How is .9 repeating equal to 1?

Answer 1

We can show this in many ways. Here are 4 different ways.

1.
Let's say we do not know what .9999999999999 repeating for ever equals. I have no way to put the bar over the 9 here so I am going to write .9... to mean .9 repeating forever

Now let's call whatever it equals x. So we have x=.9...
Since this is an equation, we can multiply both sides by 10 and we have
10x=9.9...

We are allowed to subtract the same thing from both sides of an equations.
Since x=.9...
we subtract x from the left side and .9... from the right side. We can do this because they are the same thing. That is how we defined x, in fact. Now we have

10x-x=9.9...-(.9...)
which tells us

x=1

This is because the 1-x-9x=1x and the 9.9...-(.9...)=1


2.
Here is another way to think of it that is easier for some people.
We accept and know that 1/3 is .333 repeating forever, or as we write it .3...
Now 1/3 added to itself 3 times of 1/3 x3 is 3/3. Any number divided by itself is 1. This tells us .3...+.3...+.3...=.9... but we know that .3... is 1/3 and 1/3+1/3+1/3=3/3=1. We conclude the .9...=1

3.
Here is a slightly more advanced but very pretty explanation that uses infinite geometric series.

Find the exact value of a repeating decimal, asks the same question as "find the sum of an infinite geometric series":

Think of .9... and 9/10+9/100+9/1000 etc.
This is better written with exponents so .9...=9x10-1 +9x10-2 +9x10-3 +...

This is an infinite geometric series and there is a convergence rule that tells us how to find the sum. Since the common ratio here is between -1 and 1 or |r| < 1, we know the series converges.
In our case the common ratio, r=1/10. The first number in the series is a= 9/10.
We know that the finite sum is given by the formula, S=a(1-rn)/(1-r). If -1In this case we have:
S=a/(1-r) = (9/10)(1-1/10) = (9/10)/(9/10) = 1

Or,

The repeating decimal, 0.999... can be represented by a geometric series,
0.9 + 0.09 + 0.009 + ... with (first term) t1 = 0.9 and (common ratio) r = 0.1
Since |r| < 1, then S = t/(1 - r) (see the explanation above).
So we have:
S = t/(1 - r) = 0.9/(1 - 0.1) = 0.9/0.9 = 1



4. Here is one more way to think of it. The point here is that if you give me a number, however small, I can find an integer value n such that 1 - 0.99 ... 9 to n digits is smaller than your number. That is to say, using 0.9 repeating I can get as close to 1 as you ask. So 0.9 recurring is said to be equal to 1 in the limit. Here is a way to help see this.

1-.9=.1
1-.99=.01
1-.999=.001
If we continue this, in the limit we have 1- a very very very small number, so small that in the limit is equal 1.
We could write the limit of (1-9(1/10)n ) as n goes to infinity is 1.