2005-08-22 by
qntm

(This page is entirely factually accurate. It is neither a joke nor a satire nor a collection of fallacious proofs. All these proofs are genuine and the results are true. Thanks.)

ALL numbers are infinite decimal expansions.

For example, whenever you write "1" you are using this as a convenient shorthand for "1.0000..." in the same way that "1.0000..." is a convenient shorthand for a 1, a decimal point, and infinitely many zeros.

Similarly, "13556" is short for "13556.0000...", "1/3" is short for "0.3333...", and "pi" is short for "3.14159265...". Shorter forms are merely useful notation because it's tiresome/impossible to write out infinitely many decimal digits whenever you want to write a number.

Believe it or not, this has always been true since the moment you started doing mathematics. This is also part of the fundamental bedrock of mathematics and not something which you can argue with or debate about.

Now read on:

0.9999... = 0.9999... + 0 = 0.9999... + (0.9999... - 0.9999...) = (0.9999... + 0.9999...) - 0.9999... = 1.9999... - 0.9999... = 1.0000...

For any two *different* real numbers, you can pick a third number which is between them.

Also, any real number can be written out as a decimal expansion in at least one way.

So, if 0.9999... and 1.0000... were different numbers, then it would be possible to find a number which was between them, and write it out.

But it's *impossible* to write out the decimal expansion of a number between 0.9999... and 1.0000... .

Therefore, they cannot be different numbers.

Therefore, they are the same number.

Different sets of numbers have different properties. We're looking at real numbers, not integers. There are plenty of real numbers between 3 and 4.

Set `x` to 0.9999... and `y` to 1.0000... .

Suppose `x` < `y`. Then there is a value halfway between them, `x` < (`x` + `y`) / 2 < `y`.

We find that (`x` + `y`) / 2 = (0.9999... + 1.0000... ) / 2 = 1.9999... / 2 = 0.9999... = `x`. This is not "halfway between `x` and `y`", it is exactly equal to `x`.

This is a contradiction. So our initial supposition, `x` < `y`, must be false. Instead, `x` = `y`, i.e. 0.9999... = 1.0000... .

If the difference between two numbers is zero, then they are equal. For example, if `x` - 5 = 0, then `x` = 5.

The difference between 1.0000... and 0.9999... is:

1.0000... - 0.9999... = 0.0000... = 0

Therefore, they are equal.

No I'm not, I'm just doing a simple subtraction. Work it out yourself if you like.

No, it shouldn't. There is no end. "0.0000...1" is meaningless. The "..." means "every decimal digit is 0".

Let

x = 0.9999...

Multiply both sides by ten:

10x = 9.9999...

Subtract x from both sides:

10x - x = 9.9999... - 0.9999... 9x = 9.0000...

Divide by nine:

x = 1.0000...

"9.9999...0" is meaningless. The "..." means "every decimal digit is 9".

0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...n=∞ = Σ 0.9 * 0.1^{n}n=0n=∞ = Σa*r^{n}wherea=0.9 andr=0.1n=0 =a/ (1 -r) = 0.9 / (1 - 0.1) = 0.9 / 0.9 = 1

This uses the formula `S` = `a` / (1 - `r`) for the sum of a geometric series with initial term `a` and ratio `r`, proof of which is left to the reader.

Most of the reason why people don't understand why point nine recurring is equal to one is because they don't fully understand what a decimal representation actually *means*. Take a look at the definition of 0.9999... and things become abundantly clear:

0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ... = 9·0.1 + 9·0.01 + 9·0.001 + 9·0.0001 + ... = 9·10^{-1}+ 9·10^{-2}+ 9·10^{-3}+ 9·10^{-4}+ ...n=∞ = Σ 9·10^{-n}n=1n=N:= lim Σ 9·10^{-n}N→∞n=1 = lim ( 9·10^{-1}+ 9·10^{-2}+ ... + 9·10^{-N})N→∞ = lim ( 9·0.1 + 9·0.01 + ... + 9·0.000...0001 )N→∞ \________/Ndigits = lim ( 0.9 + 0.09 + ... + 0.000...0009 )N→∞ \________/Ndigits = lim ( 0.999...999 )N→∞ \_______/Nnines = lim ( 1 - 0.000...0001 )N→∞ \________/Ndigits = lim ( 1 - 10^{-N})N→∞ = lim 1 - lim 10^{-N}N→∞N→∞ = 1 - 0 = 1

If your lines of reasoning are correct, but the conclusion you arrive at is definitely wrong, there must be something wrong with your assumptions.

Clearly

0.9999... ≤ 1.

Assume

0.9999... ≠ 1 (*).

Then

0.9999... < 1,

so there must be some positive number `P` such that

0.9999... +P= 1.

But, for ANY positive `P`,

0.9999... +P> 1,

which is a contradiction, and definitely wrong. Therefore we are forced to conclude that the assumption (*) was incorrect, that is:

0.9999... = 1

1/3 = 0.3333... 1 = 3/3 = 3 * 1/3 = 3 * 0.3333... = 0.9999...

Proof that 1/3 = 0.3333... is left to the reader.

Use long division to find 9 / 9, but, instead of 90/9 = 10, put 81 remainder 9 each time, and see what happens.

0.9999... ----------- 9 | 9.0000... 8.1 --- 90 81 -- 90 81 -- 9...

Watch the repeating pattern.

1 / 11111 = 0.00009000090000900009... 1 / 1111 = 0.0009000900090009... 1 / 111 = 0.009009009009... 1 / 11 = 0.09090909... 1 / 1 = 0.9999... = 1

A sequence can only have one limit.

Observe that the limit of the sequence

0.9 0.99 0.999 0.9999 0.99999 ...

is

0.9999...

That is, the sequence gets closer and closer to 0.9999..., in fact, infinitely close.

But the sequence also gets closer and closer to 1.0000..., in fact, infinitely close. So 1.0000... is a limit of this sequence too.

But a sequence can only have one limit, so 0.9999... and 1.0000... must be the same.

There are NO proofs that 0.9999... and 1 are different numbers.

Anywhere.

In mathematics, "obvious" means "a proof immediately springs to mind". If you don't have a proof in mind, then unfortunately no mathematical statement you make carries any weight.

There are many ways of writing any number. You could write 1/1, or 3-2, or 1.0, or 1.00, or 1.0000... or any number of other expressions, and all of them ultimately have the same meaning, "one".

All numbers are concepts.

- Just because a number can't "exist in reality" doesn't mean it can't exist in mathematics.
- Because 1 = 0.9999..., that means that 0.9999... "exists in reality" to exactly the same extent that 1 does.

Rounding errors only occur when we *truncate* a decimal expansion after a *finite* number of digits. All of the proofs above use the "..." notation at every step, which means that we always take into account all of the infinitely many decimal digits. There is no rounding, which means there is no error.

0.9999... is a single number. It doesn't move, so it can't get closer and closer to anything. It is where it is.

0.9999... is definitely less than 2, so it can't be infinitely large.

- Humans can comprehend infinity. Mathematicians do it all the time.
Infinity obeys rules. If something obeys rules in a consistent fashion, then you can do mathematics with it. Ordinal arithmetic is a good example.

In case the connection isn't clear, what is true of infinite values is equally true of infinite decimal expansions. There are rules and procedures and they work and give meaningful results. See "The Real Proof" above for a relatively tame glimpse of this, which is actually a vast region of mathematics known as "analysis", naturally based on rock-solid fundamental axioms.

Sure it is.

They were wrong.

The proof was fallacious. Send it to me and I'll show you why.

In regular science, we have theories. A theory is proposed in order to explain observations, and can be overturned in light of new, inexplicable observations. Multiple theories and opinions may compete with one another. There are fashions. There is room for debate.

In mathematics, we have theorems instead of theories. A theorem is the *result* of a *mathematical proof*. A theorem is a fact. A theorem cannot be overturned and is not a matter of opinion. Once proven, a theorem stands for eternity. Mathematics is not ideological.

Thanks to the many proofs above, "point nine recurring equals one" is just such a theorem. So, your opinion is wrong. And sorry, but no: you're not entitled to be wrong in mathematics. That's not how it works.