# The reason pi is not rational

1. May 24, 2013

### 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.

2. May 24, 2013

### DiracPool

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.

3. May 24, 2013

### rajeshmarndi

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.

4. May 24, 2013

### pwsnafu

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: May 24, 2013
5. May 24, 2013

### rajeshmarndi

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.

6. May 24, 2013

### HallsofIvy

No, "straight lines" and "circle" have nothing to do with being "rational" or "irrational".

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.

7. May 24, 2013

### pwsnafu

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.

8. May 24, 2013

### Office_Shredder

Staff Emeritus
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

9. May 25, 2013

### lavinia

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. May 25, 2013

### phinds

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

11. May 25, 2013

### DiracPool

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. May 25, 2013

### phinds

ouch !

13. May 28, 2013

### AnTiFreeze3

I really hope that pun was intentional.

14. May 28, 2013

### HallsofIvy

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

15. Jun 8, 2013

### pierce15

Hey guys, go easy on him :)

16. Jun 8, 2013

### 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.

17. Jun 10, 2013

### 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.

18. Jun 10, 2013

### 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. Im 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. Jun 10, 2013

### pierce15

That's a really good point, modus

20. Jun 10, 2013

### 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

21. Jun 10, 2013

### 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?

22. Jun 10, 2013

### 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?)

23. Jun 10, 2013

### 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!

24. Jun 10, 2013

### Office_Shredder

Staff Emeritus
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. Jun 10, 2013

### 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