I saw this somewhere but I can't remember the reason behind it.

1/3 + 1/3 + 1/3 = 1

1/3 = 0.333333333... recurring

but then

0.333333 + 0.333333 + 0.3333333 = 0.999999999...

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- Mar 4th 2010, 11:14 AM #1

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- Mar 4th 2010, 09:48 PM #2

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Well, one can express 0.99999 (repeating) as the limit as x goes to infinity of the following:

The summation (Sigma sum indexed by i bounded from 0 to x) of 9 divided by 10 * 10^i.

If you compute that, you'll see that the limit goes to 1.

That's just one unneccessarily complex way to look at it.

- Apr 24th 2010, 10:11 AM #3
I like this problem-- $\displaystyle 1-\frac {1} {\infty}=.99999999999....$

Therefore, one over infinity equals 0 if $\displaystyle \frac {1} {3}*3$ is to equal 1.

This shows that $\displaystyle \frac {1} {0}$ is infinity, also proven by $\displaystyle \tan 90^o$ being $\displaystyle \infty$, and also being the slope of a vertical line, or $\displaystyle \frac {1} {0}$

- Apr 24th 2010, 01:59 PM #4

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$\displaystyle \frac{1}{0}$ is not infinity; it is not defined.

Usually, we say two numbers $\displaystyle x,y \in \mathbb{R}$ are different, if there exists $\displaystyle c \in \mathbb{R}, ~ c \neq 0$ such that $\displaystyle x-y=c$. In this case, what is $\displaystyle 1-0.999...$? Is it non-zero?

- Apr 25th 2010, 10:03 AM #5

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- Apr 25th 2010, 10:57 AM #6

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- Apr 28th 2010, 12:31 AM #7
The reason for this is that, theoretically, 1/3 + 1/3 + 1/3 is strictly equal to one, but in practice, as no-one can fully represent an infinitely recurring decimal, the sum of 1/3 + 1/3 + 1/3 will tend towards 1 but will never actually reach it. You can't manipulate recurring decimals algebraically without using fractions. Algebra just hasn't been designed for it.

This is one of the advantages of mathematics in my opinion : by defining some axioms and building everything upon it, it is possible to prove things without actually having to check whether it works : if the axioms remain the same throughout the proof, and the proof is valid, the experimental results will follow.