Why is Pi Irrational for Circles?

[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*