number of rational point on circumference of a circle. (1 Viewer)

juantheron · Nov 1, 2015

$Prove that the number of rational points on the Circumference of a circle$

$$having center $(\pi,e)$ is atmost $1$.$

$Given a point $a$ and $b$ is rational ,if $a$ and $b$ both are rational number$

Sy123 · Nov 1, 2015

$\\ $Suppose that there exists two points on the circle that are rational, let these points be$ \ P_1(p_1, q_1) \ $and$ \ P_2(p_2,q_2) \ $where$ \ p_1,p_2,q_1,q_2 \ $are rational$$

$\\ $The points lie on the circle$ \ (x-\pi )^2 + (y-e)^2 = r^2$

$\\ \Rightarrow \ [1]: \ (p_1 - \pi)^2 + (q_1 - e)^2 = r^2 \\ \\ \Rightarrow \ [2]: \ (p_2 - \pi)^2 + (q_2 - e)^2 = r^2$

$\\ \Rightarrow \ [1] - [2]: \ (p_1-p_2)(p_1 + p_2 - 2\pi) + (q_1 - q_2)(q_1 + q_2 - 2e) = 0$

$\\ \Rightarrow \ (p_1^2 + q_2^2) - (p_2^2 + q_2^2) = 2( \pi(p_1 - p_2) + e(q_1 - q_2))$

$\\ $So$ \ R_1 = \pi r_1 + e r_2 \ $where$ \ R_1 \ $and$ \ r_1, r_2 \ $are rational$$

$\\ $A problem arises, if one can indeed give a proof that shows that there cannot be two rationals lying on the circle, then one is saying that the number$ \ R_1 \ $cannot be rational, meaning it is irrational, meaning one has proven that$ \ r_1 \pi + r_2 e \ $cannot be rational for any$ \ r_1 \ $and$ \ r_2$

$\\ $However, it is known that knowing whether$ \ \pi + e \ $is irrational or not is an unsolved problem in mathematics, and so a proof for the proposition given in the question would reduce to a proof that$ \ \pi + e \ $is irrational, which is not heuristically possible for this maths forum$$

InteGrand · Nov 1, 2015

Sy123 said:
$\\ \Rightarrow \ (p_1^2 + q_2^2) - (p_2^2 + q_2^2) = 2( \pi(p_1 - p_2) + e(q_1 - q_2))$

[...]

$\\ $A problem arises, if one can indeed give a proof that shows that there cannot be two rationals lying on the circle, then one is saying that the number$ \ R_1 \ $cannot be rational, meaning it is irrational, meaning one has proven that$ \ r_1 \pi + r_2 e \ $cannot be rational for any$ \ r_1 \ $and$ \ r_2$

$If one did give the proof that there cannot exist two distinct rational points, wouldn't one be showing that $p_1 = p_2$ and $q_1 = q_2$? This would then make the equation quoted above say that $0=0$, and so there's no problem involving needing to know about the irrationality of $\pi+e$.$

Sy123 · Nov 1, 2015

InteGrand said:
$If one did give the proof that there cannot exist two distinct rational points, wouldn't one be showing that $p_1 = p_2$ and $q_1 = q_2$? This would then make the equation quoted above say that $0=0$, and so there's no problem involving needing to know about the irrationality of $\pi+e$.$

Yep my bad, didn't give too much thought to it

glittergal96 · Nov 3, 2015

A proof of this claim would also provide a proof of the algebraic independence of pi and e over Q.

This is an open problem.

Paradoxica · Nov 3, 2015

glittergal96 said:
A proof of this claim would also provide a proof of the algebraic independence of pi and e over Q.

This is an open problem.

Which would probably net you a fields medal automagically, seeing as you'd be well within the age limit. Furthermore, if you were still in high school, dozens of universities would probably be offering you scholarships for research in pure mathematics.

Paradoxica · Nov 3, 2015

InteGrand said:
$If one did give the proof that there cannot exist two distinct rational points, wouldn't one be showing that $p_1 = p_2$ and $q_1 = q_2$? This would then make the equation quoted above say that $0=0$, and so there's no problem involving needing to know about the irrationality of $\pi+e$.$

$Yes, it is obvious upon inspection that $p_1 = p_2, q_1 = q_2$ is a solution to the conjecture. But proving that it is the only solution is another ordeal.$

lita1000 · Nov 3, 2015

Paradoxica said:
$Yes, it is obvious upon inspection that $p_1 = p_2, q_1 = q_2$ is a solution to the conjecture. But proving that it is the only solution is another ordeal.$

Integrand never said anything about if p1=p2 and q1=q2, then LHS = RHS (from sy's post). He's saying that if LHS=RHS then p1=p2, and q1=q2. Search up if and only if type arguments on wiki for details on this. Your second line makes it seem like he wasn't talking about this.

You have completely misunderstood Integrand's post.

Don't be a hayabusaboston btw, using words like "ordeal" won't make you seem any smarter.

Paradoxica · Nov 3, 2015

lita1000 said:
Integrand never said anything about if p1=p2 and q1=q2, then LHS = RHS (from sy's post). He's saying that if LHS=RHS then p1=p2, and q1=q2. Search up if and only if type arguments on wiki for details on this. Your second line makes it seem like he wasn't talking about this.

You have completely misunderstood Integrand's post.

Don't be a hayabusaboston btw, using words like "ordeal" won't make you seem any smarter.

well then. you certainly wouldn't like me because that's how I converse. Usually. If the mood is correct, I will talk like a 7-year old with a very limited set of words.

dan964 · Nov 3, 2015

Is this solution valid?

Clarifications
1. "It follows therefore that the change in x must be positive, if x1, x2 are both rational.": This line is ignored, as it is an error and kinda of irrelevant, forgot to remove;

2. Q is the distance BC squared = (x2-x1)^2; I could have put it in earlier. If the assumption is true, it has to be rational. (I have to deal with the square of the distance because the distance may be irrational square-root)
3. pi/e is clearly irrational, since neither have any common factors. The RHS is clearly rational; as the numerator and denominator a

Paradoxica · Nov 3, 2015

dan964 said:
Is this solution valid?

Clarifications
1. "It follows therefore that the change in x must be positive, if x1, x2 are both rational.": This line is ignored, as it is an error and kinda of irrelevant, forgot to remove;

2. Q is the distance BC squared = (x2-x1)^2; I could have put it in earlier. If the assumption is true, it has to be rational. (I have to deal with the square of the distance because the distance may be irrational square-root)
3. pi/e is clearly irrational, since neither have any common factors. The RHS is clearly rational; as the numerator and denominator a

http://www2.math.ou.edu/~jalbert/courses/openprob2.pdf
Refer to problem 22.
Sorry, but if you have a proof of the irrationality of pi/e, then you will have solved a massive open problem. The proof could also lead to the proof of irrationality of the other operations involving pi and e. It is not known whether or not pi is a rational multiple of e. Therefore, your claim is invalid.

InteGrand · Nov 3, 2015

dan964 said:
3. pi/e is clearly irrational, since neither have any common factors.

$What do you mean ``common factors''? Factors are defined for integers only. It does not seem clear at all to me that $\frac{\pi}{\mathrm{e}}$ is irrational. In fact, it is an open problem according to the link posted above by Paradoxica.$

dan964 · Nov 4, 2015

InteGrand said:
$What do you mean ``common factors''? Factors are defined for integers only. It does not seem clear at all to me that $\frac{\pi}{\mathrm{e}}$ is irrational. In fact, it is an open problem according to the link posted above by Paradoxica.$

right-o

Paradoxica · Nov 4, 2015

dan964 said:
right-o

And for future note, obvious is not the same as true.

number of rational point on circumference of a circle. (1 Viewer)

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