# Ending the 0.999~ = 1 debates

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lol YT! That really summed it up! That was just what I was gonna say!

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Don't we just say that 0.999..=1 because it doesn't really matters in calculations since you can ignore the small mistake you make? Same thing in chemistry, if something is small enough compared to another value we can neglect it, which doesn't mean it doesn't exist.. I think that 0.999.. is definitely NOT the same thing as 1 (it simply isn't, same thing like 1+1=1), but since the difference is so small you can say 0.999.. is EQUAL to 1 in calculations. It all depends how you look at it or what you have to compare it to..

No, 0.999 is definately 1. It's just that if you tried to write it down, it wouldn't be.

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yeah. everything i would say has been said. any number with a decimal form containing repeating block for example 0.12345234234234234234234 can be expressed as a fraction. conversly any rational number must have a terminating or a repeating decimal form.

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Proofs nicked from my pure maths lectures

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Thanks

Unfortunately it's still going on. Look at this:

And the resulting 9 page cataclysm on another forum (the World of Warcraft off-topic forum, as it were, one of the only other forums I look at and I don't know why...):

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It's basically a very silly argument. It's unfortunate such things happen in mathematics (like the definition of 0^0, and things like that) - but for some reason they persist, and I'm not quite sure why. Life would be much easier if we spent less time arguing about trivial things such as this.

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Hi,

When you formalyl define numbers, with the Peano axioms at the basis. The irrationals are defined as the limits of a sequence of rational numbers.

So the decimal representation of an irrational number itself is also defined and in this definition 0.999999999etc is equal to one. So the equality holds by definition.

Just like 1 + 1 = 2 holds by definition.

Mandrake

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i hate definitions

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Yes, definitions definately suck. Especially in analysis.

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Here's nice little proof (well, it's not laid out formally and it uses inequalities that really should be proved first (though they're easy inequalties to prove), but I've gauged it exactly for the sort of person who will argue 0.99.... is not equal to 1. Though are alot better and more comphrehensive proofs, people who've I've shown this one seem to liek it) that I came up with:

(Firstly just about every crank whose claimed that 0.999.. is not equal to 1 claims that there are nbo numebrs inbetween 0.999.. and 1, this proof really sets out to deal with this claim)

if x = 0.999... then there are 3 possible relations that x and sqrt(x) could have

1. sqrt(x) > 1

Howver if this is the case then 0< x < 1 must be UNTRUE as for any 0 < x < 1, 0 < sqrt(x) < 1. If this is the case x > 1

2. sqrt(x) = 1

If this is the case then x = 1 is the only solution, meaning that 0.999 does indeed equal 1.

3. sqrt(x) < 1

If this is the case then there is a number between 1 and x (as if 0 < x < 1 then sqrt(x) > x); what is the decimal representation of this number? Of cousre at this point someone could argue that in this case sqrt(x) = x, but this means that x msut equal 1 or zero).

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This just looks like Achilles and the turtle again (What can exist realistically vs. what can exist mathematically).

Does .9999999 = 1 ? The answer depends on what application of the value is in question.

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Does .9999999 = 1 ?

can't believe that this question is still being asked, with all of the posts and the proofs posted surely everyone can see that it is indeed true.

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This just looks like Achilles and the turtle again (What can exist realistically vs. what can exist mathematically).

Does .9999999 = 1 ? The answer depends on what application of the value is in question.

Well from a mathematical sense it's certainly true.

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Well from a mathematical sense it's certainly true.

That's exactly what I said!

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I don't know if this helps at all, as Blike (sort of) said it.

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

In decimal form:

1.0000 = .33333 + .33333 + .33333

If you line those numbers up to add vertically,

  .3333[u]3[/u]
.3333[u]3[/u]
+ .3333[u]3[/u]
________
.9999[u]9[/u]

Adding vertically, it never rounds up. So .9 = 1

I realize this is hard to understand. When I was younger, my math teacher didn't understand me and I tried forever to prove it to him.

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That's exactly what I said!

I was trying to clarify your point, sorry if it came out otherwise

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I understand exactly what everyone has said and can agree. I guess the biggest problem is that it is ingrained in our minds for so long that 1>.9 and so anything after the .9 brings you closer to 1 but you'll never reach it, just get closer. Until you get more and more advanced, you are presented no reason to change that view. Then you get some more complex stuff like you guys all explained and you see why it was wrong.

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Think of it in this kindergarten point of view. If .999... isnt equal to 1, then what is the outcome or difference? 1 - .999... = ?

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A small number that ends with something like 0.0001 for the answer to 1-0.9999. The numbers just get smaller as the number of nines increases

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So then $.00\bar01$ is equal to zero also? Assumably by all the same proofs above.

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Nope. No matter how many zeros come before the one, it will never equal to zero, it will just get closer

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So you're taking the stance that $.999\bar9$ doesn't equal 1?

Im guessing so.

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If you're assuming that 1 - 0.999 = 0.0001, then of course it will equal zero. Here's the simplest reasoning I can think of why:

0.0001 is a decimal point followed by an infinite number of zeroes, with a 1 at the end.

But infinity has no endpoint, hence you never, ever get to the 1. So it doesn't exist.

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Nope. No matter how many zeros come before the one, it will never equal to zero, it will just get closer

You're completely misinterpreting how recursion works. It's not '0.33.. that's not it, lets add another 3!' it's infinity from the off. It doesn't change over time, so there can be no 'getting closer'.

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