*Vi Hart must be having an effect on me! After proudly explaining for over 40 years why*

*0.9999*

*... must equal 1 using the Density Property of the Reals (see my post Another Proof that*

*0.9999*

*...=1), I just had an epiphany of sorts.*

If 0.9999...=1, then (0.9999...)^2 must also equal 1 from the properties of the reals. But squaring a finite string of 9's (with or without a decimal point) produces a fascinating result:

(0.99)^2=0.9801

(0.999)^2=0.998001

(0.9999)^2=0.99980001 etc...

This sequence of decimals seems to suggest the existence of a non-real number which differs from 1 by an infinitesimal amount, so-called hyperreal numbers, leading to the non-standard analysis of

**Abraham Robinson**. Who knows where the teaching of calculus might be today if Dr. Robinson had not died at the age of 55 from the disease that took my wife 2 months ago -- pancreatic cancer.

Well, maybe it's healthy to have one' roots shaken after many years. After all, my tag line for this blog for a couple of years involved how new ideas are often at first ridiculed, then vehemently opposed and finally accepted as obvious ...

NOTE: I omitted the hyperlinks in this article. I was getting too 'hyper'!

*Sent from my Verizon Wireless 4GLTE Phone*

## 2 comments:

Clever idea. I have not seen this before. One can also cube it and 4th it. There is a race possibility here.

Take it the mth power and look at the n-th digit in the starting number, x.

So for n=3, x = .999. Now take the 100th or 1000th power.

If you take higher and higher powers holding the n-th digit fixed in the approximation, you can get all the digits up to p to be 0 for some choice of p.

The original number is 1. But by this approximation method, you always calculate an approximation of 0 for the m-th power.

Excellent!

This would appear to prove that 0.999... is not equal to 1. But now we need rigorous definitions/rules for raising an infinite decimal to a power. Your argument rested on the result of raising a finite number of 9's to larger and larger finite integer powers.

But if we define 0.999... as the limit of a sequence of partial sums then we're taking another limit of the partial sums raised to the nth as n→inf.

Post a Comment