Why is Pi Irrational for Circles?

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The discussion centers on the irrationality of Pi (π) and its classification as a transcendental number, contrasting it with the algebraic nature of the square root of two (√2). Participants assert that Pi cannot be expressed as a ratio of integers (p/q), primarily because the circumference of a circle cannot be accurately represented by the diameter using a finite polygonal approximation. The conversation also touches on the geometric implications of irrational numbers and the distinction between algebraic and transcendental numbers, emphasizing that while √2 is algebraic due to its derivation from a polynomial equation, Pi's transcendental nature prevents it from fitting into any such equation.

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We cannot put the ratio of circumference/diameter in the form p/q. In this case the circumference. Because any number of sides of a regular polygon perimeter to calculate the circumference will not fit to the circumference of a circle.

That is the number of sides of a polygon tends to infinite.

Is this the reason pi is not rational, for obvious reason the arc of a circle cannot be in proportion to any of the straight lines that make up the circle.
 
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rajeshmarndi said:
We cannot put the ratio of circumference/diameter in the form p/q. In this case the circumference. Because any number of sides of a regular polygon perimeter to calculate the circumference will not fit to the circumference of a circle.

That is the number of sides of a polygon tends to infinite.

Is this the reason pi is not rational, for obvious reason the arc of a circle cannot be in proportion to any of the straight lines that make up the circle.

That's kind of interesting. Why do you figure the square root of two is irrational? it is a straight line, the Pythagorean hypotenuse of a right triangle with sides of unity.
 
DiracPool said:
Why do you figure the square root of two is irrational?

Ok. Nice one.

√2 can be put in an equation i.e x^2 - 2 = 0 . That is , it is an algebric number and not a transcendental.

∴ like I said, pi cannot be deduced into any form or equation. It is transcendental.
 
rajeshmarndi said:
Ok. Nice one.

√2 can be put in an equation i.e x^2 - 2 = 0 . That is , it is an algebric number and not a transcendental.

∴ like I said, pi cannot be deduced into any form or equation. It is transcendental.

You haven't said that in your initial post. You haven't used the word transcendental, nor have you proved that pi is transcendental.

Edit: More importantly, you need to give a method of translating algebraic concepts (eg transcendental) into geometric ones.
 
Last edited:
pwsnafu said:
You haven't said that in your initial post. You haven't used the word transcendental, nor have you proved that pi is transcendental.

I am not proving anything here, I just want to be clear what makes pi an irrational and non-repeating non-terminating number. What Diracpool mentioned very much clear the distinction between algebraic and transcendental.

Have I understood right if I say, √2 is algebraic because it is a side of a right angle triangle and we have formula for a right angle triangle and since Pi do not fit as an exact ratio (because circumference measurement cannot be calculated on the basis of diameter) ,that is why it is transcendental.

In this sense it is understandable why pi is irrational but now seem difficult to think(logically) why √2 is irrational.
 
rajeshmarndi said:
We cannot put the ratio of circumference/diameter in the form p/q. In this case the circumference. Because any number of sides of a regular polygon perimeter to calculate the circumference will not fit to the circumference of a circle.

That is the number of sides of a polygon tends to infinite.

Is this the reason pi is not rational, for obvious reason the arc of a circle cannot be in proportion to any of the straight lines that make up the circle.
No, "straight lines" and "circle" have nothing to do with being "rational" or "irrational".

Pi do not fit as an exact ratio (because circumference measurement cannot be calculated on the basis of diameter)
That is certainly not true. C= \pi D is perfectly good formula for circumference as a function of diameter. And if you say it is not because \pi is irrational, you are arguing in circles.
 
rajeshmarndi said:
Have I understood right if I say, √2 is algebraic because it is a side of a right angle triangle and we have formula for a right angle triangle and since Pi do not fit as an exact ratio (because circumference measurement cannot be calculated on the basis of diameter) ,that is why it is transcendental.

No. √2 is algebraic because it is the root of x2-2. That's it. It has nothing to do with geometry. It's an algebraic property.

What is true (from a geometric perspective) is that it is constructable, and every constructable number is algebraic. Notice that the other direction is false: trisecting an angle requires cube roots which are algebraic but not constructable.

But none of this has nothing to do with rationality.

Also, radians are defined using the arclength of the unit circle. Your claim "circumference measurement cannot be calculated" means we can't measure angles. Which is of course not true.
 
If I take any slice of the circle, it has the exact same features of the circle that you purport to use (you can approximate it with linear segments but it's never exactly equal), but there is a slice of the unit circle whose length is exactly 1
 
rajeshmarndi said:
We cannot put the ratio of circumference/diameter in the form p/q. In this case the circumference. Because any number of sides of a regular polygon perimeter to calculate the circumference will not fit to the circumference of a circle.

That is the number of sides of a polygon tends to infinite.

Is this the reason pi is not rational, for obvious reason the arc of a circle cannot be in proportion to any of the straight lines that make up the circle.

take a piece of string of any length and curl it up into a circle. It is length is rational then so is the curcumference of the circle.
 
  • #10
lavinia said:
take a piece of string of any length and curl it up into a circle. It is length is rational then so is the curcumference of the circle.

In which case the radius/diameter will not be. SOMEBODY'S got to be irrational here :smile:
 
  • #11
phinds said:
In which case the radius/diameter will not be. SOMEBODY'S got to be irrational here :smile:

Think it about it, if you are holding a sphere, then irrationality is in the Pi of the beholder. Or the beholden, however you want to look at it.
 
  • #12
diracpool said:
think it about it, if you are holding a sphere, then irrationality is in the pi of the beholder. Or the beholden, however you want to look at it.

ouch !
 
  • #13
HallsofIvy said:
... That is certainly not true. C= \pi D is perfectly good formula for circumference as a function of diameter. And if you say it is not because \pi is irrational, you are arguing in circles.

I really hope that pun was intentional.
 
  • #14
AnTiFreeze3 said:
I really hope that pun was intentional.
So do I. But that was four days ago and now I can't remember!
 
  • #15
Hey guys, go easy on him :)
 
  • #16
PROOF: Assume that √2 is a rational number, meaning that there exists an integer a and an integer b in general such that a/b=√2
Now we assume that a/b is an irreducible fraction.
So if now it holds that:
(a^2)/(b^2)=2
Solving for a^2 we get:
a^2=2b^2
Therefore a^2 is even because it is a multiple of 2. It follows that a must be even (as squares of odd integers are never even).
Because a is even, there exists an integer k that fulfills: a=2k

Substituting 2k back into the equation we get
2b^2=(2k)^2=4k^2
Devide both sides by 2 we get
b^2=2k^2.
So using the same logic as above we see that b is a multiple of 2.

So now we have an irreducible fraction of a/b where both a and b are multiples of 2. But how can we have a irreducible yet reducible fraction? We cant. Therefore we must conclude our original assumption that √2 is rational must be false.
 
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  • #17
Yes, that is the standard proof that \pi is irrational- but it has nothing to do with the topic of this thread, whether \pi is rational or not.
 
  • #18
My first thought is that its irrational because of the base ten system. If we used a base pi system it would not be irrational, but other numbers may be. I am not trained in math... but it seems to me that rational vs irrational is completely dependent on the base system used. Is this ridiculous?
 
  • #19
That's a really good point, modus
 
  • #20
I think that the rationality of a number only considers its expression as a ratio in base 10. (My gut tells me that anything rational in base 10 is also rational in any other rational base- if this is the case, replace the "base 10" in the above sentence with "any rational base".) If that's the definition of rationality, then 1 in base ##\pi## is still irrational
 
  • #21
For radius 1/(2pi), the circumference of a circle is 1, rational, but the radius is irrational. The radius and circumference are incommensurable, can't both be measured in any unit so that one is a multiple of or proportional to the other. Am I on the right track?
 
  • #22
That reasoning is a little circular. You concluded that the radius and circumference can't have a common multiple with the implication that their ratio (2pi) is irrational. However, you justified this with the fact that the radius was irrational, which you couldn't know unless you already knew that pi was irrational, which is what you're trying to show (see the problem?)
 
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  • #23
We know π is irrational because:
e^z (z being a+bi) will always irrational for any rational a or b,
So since e^iπ=-1 because we get a rational result either π or i must be irrational, and since we know i is not, π must be irrational!
 
  • #24
cmcraes said:
We know π is irrational because:
e^z (z being a+bi) will always irrational for any rational a or b

Yeah, but it seems like you haven't really done anything except push the question of "why is pi irrational" to "why is this huge set of numbers all irrational"

In fact I claim this is even false. Counterexample: z=0
 
  • #25
I should have stated that it works unless z equal 0 in the proof, but you see what I'm getting at hopefully. There is really no use in asking WHY a number is irrational considering we developed our own number systems
 
  • #26
A better (Yet still flawed, I believe) start to the proof:

For any ##\theta \in \mathbb{Q-0}## s.t. ##e^{i\theta}=a+bi##, either a, b, or both a and b are irrational.

I think that this is incorrect because I am pretty sure the above statement depends upon the fact that pi is irrational. Go into the cis form and you'll see that the problem concerns the rationality of the sine or cosine of a nonzero rational number
 
  • #27
cmcraes said:
I should have stated that it works unless z equal 0 in the proof, but you see what I'm getting at hopefully. There is really no use in asking WHY a number is irrational considering we developed our own number systems

By why it's irrational I meant prove it's irrational
 
  • #28
Have i not? Exempting the special case z=0, my hypothesis holds true
 
  • #29
I understand that proof isn't exacly beautiful and still a bit flawed but its the simpilist way i can prove it without getting into the definitions of trig functions or calculus
 
  • #30
cmcraes said:
There is really no use in asking WHY a number is irrational considering we developed our own number systems
I agree with this. The "why" question being asked here is pointless. The "why" question asked in calculus textbooks, which is to prove that ##\pi## is irrational, is obviously a different "why" question and one that actually has meaning.
 

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