs of numbers would of course be preserved. A shift by an additive constant is an example of what in general is called an isometry, or rigid motion. Isometries intuitively preserve measure. It's isometries that are used to move the pieces around in the BanachTarski paradox. So what's the catch? Not every piece is measurable. Some sets are so weird they can't sensibly be assigned any measure at all. The Vitali set is one such. Now I see what you are trying to do. You hope that by shifting the unit interval by an irrational, some "rationals will fall off the end." Or maybe some irrationals will fall off. Am I right? Then when you intersect the shifted set with the rationals, you think you are sort of clipping off the lengths you need. But that doesn't happen. Even though you shifted all the rationals to irrationals, we only care about the DIFFERENCE between pairs of points; and that difference is invariant under rigid shifts of an interval. That's obvious both in terms of geometry and algebra. I hope this is helpful. The tl;dr is that when you shift an interval, the geometric and algebraic relationships between corresponding pairs of points are preserved. If you shift $w$ and $x$ by $r$, the distance between $w$ and $x$ must be the same as the distance between $w + r$ and $x + r$. Your second question simply involves shifting the second interval to the left; but it's the same argument. By letting $r$ be positive, negative, or zero the same proof covers both your questions. Shifting intervals in either direction preserves all geometric and algebraic relationship among corresponding pairs of points. I pray to Zeus that I have disabused you of this notion. First, there can't be "one more" of anything because the two sets are in bijection. Second, points don't fall off the real line when you shift them. You see that, right? Last edited by Maschke; December 15th, 2016 at 09:57 PM.  
December 15th, 2016, 09:45 PM  #10 
Senior Member Joined: Jun 2014 From: USA Posts: 155 Thanks: 5 
You need to read the OP, see my “remarkable claim,” and then view the .pdf containing the picture of my remarkable claim before reading this proof (all above). After all that, I would like this proof reviewed because I’m not sure if it is accurate (could again be crankery given my 'crazy' claim). 1 Let $q_1, q_2, q_3,$ … be an enumeration of $A = \mathbb{Q} \cap \mathbb{R}(0, 1)$. 2 For any $r \in \mathbb{R}(0, 1)$, let $B_r = \mathbb{Q} \cap \mathbb{R}(r, 1r)$. 3 For each $r \in \mathbb{R}(0,1) \cap \mathbb{Q}$, there exists one and only one $k_r \in A$ such that $f:A \rightarrow B_r$ is a bijection where $f(a) = a  k_r$. For each $r \in \mathbb{R}(0,1) \setminus \mathbb{Q}$, there exists two and only two elements $k_{r_1} \land k_{r_2}$ where $k_{r_1} < k_{r_2}$ such that $f:A \rightarrow B_r$ is a ‘near bijection’ where a ‘near bijection’ implies that one and only one element of $B_r$ is not covered and $f(a) = a  k_{r_i}$ where $k_{r_i}$ is either $k_{r_1}$ or $k_{r_2}$. Therefore, for each $r \in \mathbb{R}(0,1) \setminus \mathbb{Q}$, instead let $f:A \rightarrow B_r$ be a bijection where $f(a) = a – (k_{r_1} + k_{r_2})/2$. Note that $z:\mathbb{R}(0, 1) \rightarrow \mathbb{R}(r, 1r)$ is a bijection where $z(x) = x – r$. If $q \in A \land r \in \mathbb{R}(0,1) \setminus \mathbb{Q}$, then $z(q) = q – r \notin B_r$ because the difference between a rational number and an irrational number is not a rational number. Therefore, $f(q) \neq z(q) \Rightarrow (k_{r_1}+k_{r_2})/2 – r = t \land t \neq 0$. 4 If $r, s \in \mathbb{R}(0,1) \setminus \mathbb{Q}$, then $r \neq s \Rightarrow k_{r_1} \land k_{r_2} \neq k_{s_1} \land k_{s_2}$, respectively. 5 If $r, s \in \mathbb{R}(0,1) \setminus \mathbb{Q}$, then $k_{r_1} \land k_{r_2} \neq k_{s_1} \land k_{s_2}$, respectively $\Rightarrow r \neq s$. 6 Therefore, $r, s \in \mathbb{R}(0,1) \setminus \mathbb{Q} \land r \neq s \iff k_{r_1} \land k_{r_2} \neq k_{s_1} \land k_{s_2}$, respectively. 7 Let $g(k_{r_1}) = r$ for any $r \in \mathbb{R}(0, 1) \setminus \mathbb{Q}$. 8 Let $C = g(q_1), g(q_2), g(q_3),$ … = $\mathbb{R}(0,1) \setminus \mathbb{Q}$. The Continuum Hypothesis states “there is no set whose cardinality is strictly between that of the integers and the real numbers.” Where $C = \mathbb{R}(0, 1) \setminus \mathbb{Q}$, this statement is true because: $\mathbb{R} \setminus \mathbb{Q}$ = $\mathbb{R}(0,1)$ = $\mathbb{R}$ = $\mathbb{N}$ = $\mathbb{Z}$. 

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fun, questions, rationals, simple 
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