Why is Pi Irrational for Circles?

[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*