Why is Pi Irrational for Circles?

[rajeshmarndi]

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.

 

[DiracPool]

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.

 

[rajeshmarndi]

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.

 

[pwsnafu]

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.

 

[rajeshmarndi]

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.

 

[HallsofIvy]

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.

 

[pwsnafu]

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.

 

[Office_Shredder]

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

 

[lavinia]

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.

 

[phinds]

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

 

[DiracPool]

In which case the radius/diameter will not be. SOMEBODY'S got to be irrational here

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.

 

[phinds]

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 !

 

[AnTiFreeze3]

... 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.

 

[HallsofIvy]

I really hope that pun was intentional.

So do I. But that was four days ago and now I can't remember!

 

[pierce15]

Hey guys, go easy on him :)

 

[cmcraes]

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.

 

[HallsofIvy]

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.

 

[ModusPwnd]

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?

 

[pierce15]

That's a really good point, modus

 

[pierce15]

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

 

[JsStewartFan]

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?

 

[pierce15]

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?)

 

[cmcraes]

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!

 

[Office_Shredder]

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

 

[cmcraes]

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

 

[pierce15]

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

 

[Office_Shredder]

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

 

[cmcraes]

Have i not? Exempting the special case z=0, my hypothesis holds true

 

[cmcraes]

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

 

[WannabeNewton]

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.

 

[HallsofIvy]

A number of years ago, I ran across this peculiar proof (I think it was in "Math Monthly" but do not remember the author's name):
Let c be a positive real number. If there exist a function, f, such that f and all of its anti-derivatives can taken to be rational at 0 and c, then c is irrational.
(There is, of course, an arbitrary constant at each anti-derivative. "Can be taken to be rational" means we can always choose the constant so that the anti-derivative is rational at 0 and c.)

Of course, f(x)= sin(x) is 0 at 0 and \pi and all anti-derivatives can be taken to be sin(x), -sin(x), cos(x), and -cos(x), all of which have values of 0, 1, or -1 at 0 and \pi, all rational. Therefore, by this theorem, \pi is irrational.

I wish I could remember the proof. As I recall, it was the "worst" kind of indirect proof. The author uses the hypothesis (that such a function exists) to show conclusion "a", the turns around and uses the contradiction of the conclusion (that \pi is rational) to show conclusion "b" which doesn't seem to have much connection with the hypotheses but contradicts conclusion "a"!

 

[JsStewartFan]

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!

OK, I'm way down here on understanding, but wanted to ask cmcraes what "since we know i is not" meant, with explanation, please. Thanks!

 

[cookiecrumbzz]

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?

The "base 10" system means that you have the numbers 0, 1, 2,3,...,9 ie ten digits in the system. The series makes use of numbers in unique combinations of these ten digits.
The "octal" (base 8) system has 8 digits (0,1,2,...,7) and the hexadecimal system has 16 digits (0,1,2,...,9,10,A,B,C,D,E). Each "number" of these systems are made of unique combinations of the member digits.

If you had a "base pi" system, what would be the "member digits" of this counting system?
You may criticize this, because my saying this implies that I already assumed pi to be the irrational 3.14... , but WHAT exactly will you assume pi to be in this system? You need to know the (whole-numbered?) value of pi to define a number system of "base pi".

A base 10 system starts from 0 and ends at 9. A base 8 system starts at 0 and ends at 7. A "base 16" system begins with 0 and ends with E (=15. The values of A,B,...,E in this system is 10,11,...15. Whole numbers again).
A base pi system would begin from zero and continue till (pi-1). But (pi-1) must be a whole number, otherwise (pi-1) does not qualify as a "digit". In fact, you don't even know what the value of the last member is, so you can't find out what the the "spacing" (or "interval" or "gap") between two successive digits of this system is!
Moreover, you could vaguely say that the "base pi" system starts at 0 and ends at (pi-1). But for knowing the limiting value (pi-1), you have to DEFINE what (pi-1) is [HOW can you create a number system if you can't define it?] . Here, again, you have to go to the good old "base 10" system to define what pi is.

In the base 10 system, pi has non-terminating, non-recurring digits after the decimal point [again, note that DECIMAL refers to the base 10 number system]. By definition, such a number is called an irrational number. So pi is irrational.

Basically, the "flaw" here is that we depend on the "base 10" system to expand our number system to the "base pi" system. But there's nothing you can do here.
Suppose you did create a "base pi" number system from scratch. Say, "pi" is a rational number. Then, that "base pi" number system could not define what our "base 10" numbers 1,2,3,...,9 are.

You know that pi = 22/7.

But in the "base pi" system, 22 and 7 are irrational numbers. So pi would be a result of dividing an irrational number by another. Both are unequal, and quite obviously are not multiples or submultiples of each other. So pi, which you initially took to be a rational number, turns out to be an irrational number.

This basically is a proof-by-contradiction technique. Pi cannot be rational.

 

[cmcraes]

In response to JsStewartFan
I meant, we know the imaginary unit 'i' is not irrational. And since e^z will be irrational for any rational complex z, (excluding 0) The only way e^z can be a rational expression is if z is also irrational. So because we know that e^ipi= -1, and that 'i' is rational, pi must be the irrational part of z.
Hope that made sense!

 

[Curious3141]

In response to JsStewartFan
I meant, we know the imaginary unit 'i' is not irrational.

Do we? What two integers can we divide to get $i$?

$i$ is certainly algebraic, but that doesn't mean it's rational. In fact, the proof of $e^{\pi}$'s transcendence uses the fact that this is equal to $i^{-2i}$ (or equivalently, $(-1)^{-i}$), which is an algebraic number (other than 0 or 1) raised to an irrational algebraic power. By Gelfond-Schneider, that makes $e^{\pi}$ transcendental.

 

[pwsnafu]

If you had a "base pi" system, what would be the "member digits" of this counting system?
You may criticize this, because my saying this implies that I already assumed pi to be the irrational 3.14... , but WHAT exactly will you assume pi to be in this system? You need to know the (whole-numbered?) value of pi to define a number system of "base pi".
<snip>
HOW can you create a number system if you can't define it?] . Here, again, you have to go to the good old "base 10" system to define what pi is.

cookiecrumbzz, are you arguing that base-π is not a thing? Wikipedia[/PLAIN] knows all.

In the base 10 system, pi has non-terminating, non-recurring digits after the decimal point [again, note that DECIMAL refers to the base 10 number system]. By definition, such a number is called an irrational number. So pi is irrational.

That is not the definition of irrationality. It's a theorem.

Basically, the "flaw" here is that we depend on the "base 10" system to expand our number system to the "base pi" system. But there's nothing you can do here.

That is not a flaw.

Suppose you did create a "base pi" number system from scratch. Say, "pi" is a rational number. Then, that "base pi" number system could not define what our "base 10" numbers 1,2,3,...,9 are.

Of course it can. They just have aperiodic expansions.

 

[HallsofIvy]

Do we? What two integers can we divide to get $i$?

$i$ is certainly algebraic, but that doesn't mean it's rational.

cmcraes did NOT say i was rational. He said it was not irrational. The "if not irrational then rational" dichotomy applies only to real numbers. If, as is usually done, you define a rational number as "a number equal to the ratio, m/n, of two integers" (which implies any rational number is real) and define irrational as "a real number that is not rational", i is neither rational nor irrational.

In fact, the proof of $e^{\pi}$'s transcendence uses the fact that this is equal to $i^{-2i}$ (or equivalently, $(-1)^{-i}$), which is an algebraic number (other than 0 or 1) raised to an irrational algebraic power. By Gelfond-Schneider, that makes $e^{\pi}$ transcendental.

I don't see how that has anything to do with whether "i" is rational or irrational.

 

[Curious3141]

cmcraes did NOT say i was rational. He said it was not irrational. The "if not irrational then rational" dichotomy applies only to real numbers. If, as is usually done, you define a rational number as "a number equal to the ratio, m/n, of two integers" (which implies any rational number is real) and define irrational as "a real number that is not rational", i is neither rational nor irrational.

This is fair enough. $i$ is not rational. Neither is it irrational, since that would imply it's real.

I don't see how that has anything to do with whether "i" is rational or irrational.

It has everything to do with $i$ being NOT rational, because Gelfond-Schneider depends on the exponent being non-rational.

 

[cmcraes]

Sorry for any confusion i may have caused i should have clarified that e^z would be irrational (AND TRANCENDENTAL!) for any algebraic number*