Does .9-repeating equal 1?

[Lul Thyme]

Originally posted by rah
Yes, they're mathmatically the same, but to a programmer, you have to be careful because it may cause you problems since some programming languages deal with some of these issues differently. So I can say yes and sometimes NO.
example.
In SAS, (a statistical language) SAS WILL see .9999...(to as many as the max precision in the language) as equal to one and if you base triggers off the comparison, they will work just fine. But if you do .33333.... and compare it to 1/3, it will not consider them equal. (which was a common proof used in many examples here.) If I then multiplied it by 3, it did recognize it as 1. Hmmmm
Now granted the programs is not set to handle infinite precision.

So the final point is, yes, they're the same, but in THE REAL WORLD, you have to verify that the tools that you're using also believe this truth.

Computers can't be wrong

Im sorry Rah but youre wrong.
".9999...(to as many as the max precision in the language)"
is NOT .999...
it's .9999....9!
where the number of 9s depends on the precision of the language.
Same with .33..
You even mentionned the programs dont handle infinite precision which is what is the subject here.

So again, like many others in the thread, you are talking about different objects.

[Lul Thyme]

Originally posted by rah


While in reality you don't deal with .999...... much, you do deal with .333...... quite often.

In "reality", you deal with .999... just as often as .333...

[loinburger]

It's easy to precisely represent .333... in a programming language -- just represent it as a numerator over a denominator.

Irrational numbers are a bit trickier, though. Probably easiest to just represent them as an interval -- that way you might not get a precise answer, but at least you won't get the wrong answer.

[Kuciwalker]

For any useful irrational number you can probably store a formula or algorithm to generate its digits. At this point you might as well just use a computer algebra system.

[Adalbertus]

Originally posted by loinburger
It's easy to precisely represent .333... in a programming language -- just represent it as a numerator over a denominator.

Yeah. Represent .333... as 1/3 and .999... as 3/3.

[Whoha]

in floating point representation both would be represented differently.

[OzzyKP]

Wow, 231 posts, this very well may be the nerdiest thread in the history of Apolyton.

[Dis]

I'm sure we've had nerdier.

[Donegeal]

Yes we have. Here, I'll show you:

It was titled Star Wars Episode IV: A New Hope!!!






(that oughta make sure this thread makes it to 500)

[SlowwHand]

Is that where Chewy was portrayed as Gay?

[Perfection]

it's as much 1 as 3-2 is.

[Lord Avalon]

I don't let facts - or mathematical proofs - get in the way of the truth. Point nine repeating feels like a different number from one, so I say no, they're not equal.

And don't you bleeding-heart liberals get your panties in a bunch about another "equal rights" amendment for numbers.

Moving on...
[/Stephen Colbert]

[Sava]

Originally posted by KrazyHorse


Find me the number c such that 1 - 0.9999.... = c

0.00000[...]00001



0.99999... > 1

It's infinitely close to one, but it still isn't one.

[loinburger]

Originally posted by Sava
It's infinitely close to one, but it still isn't one.

One number cannot be "infinitely close" to another number, unless you're using a bizarro number system that isn't the real number system that we all know and love.

I'm not an analyst, but the quick and dirty demonstration as to why one number cannot be "infinitely close" to another number is: if r1 and r2 are distinct (i.e., not-equal) reals, then (r1 + r2) / 2 is a real that lies between r1 and r2 (hence r1 and r2 are not "infinitely close" to each other). If (r1 + r2) / 2 = r1 or, equivalently, (r1 + r2) / 2 = r2 then r1 and r2 are not distinct reals, because if (r1 + r2) / 2 = r1 then r1 + r2 = 2*r1 and then, subtracting r1 from both sides, we get r1 = r2. And, of course, if r1 and r2 are "infinitely close" to each other then (r1 + r2) / 2 = r1 OR r2, because otherwise r1 and r2 are "finitely close" to each other (i.e., (r1 + r2) / 2 = a finite interval between r1 and r2, which implies that r1 and r2 are not "infinitely close" to each other).

So, if r1 is "infinitely close" to r2, then r1 = r2.

[Kuciwalker]

0.00000[...]00001

That number doesn't exist. Proof:

Let X be the smallest number greater than zero. The set of real numbers is closed under multiplication, so if X in the set of real numbers, so is X*(1/2). However, 0 < X*(1/2) < X. Contradiction.

edit: this is essentially the same as loinburger's proof.