Two names for one number
The decimal 0.999… has nines after the point that go on forever, with no last nine. Many people feel that it must be a tiny bit less than 1. It is not. 0.999… and 1 are two ways of writing the same number, in the same way that and 0.5 are.
The gap to 1
Measure the gap between 1 and a run of nines that stops. 1 − 0.9 = 0.1, 1 − 0.99 = 0.01 and 1 − 0.999 = 0.001. Each extra nine makes the gap ten times smaller.
Now suppose 0.999… were less than 1, so that the gap 1 − 0.999… were some positive number. Written as a decimal, that positive number has a first digit that is not 0, in some decimal place, say the 6th. Then the gap is at least 0.000001.
But 0.999… is bigger than 0.9999999, which has seven nines, so its gap to 1 is smaller than 1 − 0.9999999 = 0.0000001. A gap cannot be at least 0.000001 and less than 0.0000001 at the same time. Wherever the first non-zero digit is, a run of nines one place longer gives the same contradiction: a first non-zero digit in the 20th place is beaten by 21 nines. So the gap is not positive: it is 0, and 0.999… = 1.
0.999 > your number, so it is not between 0.999… and 1
Find a number that sits strictly between 0.999… and 1.
Try to put the point between 0.999… and 1. Wherever it goes, a run of nines that stops, marked in gold, already lies beyond it, and 0.999… is bigger than every such run. No point can be placed between them.
Call it x
A little algebra gives the same answer. Let x = 0.999…, with the nines going on forever. Multiplying by 10 moves every digit one place to the left, so 10x = 9.999…, again with nines forever.
After the point, 10x still has nines that go on forever, just as x does. There is a nine in every decimal place of both, so when the two are written one above the other, every nine of 10x sits above a nine of x.
x = 0.999… and 10x = 9.999…, and after the point every place of both holds a nine.
Subtract
Subtract x from 10x. After the point, every nine takes away the nine below it and leaves 0, in every place, forever. In front of the point, 9 − 0 = 9. So 10x − x = 9.999… − 0.999… = 9.
10x − x = 9x, so 9x = 9. Divide both sides by 9: x = 1. Since x was 0.999…, this shows 0.999… = 1.
Each nine takes away the nine below it, so every decimal place of 10x − x is 0. That leaves 9x = 9, so x = 1.
Two more ways to see it
By thirds. , and 3 × 0.333… = 0.999…, because 3 × 3 = 9 in every decimal place with nothing to carry. But . So 0.999… = 1.
As a sum to infinity. , a geometric series with first term and common ratio . Its sum to infinity is .
Every terminating decimal has two names
The same fact gives every terminating decimal a second name. 0.4999… = 0.4 + 0.0999…, and 0.0999… is 0.999… divided by 10, which is 1 ÷ 10 = 0.1. So 0.4999… = 0.4 + 0.1 = 0.5. In the same way, 2.36999… is another name for 2.37.
Worked example: A Dispute About a Third of a Ribbon on a Calculator
Question A ribbon of length 1 m is cut into 3 equal pieces. Jun says that each piece is 0.333… m long, that 3 × 0.333… = 0.999…, and that a little ribbon has therefore been lost. (a) Show that 0.999… is exactly equal to 1, so that no ribbon is lost. (b) Jun's calculator shows 1 ÷ 3 as 0.3333333, with 7 decimal places. Find 3 × 0.3333333, and find how far it is from 1. Explain where this difference comes from.
1.Let x = 0.999… Multiply by 10, which moves every digit one place: 10x = 9.999…
Let x = 0.999… Multiplying by 10 moves every digit one place: 10x = 9.999… 2.Subtract x from 10x. The recurring parts are the same, so they cancel: 10x − x = 9.999… − 0.999… = 9. So 9x = 9 and x = 1.
The recurring parts cancel: 10x − x = 9, so 9x = 9 and x = 1. 3.(a) 0.999… = 1 exactly, so the three pieces make 3 × 13 = 1 m and no ribbon is lost. A second reason: the difference 1 − 0.999… would have to be less than 0.1, less than 0.01, less than 0.001 and so on, and the only number that is not negative and is less than all of these is 0.
(a) 0.999… = 1 exactly, so the three pieces make 1 m and no ribbon is lost. 4.The calculator stops after 7 decimal places, so it shows 0.3333333, which is slightly less than 13. Multiply: 3 × 0.3333333 = 0.9999999.
The calculator stops after 7 decimal places: 3 × 0.3333333 = 0.9999999. 5.(b) 1 − 0.9999999 = 0.0000001, so the product is 0.0000001 m less than 1 m. The difference comes from the digits that the display cut off, and not from the ribbon. With nines that never end, there is no difference at all.
(b) 1 − 0.9999999 = 0.0000001. The difference comes from the digits that the display cut off.
Answer: (a) 9x = 9, so x = 0.999… = 1 exactly; (b) 3 × 0.3333333 = 0.9999999, which is 0.0000001 less than 1, because the display was cut off after 7 decimal places
Common mistakes
- Saying that 0.999… is the number just before 1. There is no number just before 1: between any two different numbers lies their mean, and no decimal lies between 0.999… and 1. The two are the same number.
- Treating the display 0.3333333 as equal to 13. The display has 7 digits and 13 has digits that never end. The difference of 0.0000001 in part (b) belongs to the shortened decimal only.
The usual mistakes
Calling 0.999… the number just before 1. No number is just before 1: between any number below 1 and 1 itself lies their mean, which is nearer to 1 still. And the gap to 1 shows that 0.999… is not below 1 at all.
Thinking that 10x has one nine fewer after the point, so that 10x − x comes out a little less than 9. That would need a last nine, and there is none: 10x has a nine in every decimal place, just as x does.
Trusting a calculator. A screen showing 0.9999999 has cut the nines off after 7 places. That number really is less than 1, by 0.0000001. The endless decimal 0.999… is not.