My Math Forum Rational Points on a circle

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June 22nd, 2012, 10:05 AM   #11
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Re: Rational Points on a circle

Quote:
 Originally Posted by ghostwalker For some reaons I can't edit - nor find what the minimum number of posts is in order to do so: Ho, hum. .
That information is top secret! Also, I think it was recently doubled from ten. Shhhhh

June 22nd, 2012, 10:08 AM   #12
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Re: Rational Points on a circle

Quote:
Originally Posted by The Chaz
Quote:
 Originally Posted by ghostwalker For some reaons I can't edit - nor find what the minimum number of posts is in order to do so: Ho, hum. .
That information is top secret! <deleted: top secret.>
"I won't let on", he said, gradually building up his number of posts.

 June 22nd, 2012, 10:14 AM #13 Member   Joined: Oct 2011 Posts: 81 Thanks: 0 Re: Rational Points on a circle Let's assume for eg. there are only two rational points on the circle, so the chord passing through them will have a rational slope and perpendicular bisector from centre will also have rational slope. But in reality, perpendicular bisector has irrational slope, that is a contradiction, so (assumption was wrong) even two points on a circle can't be rational.(right?)
June 22nd, 2012, 10:31 AM   #14
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Re: Rational Points on a circle

Quote:
 Originally Posted by guru123 Let's assume for eg. there are only two rational points on the circle, so the chord passing through them will have a rational slope and perpendicular bisector from centre will also have rational slope. But in reality, perpendicular bisector has irrational slope, that is a contradiction, so (assumption was wrong) even two points on a circle can't be rational.(right?)
[color=purple]That's the gist of it.[/color]

I have rather a lot of caveats with the wording and am making the odd assumption when reading that. But to go into them would deflect from the main point of the post; and actually serve no purpose. An elementary introduction to logic, or propositional calculus, would be a useful read, and cover all you need I think, if you feel so inclined.

 June 22nd, 2012, 11:32 AM #15 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Rational Points on a circle let $(x_0,y_0)$ be a rational point, that is $x_0 ,y_0 \in \mathbb Q$ Make a circle with center$(0, sqrt 2)$ passing this point, and assume it should pass some other rational point: $(x,y)$ $x,y \in \mathbb Q$ We know, $x_0^2+$$y_0-sqrt 2$$^2=x^2+$$y-sqrt 2$$^2\text{ (1)}$ Rearrange (1), we obtain: $(x-x_0)(x+x_0)=(y_0-y)(y+y_0-2sqrt 2)\text{ (2)}$ We know the truth that: $\text{rational} +\text{rational}= \text{rational}$ $\text{rational} \times\text{rational}= \text{rational}$ $\text{rational} +\text{irrational}= \text{irrational}$ $\text{rational} \times \text{(infinite non-repeating decimal such as}sqrt 2)=\text{rational} \text{ iff. the rational =0}$ So the left of (2) must be rational, but the factor $(y+y_0-2sqrt 2)$ in the right should be irrational. In order to make the right of (2) be a rational (and make left=right), you should let $y_0=y$, that will result in right=0 and left =0 too. Therefore $x=x_0$ or $x=-x_0$, or the circle passes two rational points at most: $(x_0,y_0)$ and $(-x_0,y_0)$.
 June 22nd, 2012, 12:51 PM #16 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Rational Points on a circle Q2 It is true while the fixed line is the perpendicular bisector to some segment, because any point on the fixed line has equidistance from both endpoints of the segment. When one end point of the segment is a fixed point, the other end point is the other fixed point in this case.
June 22nd, 2012, 01:25 PM   #17
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Re: Rational Points on a circle

Quote:
 Originally Posted by stainburg let $(x_0,y_0)$ be a rational point....
Nice proof.

June 22nd, 2012, 01:37 PM   #18
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Re: Rational Points on a circle

Simple proof bout this:
Quote:
 Originally Posted by stainburg $\text{rational} +\text{rational}= \text{rational}$ $\text{rational} \times\text{rational}= \text{rational}$ $\text{rational} +\text{irrational}= \text{irrational}$ $\text{rational} \times \text{(infinite non-repeating decimal such as}sqrt 2)=\text{rational} \text{ iff. the rational =0}$
We know a rational number $A$ can be formed as $A=\frac{m_A}{n_A}$. $m_A,n_A \in \mathbb Z$, $n_A\neq 0$

Now define another rational number $B=\frac{m_B}{n_B}$. $m_B,n_B \in \mathbb Z$, $n_B\neq 0$.

So

a) $A+B=\frac{m_A}{n_A}+\frac{m_B}{n_B}=\frac{m_An_B+m _Bn_A}{n_An_B}$ , since $(m_An_B+m_Bn_A), (n_An_B)\in \mathbb Z$, $A+B \in \mathbb Q$.

b) $A\times B=\frac{m_A}{n_A}\times \frac{m_B}{n_B}=\frac{m_Am_B}{n_An_B}$ , since $(m_Am_B), (n_An_B)\in \mathbb Z$, $A\times B \in \mathbb Q$.

c)Suppose $C\in \mathbb R-Q$, if $A+C\in \mathbb Q$, $D=A+C=\frac{m_D}{n_D}$. $m_D,n_D \in \mathbb Z$, $n_D\neq 0$. But according to a), we know, $C=D-A\in \mathbb Q$, that is contradiction of $C\in \mathbb R-Q$. So $D \in \mathbb R-Q$ too.

d) Similar as c), let $A\in \mathbb Q$, $C\in\mathbb R-Q$, if $B=A\times C\in \mathbb Q$, $A=B=0$

 June 22nd, 2012, 01:38 PM #19 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Rational Points on a circle Thanks, ghostwalker!

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# How to find maximum rational points on a circle with irrational centre

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