Real analysis: If 0.5=0.499..... ,then 0=0.00.....1.?

Question

this mean that the neibourhood of zero is zero, doesnt it ruin the basis of real analysis. is still there any meaning of infinismally small. what is about epsilon scattering every where in proofs. shouldnot it have any value less than 0.00...1

Answer 1

There's a better way of looking at it.

For every real number x in [0,1) you define D(x, n) as the nth decimal of x. This is defined for every natural number n.

For x=1/3, we have D(1/3, n) = 3 for every n. The number 0.3333... (the dots are important, they indicate that the nth decimal is a 3 for every n) is not "very close" to 1/3, it does not "tend toward" 1/3. It is EXACTLY the number 1/3. Show me a number other than 1/3 and I will show you that its decimal representation is not 0.33333...

For 0, we have D(0,n) = 0 for every n. There is never a '1' decimal. If you give me an arbitrarily small number 0.00000...1, the digit 1 is at a finite position p. I can then reply with a number such that D(x,n)=0 for all n<=p, and D(x,p+1) = 1. That number is greater than zero and smaller than the number that you gave me.

The 0.5 = 0.49999... thing is just a quirk of decimal notation. We can formally prove the equality as follows:

Let a = 0.5, i.e. D(a, 1) = 5 and D(a,n) = 0 for all n>2.
Let b = 0.4999..., i.e. D(b, 1) = 4 and D(b,n) = 9 for all n>2.

Proof by contradiction that a=b: Let's suppose on the contrary that a > b. This means there must be a number x between a and b.

Let's figure out the digits of x. Since x is less that a=0.5, its first decimal must be a 4. D(x,1) = 4.

The second decimal of x must be 9 since x > b `> 0.49.
The third decimal of x must be 9 since x > b >0.499.
And so on, the nth decimal of x must be a 9 because x > b > the number y such that D(y,1)=4, D(y,i)=9 for i=2 to n and D(y,i) = 0 for i>n.

Therefore x has exactly the same digits as b=0.499999... (I just showed that for every n, the nth digit of x is the same as the nth digit of b), and so x=b. Contradiction, ergo a = b.

That's the difference between 0.499999... vs. 0.5 and 0.0000...1 vs. 0. In the latter case, it is possible to find a number that is between the two.

Answer 2

0 is unique in this context. You "prove" that 5 = 4.999999999..... by using a series of division errors:
...4.9
1)5.000
.. 4
.. 1.0
..... 9
..... 10
and so on. Actually, there is always a
0.000000 ...... ad nauseum ...... 00001 difference between this quotient and 5, just as there is a difference of 0.000000 ...... ad nauseum ...... 00001 between the number and 0. The limits are 5 and 0, but the numbers never really are 5 and 0.

Answer 3

What does "0.00...1" mean?

Answer 4

0.0000....1 would be 1/infinity or zero.

Answer 5

.009 I think