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Anonymous Wed 17 Jun 06:09:05 2020 No.11807269 View Reply Original Report

Quoted By: >>11807280 >>11807480 >>11807658 >>11807887

Can you prove this problem?

Is tan1° irrational number or rational number?

from entrance examination of Kyoto Uni.

Anonymous

Anonymous Wed 17 Jun 2020 06:17:04 No.11807280 Report

Quoted By: >>11807288 >>11807480

>>11807269
yes, irrational. by tangent square identity, tan is root (sec^2 - 1). Now we look at at sec1, or 1/cos1. After a somewhat involved proof involving algebraic geometry, we find that cos1 is not even constructible, let alone rational. Thus, inverting it, squaring it, subtracting 1, then rooting it, we are still left with a number that is not constructible, and thus must be irrational.

Anonymous

Anonymous Wed 17 Jun 2020 06:20:13 No.11807288 Report

Quoted By: >>11807303

>>11807280
it should be noted that your entrance exam is probably looking for a simpler proof, and using upper division math would look suspicious.

Anonymous

Anonymous Wed 17 Jun 2020 06:26:24 No.11807303 Report

Quoted By: >>11807309

>>11807288
>using upper division math would look suspicious
In Japan the nail that sticks out gets hammered down

Anonymous

Anonymous Wed 17 Jun 2020 06:29:19 No.11807309 Report

Quoted By:

>>11807303
hey bud, I just do math.

Anonymous

Anonymous Wed 17 Jun 2020 06:59:28 No.11807362 Report

Quoted By: >>11807493 >>11807505 >>11807521 >>11809025 >>11809043

Ok I figured it out. If a tangent of some angle theta is rational, then the tangent of any multiple of the angle must be rational too.

This follows since we can write
$e^{i\theta}=r(a+ib)$
where a and b are integers (and r need not be rational), and then if we take any integer power the real and imaginary parts (disregarding the common factor involving r) will just be some polynomial in a and b and thus also integers.

But tan of 30 is obviously irrational so that means tan of 1 can't be rational.

Anonymous

Anonymous Wed 17 Jun 2020 07:58:47 No.11807480 Report

Quoted By:

>>11807269
Similar argument to >>11807280
2sin(1°) is the sum of e^+-i*2pi/360, so it's an algebraic integer. If sin(1°) is rational, then 2sin(1°) is rational too, but the only rational algebraic integers are integers and simple computation will show you that 2*sin(1°) is not an integer. So 2sin(1°) is not rational and neither is sin(1°).

I have no fucking clue how high school students are supposed to know this and I wouldn't be able to solve this question from memory. Really mean question for an entrance exam.

Anonymous

Anonymous Wed 17 Jun 2020 08:06:27 No.11807493 Report

Quoted By: >>11807516

>>11807362
Thanks!!! I solved in your opinion!!!

Anonymous

Anonymous Wed 17 Jun 2020 08:08:47 No.11807499 Report

Quoted By:

$\tan (1^o) = \tan (\frac{\pi}{180}) \approx \frac{\pi}{180}$
Looks irrational to me.

t. engineer

Anonymous

Anonymous Wed 17 Jun 2020 08:13:55 No.11807505 Report

Quoted By: >>11807512

>>11807362
>where a and b are integers
for z = a + ib, a and b are by definition real, if you at the start assume that they are integers you don't even need to take powers because, also by definition, a = cos and b = sin then tan = sin/cos is rational. but your assumption wrong

Anonymous

Anonymous Wed 17 Jun 2020 08:17:18 No.11807512 Report

Quoted By: >>11807643 >>11807648

>>11807505
You misunderstood the logic of the proof completely. It's a proof by contradiction

Anonymous

Anonymous Wed 17 Jun 2020 08:18:28 No.11807516 Report

Quoted By:

>>11807493
Oh damn, I hope you are not actually cheating on your entrance exam lol

Anonymous

Anonymous Wed 17 Jun 2020 08:21:37 No.11807521 Report

Quoted By: >>11807530

>>11807362
But tan of 45 is rational, wouldn't that imply that tan of 1 is rational too? I don't think I understand your argument.

Anonymous

Anonymous Wed 17 Jun 2020 08:26:09 No.11807530 Report

Quoted By:

>>11807521
I showed that if tan theta is rational then any integer multiple n*theta also has a rational tangent. You can't divide by n, you have to multiply

Anonymous

Anonymous Wed 17 Jun 2020 08:37:15 No.11807546 Report

Quoted By: >>11807555 >>11809024 >>11809046

Suppose $\tan 1^{\circ}$ is rational. We know that $0<\tan 1^{\circ}<1$ .
Suppose $\tan k^{\circ}$ is rational for some natural number $k<45$ . We know that $0<\tan k^{\circ}<1$ .
Then $\tan (k^{\circ}+1^{\circ})=\dfrac{\tan k^{\circ}+\tan 1^{\circ}}{1-\tan k^{\circ}\tan 1^{\circ}}$ is rational
because $\tan 1^{\circ}$ and $\tan k^{\circ}$ are both rational and $0<\tan k^{\circ}\tan 1^{\circ}<1$ .
By induction, $\tan n^{\circ}$ is rational for all natural numbers $n\le45$ .
But $\tan 30^{\circ}=\dfrac{\sqrt{3}}{3}$ is irrational, which is a contradiction. So $\tan 1^{\circ}$ is irrational.

Anonymous

Anonymous Wed 17 Jun 2020 08:40:51 No.11807555 Report

Quoted By:

>>11807546
this is a good solution

Anonymous

Anonymous Wed 17 Jun 2020 09:18:16 No.11807643 Report

Quoted By:

>>11807512
so if for theta sin/cos is rational then sin and cos are integers, and by the polar representation 30*theta would yield a a polynomial such that the ratio of it's real to imaginary part would be tan(30*theta) and since both the real and imaginary parts are multiples of an integer it follows that tan(30*theta) is also rational. but since we know that tan(30*theta) is irrational then tan(theta) can't be rational.

Anonymous

Anonymous Wed 17 Jun 2020 09:20:27 No.11807648 Report

Quoted By:

>>11807512
so if for theta tan(theta) is rational then a and b are integers, and by the polar representation 30*theta would yield a a polynomial such that the ratio of it's real to imaginary part would be tan(30*theta) and since both the real and imaginary parts are multiples of an integer it follows that tan(30*theta) is also rational. but since we know that tan(30*theta) is irrational then tan(theta) can't be rational.

Anonymous

Anonymous Wed 17 Jun 2020 09:23:48 No.11807658 Report

Quoted By:

>>11807269
Likely yes, because there are a lot more irrational numbers than rational numbers
>t. engineer
>That'll be $35 for my hourly consultation

Anonymous

Anonymous Wed 17 Jun 2020 11:48:02 No.11807887 Report

Quoted By:

>>11807269
tannu onuu isa illationar desu ne

Anonymous

Anonymous Wed 17 Jun 2020 18:58:18 No.11809024 Report

Quoted By: >>11809025

>>11807546
Best one itt

Anonymous

Anonymous Wed 17 Jun 2020 18:59:16 No.11809025 Report

Quoted By:

>>11809024
It's the same as this one
>>11807362

Anonymous

Anonymous Wed 17 Jun 2020 19:06:44 No.11809043 Report

Quoted By: >>11809045

>>11807362
You can't write a general complex number in that form.

Anonymous

Anonymous Wed 17 Jun 2020 19:07:35 No.11809045 Report

Quoted By: >>11809049 >>11809050

>>11809043
It's not a general one. It has a rational. Ffs the brainlets in this thread...

Anonymous

Anonymous Wed 17 Jun 2020 19:07:46 No.11809046 Report

Quoted By:

>>11807546
Based

Anonymous

Anonymous Wed 17 Jun 2020 19:08:36 No.11809049 Report

Quoted By:

>>11809045
It's phase has a rational tangent I mean

Anonymous

Anonymous Wed 17 Jun 2020 19:08:47 No.11809050 Report

Quoted By:

>>11809045
What's a rational number

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