# Divided by itself equals 1?

1. Sep 2, 2005

### The Rev

Is it a reasonable statement to say either:

1) $$\frac{\infty}{\infty} = 1$$ ?

2) $$\frac{0}{0} = 1$$ ?

$$\phi$$

The Rev

2. Sep 2, 2005

### lurflurf

No. Those are not well defined operations, and infinity is not a number. In classical analysis this problem is dealt with by considering limits.
Such as
$$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$$
in the case when if f and g both go to 0 (or infinity) as x goes to a
further analysis is needed to decide the limit
These are called indeterminate forms since the rule about a limit of a combination of functions being equal to the individual limits combined the same way does not apply.
There are several forms of indeterminate forms that commonly occur
0/0
infinity/infinity
0^0
1^infinity
infinity-infinity
0*infinity
1/0-1/0

3. Sep 2, 2005

### The Rev

Is this how they cancel out infinities in renormalization in QM?

4. Sep 2, 2005

### matt grime

Is what how they cancelled out infinities in renormalization? WIth infinite stuff physicists can and do simply ignore things (nothing to do with cancelling), and annoyingly it all seems to work. Hre is a link explaining it in non-specific language:

http://math.ucr.edu/home/baez/renormalization.html

5. Sep 2, 2005

### The Rev

Well, he lost me at Lagrangian, but I did get the feeling that renormalization isn't about cancelling out infinities with simple division. (I know nothing of this higher level math, yet.)

I was impressed, however, that singer Joan Baez is also an expert on Quantum Mechanics.

$$\psi$$

6. Sep 2, 2005

### matt grime

gosh, i bet jon's never head that one before. if i ever bump itno him again i'll be sure to pass on the joke.

7. Sep 3, 2005

### Berislav

I think they're cousins.