## Thursday, May 3, 2012

### Questioning 0.9999...=1 or Heres to you Mr. Robinson

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:

Anonymous said...

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.

Dave Marain said...

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.