How Can 0.999... be equal to 1?

Maybe it's just that it feels wrong that something as nice and neat and well-behaved as the number 1 could also be written in such a messy form as 0.9999.... Whatever the reason, many students (me included) have, at one time or another, felt uncomfortable with this equality.

One of the major sticking points seems to be notational, so let me get that out of the way first.

What's the difference between 0.999 and 0.999...?

When I say "0.9999...", I don't mean 0.9 or 0.99 or 0.9999 or 0.999 followed by some large but finite (that is, limited) number of 9's. The ellipsis (that is, the "dot, dot, dot") after the last 9 means "this goes on forever in the same manner". In other words, the dot, dot, dot, says that 0.9999... never ends. There will always be another 9 to tack onto the end of 0.9999.... So don't object to 0.9999... = 1 on the basis of "however far you go out, you still won't be equal to 1", because there is no "however far" to go out to; you can always go further. There is no end; there are always more 9s.

"But", some say, "there will always be a difference between 0.9999... and 1." Well, sort of. Yes, at any given stop, at any given stage of the expansion, for any given finite number of 9s, there will be a difference between 0.999...9 and 1. That is, if you do the subtraction, 1 − 0.999...9 will not equal zero.

But the point of the "dot, dot, dot" is that there is no end to the 9s; 0.9999... is inifinte. There is no "last" digit. So the "there's always a difference" argument betrays a lack of understanding of the infinite. (That's not a criticism, per se; infinity is a messy topic.)

How can an infinite series prove that 0.999... = 1?

The number 0.999... can be expanded and then converted into an infinite series. If the common ratio is less than 1, then there is a formula for finding the expansion's "closed form" (that is, the formula can convert the expansion into a fraction of some form).

The number 0.9999... can be expanded as:

In other words, each term in this endless summation will have a 9 preceded by some number of zeroes. This may also be written as:

That is, this is an infinite geometric series with first term a = ⁹/₁₀ and common ratio r = ¹/₁₀. Since the size of the common ratio r is less than 1, we can use the infinite-sum formula to find the value:

So the formula proves that 0.9999... = 1.

Note: Technically, the above proof requires that some fairly advanced concepts be taken on faith. If you study "foundations" or mathematical philosophy (way after calculus), you may encounter the requisite theoretical constructs.

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How does 0.333... = 1/3 prove that 0.999... = 1?

If you haven't already learned that ¹/₃ = 0.333... in decimal form, you can prove this easily by doing the long division:

...and so on and so forth, ad infinitum.

So ¹/₃ + ¹/₃ + ¹/₃ = 3( ¹/₃ ) = 1. Reasonably then, 0.333... + 0.333... + 0.333... = 3(0.333...) should also equal 1. But 3(0.333...) = 0.999.... Then 0.999... must equal 1.

What about the 1 at the end of the subtraction?

When you subtract a number from itself, the result is zero. For instance, 4 − 4 = 0. So what is the result when you subtract 0.999... from 1? For shorter subtractions (that is, for the subtractions that involve a finite number of 9s), you get:

Then what about 1.000... − 0.999...? You'll get an infinite string of zeroes. "But," you ask, "what about that '1' at the end?" Ah, but 0.999... is an infinite decimal; there is no "end", and thus there is no "1 at the end". The zeroes go on forever. And 0.000... = 0.

Then 1 − 0.999... = 0.000... = 0, and 1 = 0.999....

But aren't they really two different numbers?

If two numbers are different, then you can fit another number between them, such as their average. But what number could you possibly fit between 0.999... and 1.000...?

How can algebra prove that 0.999... = 1?

The expression 0.999... is some number; it has some value, and we'll call this numerical value x, so 0.999... = x. Now multiply this equation by ten:

Subtract the first equation from the second equation:

Solving, we get x = 1, so 0.999... = 1.

"But," you ask, "when you multiply by ten, that puts a zero at the end, doesn't it?" For finite expansions, certainly; but 0.999... is infinite. There is no "end" after which to put that alleged zero.

But won't 0.999... always be a little bit smaller than 1?

A common objection is that, while 0.999... gets "arbitrarily" close to 1, it is never actually equal to 1. But what is meant by "gets arbitrarily close"? It's not like the number is moving at all; it is what it is, and it just sits there, looking at you. It doesn't come or go; it doesn't move or get close to anything.

On the other hand, the terms of the associated sequence, 0.9, 0.99, 0.999, 0.9999, ..., do get arbitrarily close to 1, in the sense that, for each term in the progression, the difference between that term and 1 gets smaller and smaller as the number of 9s gets bigger. No matter how small you want that difference to be, I can find a term where the difference is even smaller.

This "getting arbitrarily close" process refers to something called limits. You'll learn about limits later, probably in calculus. And, according to limit theory, "getting arbitrarily close" means that they're equal: 0.999... does indeed equal 1.