Sunday, March 8, 2009

Division by Zero

We all know that division by zero is undefined. But what does it mean? Firstly, it is not "completely" undefined. For example, it is possible to say that:


1/0=lim1/x, x->0

This particular definition is not very good, because the limit doesn't exist. But we can always suppose that we look only on positive x and the the limit is defined to be infinity. Alternatively, it is possible to use geometric series. The formula for the sum of the geometric series says that:

1+q+q^2=q^3+...=1/1-q,

If we will take q to be 1 we will get: 1/0=1+1+1+1+....=infinity

As you can see I just wrote two different definitions to division be zero. However, this doesn't solve anything. The problem is that infinity is not a number. We could just as well say that 1/0=watermelon. Mathematically it is basically the same. Therefore when we divide by zero we no longer deal with numbers, and this what makes such division undefined.
Now, what about other definitions? A very long time ago, when "zero" appeared in mathematics there was an attempt to define division be zero by simply stating that division by zero gives a fraction whose denominator is zero. But this is again not a number.
We also cannot define 1/0=a where a is a number. The reason for this is that in this case we get:

1=0*1/0=0*a=0

And if this happens we get that the only number that we have is 0, because all the "other" numbers are equal to it. Obviously this is not an interesting situation. Because of this it is necessary that the devision by zero is not defined.

Lets look on some other example of undefined identities. The first one is 0^0. The reason this one is undefined is very simple:

0^0=0^(1-1)=0^1*0^-1=0/0=0*1/0

And we are back to division by zero. It is important to note that it is sometimes assumed that 0^0=1, but this is mostly done as an alternative to saying that in a polynomial x^n in which n can be zero x is not zero.
A more complex example is (-3)^x where is a real number (that is, x is not rational). Obviously, there is nothing special in number (-3), this expression is undefined for all negative numbers. For positive numbers the definition uses limits. For example:

3^x=lim(3^q), limq=x, q- rational

This definition works well for positive numbers, but for negarive numbers we have the problem that the square root is not real. Because of this if we don't allow complex numbers we get many undefined points in the series, and if we allow such numbers the series doesn't converge.

As a bonus, here is a little proof that 2=1. Can you see where is the error?
a=b.
a^2=ab ,
a^2+a^2=a^2+ab,
2a^2=a^2+ab,
2a^2-2ab=a^2-ab,
2(a^2-ab)=1(a^2-ab)
2(a^2-ab)/(a^2-ab)=1(a^2-ab)/(a^2-ab)
2=1

6 comments:

Lucas Lindström said...

(a^2 - ab) = 0, so division by zero in the second last step?

Lucas Lindström said...

(a^2 - ab) = 0, so division by zero in the last step?

Anatoly said...

Hello Lucas,
You are correct, the mistake is in the last step. In this step we supposed that (a^2 - ab)/(a^2 - ab)=1, but (a^2 - ab)/(a^2 - ab)=0/0=0^0.
There is also a mistake in the second to last step, but the problem in that step is that we divide both of the sides by zero. Therefore, if we look at it as a limit, it is possible to claim that this step is still correct.

Lucas Lindström said...

I accidentally double-posted, didn't notice comments needed approval before it was too late. So in case you misunderstood, I didn't actually try to point out two different errors.

Your blog is very interesting by the way. Keep up the good work. =]

aNn mArShMaLow said...

0/0 is indeterminate form.
can't simplify that rite :P

Burn said...

Here is an interesting perspective on all of this:

http://beehivecomputer.com/mikesblog.htm