Two of these are pretty much exactly the same, only the first is much more intuitive than the second. I can hear the skeptics yelling "Bring it on!", so here it goes.

#### The First Way Of Proving 0.999... = 1

Everybody knows that`1/3 = 0.333...`

. Just multiply both sides by 3 to get `3*1/3 = 3*0.333...`

which results in `1 = 0.999...`

.

#### The Second Way Of Proving 0.999... = 1

Almost exactly the same as the first, only I'll be using a letter in this proof and people are afraid of letters when it comes to math. Intimidation is by far my favorite proof technique...

```
(1): x = 0.999...
(2): => 10x = 9.999...
```

Now `(2)`

minus `(1)`

gives `10x-x = 9.999... - 0.999...`

, so `9x = 9.000`

and thus `x = 1`

.

#### The Third Way of Proving 0.999... = 1

For the third proof, we'll need the formula for the sum of an infinite, converging, geometric series. Like Wikipedia says, the formula for this is (when`-1 < r < 1`

)
` S = \sum_{k=0}^\infty ar^k = \frac{a}{1-r}.`

Now, let's rewrite 0.999...

`0.999... = 0.9 + 0.09 + 0.009 + \cdots = {9 \over 10} + {9 \over 100} + {9 \over 1000} + \cdots = \sum_{n=1}^\infty \frac{9}{10^n}= \sum_{n=1}^\infty 9 \times \frac{1}{10^n}`

Notice how the sum index in the last equation starts from 1 and the one in the Wikipedia formula starts from 0? No problem, we'll merely have to manufacture the first term out of thin air...

`0.999... = \sum_{n=1}^\infty 9 \times \frac{1}{10^n} = 9 - 9 + \sum_{n=1}^\infty 9 \times \frac{1}{10^n} = 9 \times \frac{1}{10^0} - 9 + \sum_{n=1}^\infty 9 \times \frac{1}{10^n}`

`= -9 + \sum_{n=0}^\infty 9 \times \frac{1}{10^n}`

So `a = 9`

and `r = 1/10`

(which is between -1 and 1, so no problem there).

`0.999 = -9 + \sum_{n=0}^\infty 9 \times \frac{1}{10^n} = -9 + \frac{a}{1-r} = -9 + \frac{9}{1-{1 \over 10}} = -9 + \frac{9}{9 \over 10} = -9 + 10 = 1`

There are undoubtedly other ways to prove 1 = 0,999... but these three are my favorites and should be enough to convert even the strongest of unbelievers. Isn't math lovely?

The first way- 0.333... repeating doesn't equal one third, exactly. It is slightly less.

ReplyDeleteThe second way- Why did you make 10 groups of 0.999... When you took away one group? Then it would be 9 groups of 0.999... Which equals 8.999...9991. ... being infinity.

The third way- You are just rewriting the first and second ways on a bigger scale.

The first way- It does. The nines keep repeating infinitely long. Written another way:

ReplyDelete0.9999... = \lim_{x \to \infty}{\sum_{i=0}^{x}{\frac{9}{10*(i+1)}}}

How large would you say the difference between 1/3 and 0.999... is? As x goes to infinity, it will become zero.

The second way- You're claiming any real number has a "last" digit? If I were to say that the last digit of pi was an eight, you'd surely thing I'd have gone insane? There is no last digit. That's what it means to be infinite.