Is 1/3 Really Equal to 0.333...? Find Out the Easier Way with This Trick!

[HallsofIvy]

Exactly what a thousand people have said before. And obvious IF you assume that 10*(0.99999...)= 1.00000... Have you proved that the simple arithmetic operations you apply every day to numbers with finite number of decimal places also apply to those with unterminating decimal expansions?

 

[HallsofIvy]

if the probability of something not happening, such as tossing a potentially infinite number of heads in a row is .99999 repeating, hiw could you say that it is 1 since there is still a chance of it happening

I don't understand what you mean by "a potentially infinite" number of heads. It's impossible to toss and infinite number of heads because you don't have infinite time in which to toss an infinite number of coins whatever they come up to!

However, there is a fundamental assumption here that is not true- it is only in "discrete" (finite) probability that "probability 1" means"certain to happen" or "probability 0" means "impossible". Take for example the experiment "choose a number between 0 and 1 inclusive" where we take the probability distribution to be the uniform distribution- every number is equally likely to be chosen. The probability of any specific number being chosen is 0- but clearly some number has to be chosen. The probability of choosing 1/2 or 1/3 or \pi or e is 0 but since this is a uniform distribution, they are as likely to be chosen as any other number. The probability that "any number except 1/2 will be chosen" is 1 but certainly it is possible that that will not happen- that 1/2 will be chosen.

 

[Zurtex]

if the probability of something not happening, such as tossing a potentially infinite number of heads in a row is .99999 repeating, hiw could you say that it is 1 since there is still a chance of it happening

If you toss an coin an infinite number of times, you will eventually get a tails, regardless if you get an infinite number of heads.

But that's really side tracking from the subject of the thread.

 

[Hurkyl]

If you toss an coin an infinite number of times, you will eventually get a tails, regardless if you get an infinite number of heads.

No, you will get a tails with probability 1.

Actually, there's a fundamental mathematical problem with this whole issue anyways: it's a nontrivial task to figure out how one can even apply the concept of probability to the set of all possible outcomes.

 

[Sydius]

Uhm… maybe someone already said this, but 0.333… comes infinitely close to 1/3rd, but is not 1/3rd. 0.999… repeating comes infinitely close to 1, but is not 1.

Infinitely close, for all practical purposes, might as well be “equal”, which is why they say “1/3 = 0.333…” instead of “1/3 (squiggly equal sign) 0.333…” – it would just confuse students and make middle and high school math teachers get stressed out trying to explain something, that, for all practical purposes, does not matter. Regardless, though, being infinitely close to something does not make it that something and therefore saying “1/3 = 0.333…” is infinitely close to the truth, but not the truth.

 

[TD]

Maybe you didn't read this thread properly.

0.3333 \approx 1/3 and 0.333333333 \approx 1/3 is even better but 0.3333... = 1/3 is exactly the same, at least assuming that you mean an infinite expansion of decimal 3's.

 

[Sydius]

Maybe I am not reading it properly, but 0.3333… (With an infinite expansion of decimal 3’s) is NOT = to 1/3, it is just infinitely close.

Anytime you get a repeating decimal, you actually have an answer which is infinitely close to the true answer, but not the true answer. The repeating 3 is just a way to express “this is infinitely close to 1/3rd” – not that it *IS* 1/3rd…

Just like 0.9999… is infinitely close to 1, but not 1.

0.333… is infinitely close to 1/3, but not 1/3.

Maybe that is what you are saying, and I am just missing it. :-|

 

[TD]

No, that's what you're saying, but it's not correct.
I suggest you read the first 2 pages of this thread first

 

[Sydius]

It is true. Maybe we are just arguing about the applicability of the “equal” sign, but in the strictest term of the word, ONE divided by THREE does *NOT* EQUAL 0.3333 (Repeating)

Just because a number is infinitely close to another number, does not make them the same.

If you wrote a function that simply kept adding a “3” to the decimal, the LIMIT of that function would be 1/3rd, but it would NEVER reach 1/3rd, even if it went on for infinity.

Infinitely close to x != x

 

[Gokul43201]

...but in the strictest term of the word, ONE divided by THREE does *NOT* EQUAL 0.3333 (Repeating)

Yes, it DOES.

Just because a number is infinitely close to another number, does not make them the same.

Yes, it DOES.

 

[TD]

I'm affraid this is useless, but would you want to comment here beginning with "Uhm… maybe someone already said this," if you clearly didn't read the previous replies yet.

 

[Sydius]

*cries* I don’t get it!

So, if any function comes infinitely close to a number, than that function(infinity) = that number?

 

[HallsofIvy]

Maybe I am not reading it properly, but 0.3333… (With an infinite expansion of decimal 3’s) is NOT = to 1/3, it is just infinitely close.

Anytime you get a repeating decimal, you actually have an answer which is infinitely close to the true answer, but not the true answer. The repeating 3 is just a way to express “this is infinitely close to 1/3rd” – not that it *IS* 1/3rd…

Just like 0.9999… is infinitely close to 1, but not 1.

0.333… is infinitely close to 1/3, but not 1/3.

Maybe that is what you are saying, and I am just missing it. :-|

No, you are just completely wrong. 1/3 means (by definition of a "place ten numeration system") the sum of the infinite geometric series
\frac{3}{10}+ \frac{3}{100}+ \frac{3}{1000}+ ... and that, as any good high school graduate should know is
\frac{\frac{3}{10}}{1-\frac{1}{10}}=\frac{\frac{3}{10}}{\frac{9}{10}}= \frac{3}{9}= \frac{1}{3}.

0.333... is exactly equal to 1/3.

Similarly, 0.999... means the sum of the infinite geometric series
\frac{9}{10}+ \frac{9}{100}+ \frac{9}{100}+ ...
which is
\frac{\frac{9}{10}}{1-\frac{1}{10}}= \frac{\frac{9}{10}}{\frac{9}{10}}= 1

0.999... is exactly equal to 1.0

 

[Sydius]

I'm affraid this is useless, but would you want to comment here beginning with "Uhm… maybe someone already said this," if you clearly didn't read the previous replies yet.

I read them, but probably did not understand them. (I guess that is obvious now)

 

[TD]

Ok, that's no problem then

The things is, you shouldn't see this 'geometrically' where the 0.9, 0.99, 0.999 etc gets closer and closer to 1 but never there. Purely analytically, this is a limit (infinite number of 9's) and that limit doesn't "approach" 1, it's "equal" to 1.

 

[Sydius]

Okay, okay, sorry – lack of comprehension on my part. That, and math teacher telling me I was right when I was not (sigh)… I apologize.

I get the math, but still do not understand it in my mind, how a number infinitely close to another number is that number…

 

[Sydius]

Ok, that's no problem then

The things is, you shouldn't see this 'geometrically' where the 0.9, 0.99, 0.999 etc gets closer and closer to 1 but never there. Purely analytically, this is a limit (infinite number of 9's) and that limit doesn't "approach" 1, it's "equal" to 1.

Ohhhhhhhh! I see now. That makes sense :) Thanks...

 

[Sydius]

Just to be sure, though…

When you see x = some-repeating-decimal, it actually means x = limit(some-function-that-would-generate-that-repeating-decimal)?

 

[HallsofIvy]

When you see "x= some decimal, repeating or not" it actually means that x is the sum of the infinite series implied by that decimal:

if x= 0.abc... then x= (a/10)+ (b/100)+ (c/1000)+ ...

 

[Hurkyl]

When you see x = some-repeating-decimal, it actually means x = limit(some-function-that-would-generate-that-repeating-decimal)?

Yes, that is one way to define the decimals.


Incidentally, the decimals can instead be defined without resorting to calculus simply by specifying how to do the arithmetic with infinite strings of digits. (which includes things like 0.999~ = 1)

 

[HallsofIvy]

Then you are doing better than I could! I couldn't figure out what was meant by "some-function-that-would-generate-that-repeating-decimal"!

 

[Hurkyl]

I presume he means something like the sequence of partial sums of that series.

I.E. f(1) = 0.9, f(2) = 0.99, f(3) = 0.999, ...

 

[Sydius]

Yup, that is what I meant. I get it now. Thanks.

 

[moose]

I guess I will never believe .999 repeating = 1

I mean, if something asymptotes at 1, it gets infinitely close to 1, but it never actually reaches one... you can have .999 repeating but not 1

.99999 repeating is infinitely close to one, but how can it be one? I have read all these proofs, but for example, when you take the sum of the infinite series, it's always defined as approaching that number but not actually being that number soooo, i don't know I will just always find there is something wrong in my head...

 

[Hurkyl]

In the real numbers, the only infinitessimal number is zero. Therefore, if x is infinitely close to y, then x - y = 0, and therefore x = y.

but how can it be one?

The same way the fraction 2/2 can be one -- their numerical values are equal.

the sum of the infinite series, it's always defined as approaching that number

No, it's not. The sum of an infinite series is nothing more than a number. A number cannot be "approaching" some other number.

It's the sequence of partial sums that is approaching something. It is true that no partial sum of the series \sum_{i=1}^{\infty} 9 * 10^{-i} will ever be equal to one. It is true that this sequence of partial sums becomes arbitrarily close to one. (Meaning that if you pick a positive number ε, then there is a partial sum whose distance to one is less than ε)

But the sum of the series \sum_{i=1}^{\infty} 9 * 10^{-i} is one.

 

[polyb]

Having randomly bopped into this one I think I may have a small point to make of this issue. You guys have done some pretty interesting twists and turns with your logic but you have seem to missed something very simple. So bear with me:

As stipulated, if:

1/3=0.33333...

and

2/3=0.66666...

then by a conclusion we say,

1/3+2/3=0.99999...

Which looks like

1=0.99999...

Which plenty of arguments were given to make this so, but here is my issue:

If

1=0.99999...

then

1-0.99999...=0 ; which is obviuously not true. So what are we missing? Perhaps the notion of an infinitesimal is applicable. If we say this:

1=0.99999...+dx,where dx is an infinitesmal number, then the equation holds true and does not leave us with the earlier contardiction.

But if this considered true, then might it better to say,

1/3=0.33333...+dx?

I don't know if you call this a definition but I did find this interesting enough to pass along.

Thoughts?

 

[arildno]

1-0.99999...=0 ; which is obviuously not true.

Wrong; it is true.

 

[HallsofIvy]

1-0.99999...=0 ; which is obviuously not true. So what are we missing?

Perhaps you meant to type "which obviously IS true"? If not then what YOU are missing (I don't know about anyone else) is that 1- 0.999...= 0 is completely true. "1" and "0.999..." are just different ways of expressing exactly the same thing. Their difference is exactly 0.

 

[arildno]

Consider the following, polyb:
1/2, 2/4 and 3/6 are merely different ways of expressing the same number. You don't have any problems with that, do you?

 

[russ_watters]

1-0.99999...=0 ; which is obviuously not true. So what are we missing? Perhaps the notion of an infinitesimal is applicable. If we say this:

1=0.99999...+dx,where dx is an infinitesmal number, then the equation holds true and does not leave us with the earlier contardiction.

But if this considered true, then might it better to say,

1/3=0.33333...+dx?

I don't know if you call this a definition but I did find this interesting enough to pass along.

Thoughts?

The part you are missing is that the "..." covers that "+dx" term that you are trying to add.