Does 0.9999... = 1?

[Maurepas]

This is apparently a major issue among many a forum populace, but for the life of me I can't figure out why. To me, and my admittedly not quite complete mathematical skills, .999...repeating would require an additional .111...repeating to equal 1, yet, for some reason this is a major debate.

And to both satisfy my own curiosity, as well as meet my quota of at least 1 thread creation per year so I can say I contribute, lol, I ask you, the denizens of NSG:

Does 0.9999...Repeating = 1?

[AiliailiA]

Yes.

Next!

[Kemal Ataturks left sock]

I was gonna post this yesterday after reading a Cracked article. Decided against it, but may as well throw in that 0.999... does not equal 1.

[South Norwega]

Oh, that Cracked article, yes?

I'm not really convinced that it matters, the difference is so great that it doesn't actually exist, given there is nothing that can be added to 0.999... to make it equal one, as the 9s are infinite and you'd need infinite-1 zeroes and then a 1 on the end, after the not quite infinite zeroes, which destroys everything.

[Parhe]

They are the same thing, there exists no difference between the two except how it is written.

[AiliailiA]

I was gonna post this yesterday after reading a Cracked article. Decided against it, but may as well throw in that 0.999... does not equal 1.

1 - 0.9 = 0.1
1 - 0.99 = 0.01
1 - 0.999 = 0.001
...

Look at the right hand side. For each step n (n=1, n=2, n=3 in the first three terms) the RHS is 1 ÷ 10ⁿ

as n --> ∞, (1 ÷ 10ⁿ) --> 0

The difference between 0.999... and 1 is 0. Therefore they are equal.

[Timurid Empire]

If 0.9999 was equal to 1, it would just be 1 wouldn't it?

[Extollo]

This is apparently a major issue among many a forum populace, but for the life of me I can't figure out why. To me, and my admittedly not quite complete mathematical skills, .999...repeating would require an additional .111...repeating to equal 1, yet, for some reason this is a major debate.

Not quite. To be precise, it would require matching infinite .01 to the infinite .99. There's problem with that.

[Strawberry milk shakes]

It is the summation of 9/10^n from n=1 to infinity. The sum ((9/10)/(1-1/10)) is 1. So yes they're equal.

[Extollo]

If 0.9999 was equal to 1, it would just be 1 wouldn't it?

Similarly, .25 and 1/4 are the same.

[Timurid Empire]

If 0.9999 was equal to 1, it would just be 1 wouldn't it?

Similarly, .25 and 1/4 are the same.

You cheeky bastard!

[Carcharhinidae primari]

This is apparently a major issue among many a forum populace, but for the life of me I can't figure out why. To me, and my admittedly not quite complete mathematical skills, .999...repeating would require an additional .111...repeating to equal 1, yet, for some reason this is a major debate.

Not quite. To be precise, it would require matching infinite .01 to the infinite .99. There's problem with that.

yes... though i think there comes a point (when the number of decimals exceeds the amount of particles in the known universe for instance), when it becomes pointless to differentiate. the difference is no longer 'significant', in it's scientific meaning.

[Extollo]

Also, FWIW, the Wikipedia article on it is informative.

[Extollo]

Not quite. To be precise, it would require matching infinite .01 to the infinite .99. There's problem with that.

yes... though i think there comes a point (when the number of decimals exceeds the amount of particles in the known universe for instance), when it becomes pointless to differentiate. the difference is no longer 'significant', in it's scientific meaning.

Yes, that was sort of my point.

[Sucrati]

Yes, if you happen to round to the biggest integer, by itself, it really isn't but it is commonly taken as equal.

Now, that's like asking, does Pi really equal 3.14? No, it does when rounded but the number for Pi is VERY long and we don't care and/or are too lazy to count that high.

[Procrastica]

If 0.9999 was equal to 1, it would just be 1 wouldn't it?

Using that argument, if 2-1 was equal to 1, it would just be 1, wouldn't it?
We sometimes use different ways of writing the same thing to make different points.
Sometimes it's convenient to think of 1 as 0.999...

[Great Malema]

I think I'm having a George Bush moment here.

[Carcharhinidae primari]

Yes, if you happen to round to the biggest integer, by itself, it really isn't but it is commonly taken as equal.

Now, that's like asking, does Pi really equal 3.14? No, it does when rounded but the number for Pi is VERY long and we don't care and/or are too lazy to count that high.

or our measurements reach a point where the accuracy starts to degrade. which is the whole point of the scientific notation.

[OMGeverynameistaken]

If .999999...=1, then is Pi exactly 4?

[Extollo]

Similarly, .25 and 1/4 are the same.

You cheeky bastard!

I tend to be!

Basically, just a reminder that in math there is often more than one way to skin a cat.

[AiliailiA]

Yes, if you happen to round to the biggest integer, by itself, it really isn't but it is commonly taken as equal.

No.

0.999... isn't an approximation (any more than any other repeating decimal got by an arithmetic process). It has a precise value and that value is 1.

---

Here's one I don't get:

Repeating decimals like 1.0909... can be expressed as a fraction of two integers (that one is 12 ÷ 11). So what's the fraction for 0.888... ?

EDIT: D'oh. viewtopic.php?p=7237628#p7237628

0.888... = 8 x 0.111... and see Dazchan's post.

[Risottia]

This is apparently a major issue among many a forum populace, but for the life of me I can't figure out why. To me, and my admittedly not quite complete mathematical skills, .999...repeating would require an additional .111...repeating to equal 1, yet, for some reason this is a major debate.

Because it's wrong.

0.99999..... + 0.11111... is GREATER than 1.
Why?
0.99 + 0.11 = 1.10 , right? But 0.99999... > 0.99, and 0.11111... > 0.11 . Hence 0,99999... + 0.11111... > 1.10 . (That's because the set of the real numbers is an "ordered field", quod wiki).
And 1.10 > 1 . QED.

Does 0.9999...Repeating = 1?

Yes.


To have a hint about why 0.99999... = 1, just take this reasoning (not a proof though... wiki about "Dedekind cuts" and "cardinality of R" for formal proofs).
1 - 0.9 = 0.1
1 - 0.99 = 0.01
1 - 0.999 = 0.001
etc, that is:

Let n be a positive integer; let p be a rational number written as 0.(digit "9" repeated n times). Then 1 - p = 0.(digit "0" repeated n-1 times and a final digit "1") = 10^-n (10 to the negative n-th power)

Now, when you write 0.9(repeating) or 0.1111..., you mean that the number of digits after the decimal point (aka, n) goes to infinity. Hence, if you take p as 0.99999(repeating),

1 - p = lim_n->infinity(10^-n)

which is, of course, 0 (take a pie, divide it in an ever-increasing number of slices: the slices become thinner and thinner, aka the size of the slices goes towards 0).

So we've found that, for an ever-increasing number of digits n,

1 - p = 0, that is, 1 = p.

But p was defined as 0.99999..., hence 0,99999(repeating) = 1 .

[Extollo]

So, really, in relation to another popular thread:

.999... does not simply walk into Mordor.

[Meridiani Planum]

This is apparently a major issue among many a forum populace, but for the life of me I can't figure out why. To me, and my admittedly not quite complete mathematical skills, .999...repeating would require an additional .111...repeating to equal 1, yet, for some reason this is a major debate.

No, adding .111... to .999... would yield 1.111...

Even adding 0.111 (non-repeating) to 0.999 (non-repeating) yields 1.11. Do the math.

Does 0.9999...Repeating = 1?

Yes, it sure does. But keep in mind one very important point. You cannot represent 0.999... in terms of writable symbols. It's a conceptual thing.

[Kemal Ataturks left sock]

1 - 0.9 = 0.1
1 - 0.99 = 0.01
1 - 0.999 = 0.001
...

Look at the right hand side. For each step n (n=1, n=2, n=3 in the first three terms) the RHS is 1 ÷ 10ⁿ

as n --> ∞, (1 ÷ 10ⁿ) --> 0

The difference between 0.999... and 1 is 0. Therefore they are equal.

You lost me, sorry.