Is It True That 1.230000... = 1.230000...? Exploring the Proof

[ILikePizza]

So, Friday in class, my teacher was talking about how .99999... = 1.00000...

So, this makes sense and all, but this is my question:
How do you know that 1.230000... = 1.230000...?

I guess that my real question is that if
a1.a2a3a4...aN = d1.d2d3d4...bN (finite number of digits)

then all the corresponding digits are the same.


I thought that this was true, but then again, I thought .999... < 1.000.

So. Is it true? Is it provable in the same way .999... = 1? If so, what is it?

 

[statdad]

This question has a long history on the internet, where you will find thousands (at least) of people who insist that mathematicians are wrong and that $$0.999\cdots \ne 1$$, because (pick one of these)

• Mathematics doesn't deal with infinity
• You can't have an infinite series of nines
• It's obvious

I'm not sure of your mathematics background. The classical way to justify this is to appeal to a geometric series, as shown below.

$$\begin{align*} 0.9999 \cdots & = \sum_{i=1}^\infty \left(\frac 9 {10} \right)^i\\ & = 0.9 + 0.09 + 0.009 + \cdots \end{align*}$$

This is a geometric series with $$a = 0.9, r = 0.1$$, so the sum is

$$\frac{a}{1-r} = \frac{0.9}{1-0.1} = \frac{0.9}{0.9} = 1$$

You may see all of this condensed down to the following steps.

$$\begin{align*} x & = 0.9999999 \cdots \tag{Line 1} \\ 10x & = 9.999999 \cdots \tag{Line 2}\\ \intertext{Subtract Line 2 - Line 1} 9x & = 9 \Rightarrow x = 1 \end{align*}$$

Hidden behind these ''simple'' steps is all of the mathematics of the geometric series approach.

Essentially, this is a special case of the fact that every number that is not an infinite decimal has two decimal forms - examples are

$$\begin{align*} 0.99999 \cdots & = 1 \\ 4.23 & = 4.2299999 \cdots \\ 5.222 & = 5.221999999999 \end{align*}$$

Regarding your ''does $$1.2300000 = 1.2300000$$?'' question - yes - the additional zeros on the end of a decimal contribute nothing to the value of that decimal.

For your second question: yes: if you have a finite decimal representation of a number, that representation is unique.

 

[JG89]

Suppose that you have a single-element set, S, with that number in it. Say 1. Then SupS = 1
because 1.000... <= 1. Suppose that you changed the <= sign to just <, then you have
0.999... < 1. And so the supremum of that set is still 1, even though the element in it is 0.999... It is easy to prove that two single-element sets contain the same element if and only if both sets have the same supremum. Since both of those sets have the same supremum, they both contain the same element.

 

[statdad]

No, even in the context of your complete quote, this statement

then you have
0.999... < 1.

is not true. You seem to agree that $$0.999 \cdots = 1$$, so I'm not sure how you reconcile the two comments.
In addition, if you begin with a one element set, it makes no sense to suddenly bring another number into play.

 

[ILikePizza]

I don't think you guys are reading what i said... I now *know* that .9999 = 1, but my question is:

"I guess that my real question is that if
a1.a2a3a4...aN = d1.d2d3d4...bN (finite number of digits)

then all the corresponding digits are the same."

Is this true or not? Do you have a proof?

 

[statdad]

I think you missed the final line of my comment:
"For your second question: yes: if you have a finite decimal representation of a number, that representation is unique."

As an outline of a proof: suppose to the contrary that you can find two different representations (with finitely many digits) for a number. To make the discussion simple, suppose this number is smaller than ten. There must be a first position where the digits are different.

• 
  * Can it be the digit in the ones position? No, because if those are different, you have different numbers. Since the ones digit must be the same, we can move on
  * Can it be the tenth's digit? No, for the same reason above

Continue with this line of argument.

 

[JG89]

No, even in the context of your complete quote, this statement

is not true. You seem to agree that $$0.999 \cdots = 1$$, so I'm not sure how you reconcile the two comments.
In addition, if you begin with a one element set, it makes no sense to suddenly bring another number into play.

You're right, the inequality should be 0.999...<= 1. But it is still easy to prove that two single-element sets contain the same element if and only if they have the same supremum, which would give equivalence to 1 and 0.999...

 

[mistermath]

I want to say something like..
x and y in R are equal if for every epselon > 0, | x - y | < epselon.

Meaning, you can make those numbers as close to each other as you'd like.

After all, is there a difference between 2 "different" things, if no one can ever tell them apart?

 

[JG89]

Here is a proof that two single-element sets contain the same element if and only if both sets have the same supremum:

The left to right implication is easy. Suppose there is a set A = {a} and another set B = {b}.
Suppose that SupA = c. Then a <= c, but since a = b, then a = b <= c. So SupB = c.

For the right to left implication: Suppose that both sets have the same supremum, c. I'll show through contradiction that both sets contain the same element. Let's say that the elements are different, then we have either a < b or a > b.

Case1: a < b. If a < b then we also have a < b <= c since SupA = SupB = c. However, that inequality says how there is an upperbound for the set A that is less than the supremum, which is a contradiction.

Case2: a > b. If b < a then we also have b < a <= c since SupA = SupB = c. Again, the inequality says that there is an upperbound for B that is less than SupB, which is impossible. Again we have a contradiction.

This proves the right to left implication. QED.

---------------

So if we have two single element sets, {1} and {0.999...}, then since they both have the same supremum, they both contain the same element, which gives equivalence to 1 and 0.999...

 

[mistermath]

JG89, that's a really cool proof :)