I’ve been told by one of my Math Wizard friends that it’s possible to make pi equal zero, as well as any other number. I haven’t seen him actually do it yet, but the possibility intrigues me. Any of you Math Dopers out there know what he’s talking about? Care to give a demonstration?
Whaaa? pi is by definition not 0. Here, I can make pi 0:
Define “pi” to be 0.
done.
Too bad in most notation pi is the ratio of a circumference to the diameter. The only way this would be 0 is if you had a circle with a 0 circumference and non-zero diameter. Show me that, Ill mail you a dollar.
Though maybe there’s some weirdness in non-euclidean geometry Im not aware of…
Multiply it by zero? You said your friends can make any other number zero. How exacty would you make 7 equal to zero (7=0)?
Give him a pie and he’ll eat it. Then it’s zero.
(shrug)
Possibly by solving a pair of linear equations that are inconsistent.
Or having a circle be defined in a non-euclidian space, so it effectively covers the surface of a globe. Imagine a circle that is on the 80th north parralel, covering the land north of that. Now extend the circle southwards. After you cross the equator, the circumference shrinks, but the diameter still grows. When the circle envelops the whoel globe, the circurference will be zero, diameter will by 28K miles or so. Then you can pull all kinds of stupid math tricks.
Or you could use this old trick… assume a=b
- a=b
- ab=b^2 (mult by b)
- ab-b^2=a^2-b^2 (subtract b^2)
- b(a-b) = (a+b)(a-b) (factor)
- b=a+b (divide out the common factor)
- b=b+b (we assumed they’re equal, now substitute
- b=2b
- 1=2 (divide by b)
This is a fun proof (find the flaw) that comes up a lot. Now go further. - 0=1 (subtract 1)
- pi=pi (duh)
- pi1=pi0
- pi=0
Your friend may have similar tricks in mind. As far as I know though, there is no mathematically legitimate way to show that pi=0 through mathematics using standard definitions and procedures.
One flaw in your equation is in line (3):
if you subtract b^2 from both sides of
(2) ab=b^2
you would get
ab-b^2=b^2-b^2 (which is ab-b^2=0)
NOT
your line: 3) ab-b^2=a^2-b^2 (subtract b^2)
I think the ‘proof’ you may be looking for is:
Theorem 1: 1=2
a=b postulate
a^2=ab (multiply both sides by a)
a^2-b^2=ab-b^2 (subtract b2 from both sides)
(a+b)(a-b)=b(a-b) (factor)
a+b=b (divide by (a-b))
b+b=b (as a=b, substitute)
2b=b (b+b=2b)
2=1 (divide by b)
QED
Or many more of similar relation.
No time for a detailed explanation, but that’s not how a circle in spherical geometry would look. Search for more information.
This is absolutely correct. The only way to show that [symbol]p[/symbol] = 0 is to do some fakery.
Arrgh… hermn8r is absolutely right. Lousy stupid early morning posts. Should have subtracted a^2 not b^2, then proceed from there.
exp(2pi×i) = 1 = exp(0)
Taking the log of both sides gives…
2pi i = 0
Divide both sides by 2i.
pi = 0
Well, I have a wonderful proof for pi=0, but this thread is too small to contain it… 
muttrox
Is the problem step 5 where you divide by zero?
Yep, that’s the fakery.
Curious to hear what’s the OP’s friends trick was.
Achernar - where is the trickery in yours? Is it something to do with log being a multi-valued function in the complex domain?
I think its simpler than that, DarrenS.
If 2pi i = 0, and we know 2 and pi are both =/= 0, then i must = 0. Therefore, you cannot divide both sides by i (or 2i).
Well, think about it. Have you ever seen a pie that doesn’t look like O?
DarrenS is correct. i[sup]2[/sup] = -1, and 0[sup]2[/sup] != -1, so i != 0.
ultrafilter, you’re assuming that 0 is not equal to -1. I’m going to need a proof for that “minor” detail you just glossed right over.
And yeah, DarrenS is totally right.
All right. I’m going to assume that you’ll give me all the properties of natural numbers that I need.
Recall that an integer is defined to be an equivalence class of ordered pairs of natural numbers. In this case, (a, b) = (c, d) iff a + d = b + c. Also remember that [(a, b)] + [(c, d)] = [(a + c, b + d)].
First, you’ll note that the integer 0 is the class [(a, a)]. This is easy to see: [(b, c)] + [(a, a)] = [(a + b, a + c)]. (a + b, a + c) = (b, c) means that (a + b) + c = (a + c) + b, and due to the commutative and associative properties of addition of natural numbers, that’s true. You don’t want me to prove those. 
So what’s the representation of 1? Well, you may know that the idea of writing c as [(a, b)] is to make a - b = c in the informal sense. That means that [(a + 1, a)] is 1. You could show that it satisfies the multiplicative identity property if you want to, but let’s just agree that I’m right.
This also means that [(a, a + 1)] is the additive inverse of [(a + 1, a)]. This is easy to show: [(a, a + 1)] + [(a + 1, a)] = [(2a + 1, 2a + 1)], which is zero!
Now for the meat of it. Is [(a, a + 1)] = [(a, a)]. No, of course not! (0, 0) is in the second, but not the first, and (0, 1) is in the first but not the second. Additionally, a + a != a + a + 1 (this is using set equality on the naturals).
Is that good enough?
How about a “circle” with zero circumference and zero diameter. That would give pi=0.