$\pi$ is not transcendental? Say what?

Logic Level 3

Which step did I commit a fallacy?

Step 1 :

With my calculator, I get $\sin \left (2017 \sqrt[5]{2} \right) = -1 = \sin \left (-\frac {\pi}{2} \right )$

Step 2 :

Because the $f(x) =\sin(x)$ has period $2\pi$ ,

$2017 \sqrt[5]{2} = -\frac {\pi}{2} + 2\pi n$ , for a certain integer $n$

Step 3 :

Rearrange the equation and set $\pi$ as the subject

$\LARGE \pi = \frac {2017 \cdot 2^{6/5} }{4n - 1}$

This means $\pi$ is not a transcendental number but an algebraic number.

Step 2 Step 3 Step 1 No fallacy committed. Math is weird.

8 solutions

Trevor B.
Mar 8, 2014

If you try to find $\sin2017\sqrt[5]{2}$ with a calculator or WolframAlpha, you will see that it is equal to $-1.$ However, if it is in fact equal to $-1,$ then it is equal to $2k\pi+\frac{3\pi}{2}.$ This value is always divisible by $\frac{\pi}{2},$ so you would expect $\frac{2017\sqrt[5]{2}}{\frac{\pi}{2}}$ to be an integer. However, you will find that is equal to $1475.000000004168...$ which is not an integer. Therefore, $2017\sqrt[5]{2}\neq\frac{\pi}{2}k$ for integer $k.$ $\sin2017\sqrt[5]{2}$ is very $\textit{very}$ close to $-1,$ but it is not equal to $-1$ exactly. The mistake is in $\boxed{\text{Step 1}}$

If you try to find $\sin 2017 \sqrt[5]{2}$ with [...] WolframAlpha, you will see that it is equal to $-1$

I'd say otherwise. But otherwise correct.

Ivan Koswara - 7 years, 3 months ago

This is what I tried.

Trevor B. - 7 years, 2 months ago

...click "More digits", of course...

Ivan Koswara - 7 years, 2 months ago

@Ivan Koswara – Click radian mode, and I see you got -1.000... but I typed in 1/5 as the power, and I got a lot of digits... weird...

Kevin Shen - 7 years, 2 months ago

It is also nice to know that according to wolfram the number $\sin(2017 \times 2^{1/5})$ is also a transcendental number.

Lucas Tell Marchi - 7 years, 2 months ago

True that, but it must also consider that the calculator do not count every decimal point on π, therefore the radian that was put in is concidentally the same with π. I have checked with a calculator that the number is actually the same with π up to the 11 decimal points. The calculator might only count up to 11 or up to 10, which makes the calculator mistakes it as a π.

Jeremy Intan - 7 years, 2 months ago

That's a tricky error... I would have never realized.

Harshal Sheth - 7 years, 3 months ago

You should have been dubious when $\sin x$ was equal to $-1$ with an argument that was not some variation of $\pi$ .

Michael Tong - 7 years, 3 months ago

skeptical, not dubious

Cody Johnson - 7 years, 2 months ago

because calculators also do approimation after a certain place after point

Adarsh Pandey - 7 years, 2 months ago

sin(-pi/2) = zero, not -pi/2, I believe that is also wrong.

Gurvir Singh - 7 years, 2 months ago

SIN(THETA) = SIN(PHI) DOES IT IMPLY THAT THETA EQUALS PHI

Sheshgiri Prabhu - 7 years, 2 months ago

Nice shouting skills, pal.

Jake Lai - 5 years, 11 months ago

Ivan Koswara
Mar 8, 2014

$\sin (2017 \sqrt[5]{2})$ only approximates $-1$ , not exactly equal to $-1$ . Indeed, a trip to WolframAlpha tells that it differs on the 17th digit. Most calculators, even floating point types on programming languages (64-bit floating data types), can only carry about 16 decimal digits, so the result is rounded down to $-1$ .

In any case, "with my calculator" is not a rigorous step, so Step 1 is incorrect.

रिघत यौ अरे

San Malu - 7 years, 2 months ago

Chew-Seong Cheong
May 4, 2015

Logically Step 1 is false, otherwise it will be taught in schools and Brilliantant Pi Han Goh would have received his Nobel prize for this amazing discovery.

BEST ANSWER RIGHT HERE! +1

Pi Han Goh - 6 years, 1 month ago

LOL...nice answer @Chew-Seong Cheong !!!

Noel Lo - 5 years, 10 months ago

Mayyank Garg
Mar 13, 2014

sin of the measure given = 0.3919424427...... which contradicts given info...so step 1

Your calculator is configured to degrees. We are talking about radians here.

Lucas Tell Marchi - 7 years, 2 months ago

Ok...thnks dude

MAYYANK GARG - 7 years, 2 months ago

Milly Choochoo
Mar 8, 2014

Assuming you've read the other solutions, we can now come to a general consensus.

Transcendental functions can only produce algebraic results when taking transcendental arguments.

*A transcendental number is basically a number that can't be made using rational numbers and basic algebra. Logarithms, exponentials with e (2.718...), and trigonometric functions aren't basic algebra. They are transcendental functions.

Noel Lo
Jul 30, 2015

$\boxed{Step 1}$ is wrong as it is NOT an exact equation, it is only an approximation.

Min-woo Lee
Dec 31, 2014

Trigonometric function is a transcedental function. 2017*2^(1.5) is an algebraic number. Therefore the statement in step 1 is wrong.

Eugene Sebilo
Mar 22, 2014

Step 1 : Tried with a calculator. $ 2017\sqrt [ 5 ]{ 2 }$ is not equal to $\sin ^{ -1 }{ (1) } ={ 90 }^{ \circ }=0.5\Pi $ . Condition were set as to what is the fallacy... so choosing only one answer (since it should be picked only once) leads to.... you know.

π \pi π is not transcendental? Say what?

8 solutions

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$\pi$ is not transcendental? Say what?