Thread: .999~=1!

To prove it in a different way, you can actually use the formula for calculating the sum of an infinite geometric progression. The initial term would be 0.9 and the ratio would be 0.1:

S = (0.9 + 0.09 + 0.009 + ....)

S = a1 / (1 - q)

S = 0.9 / (1 - 0.1)

S = 0.9 / 0.9

S = 1

Yesh that's a nice way, or you can even go back to the basics of proving the GR formula:

S = 0.9 + 0.09 + 0.009 + ...
0.1S = 0.09 + 0.009 + 0.0009 + ...
S - 0.1S = (1 - 0.1)S = 0.9S = 0.9
=> S = 1

It really comes down to math vs. logic. Math tells us that it does equal 1, but logic tells us just the oposite.

As Kromoh said, you're talking about intuition, and the problem is that intuition is subjective, unlike mathematics. As a matter of fact I find it totally intuitive that 0.999... is equal to 1, as the number is infinitely close to 1 and hence equal; I do not understand anybody who finds the opposite intuitive.

That's why we have mathematical proof anyway. Sometimes results are surprising.

I wasn't expecting serious answers in SB -_-

Originally Posted by hellohihello

I wasn't expecting serious answers in SB -_-

I may be guilty on that one +.+

0.999~ = 0.9 actually. :V

You can't comb the hair of a spherical dog without introducing a part

When I saw the post I though "well, that's all wrong. It can't be"
Then I actually read some of the posts, and it makes perfect sense. Why do people continue to deny it? >.<

The complex numbers are isomorphic as a ring to R[x]/(X^ + 1)

A vector space is just a ring homomorphism from a field into the endomorphism ring of an abelian group

A torus has zero global curvature

You can't comb the hair of a spherical dog without introducing a part

This is also true of spherical cows.

Another interesting fact about spherical cows is that you can split the cow into a finite number of pieces, reassemble it, and you'll have two new spherical cows of exactly the same volume as the first one.

This assumes that the cows are continuous.

Originally Posted by khh

When I saw the post I though "well, that's all wrong. It can't be"
Then I actually read some of the posts, and it makes perfect sense. Why do people continue to deny it? >.<

And that's how math works. Isn't it pretty? You really can't disagree.

Originally Posted by Xei

This is also true of spherical cows.

Another interesting fact about spherical cows is that you can split the cow into a finite number of pieces, reassemble it, and you'll have two new spherical cows of exactly the same volume as the first one.

This assumes that the cows are continuous.

And that you want to work your way through all those homeomorphisms

Actually I disagree with the axiom of associativity.

Prove it now, dick.

And that you want to work your way through all those homeomorphisms

And udders. :V

let's see, associativity....that's (a + b) + c = a + (b + c). It's gonna take me a second and I don't have my book that covers the fundamentals so I might not be able too. I'll do it in pieces.

we start with the symbols { and }. We introduce an operation, lets use S, which takes the given symbol, x, and returns {x} so that S(x) = {x}

We call S the increment operator and specify the rule that every { must be closed by a }

Fuck it. I haven't looked at that shit in ages. The problem is that to prove it (I'm sure you were being sarcastic) you have to construct the naturals from sets and that is frankly a pain in the ass that a lot of mathematicians couldn't do off the top of their head

I am a little embarrassed though....

I thought associativity is unprovable...

Perhaps not, dunno. Ask me in three years when I've got a title. :V

You studying maths Xei? Didn't know that.

It works like this. We collect a set of axioms and prove things assuming them. The axioms that describe the integers are collected together and called the ring axioms. The ones that describe the rationals and reals are called the field axioms. A set that satisfies them is called a ring or a field respectively. (only a ring doesn't necessarily need to be commutative in multiplication, only addition. If it is then it's specifically called a commutative ring)

Some other examples of rings are the ring of polynomials in the coefficient X under polynomial addition and polynomial multiplication, the ring of N x N matrices under matrix addition and multiplication, etc. If you want to apply the theorems that apply to a ring, field, group, etc., you have to prove that the set that you want to apply it to satisfies those axioms.

It is almost mandatory to assume that the number systems do satisfy them to learn basic math but it is necessary to construct them as I began to do above once you start to really get into it. You do it once in your undergraduate education (maybe) and then forget about it and take them as axioms It's pretty dull and boring honestly.

A really good book is "Linear Algebra: An introduction to Abstract Mathematics" It doesn't prove that the axioms apply to the number systems but he does it for matrices ( and polynomials I believe) and it is a great introduction to the concepts. I think that you would get a lot out of it. He only assigns one computation in the whole book. All the rest of the problems are proofs.

If you want to see the numbers built up from scratch, check the book "Introduction to Algebra" by peter cameron. It has one chapter about it. The rest is algebra. He's at oxford and it's on oxford press. I'm not sure if it's still in print but you should be able to find a used copy. It's one of the best introductions to algebra in existence. I have no idea why it went out of print. If you can't find it, try "Undergraduate Algebra" by serge lang.

Algebra is my big thing. I love the stuff.

Thanks PS, I need to get some summer reading done actually. I prefer algebra to most other aspects of mathematics, although I understand it's a totally different beast in higher education. I think the thing I've enjoyed most so far was learning about the Maclaurin series and how they can prove euler's identity and then how you can use that for integrals and trigonometric sums and things.

You studying maths Xei? Didn't know that.

Yeah and planning on mastering in neuroscience. Cambridge or Imperial College London, gonna be fun.

Neuroscience, sweet. I'm gonna study medicine, specialize on psychiatry, then post-graduate on neuroscience. Maybe we'll meet one day ^^

Originally Posted by PhilosopherStoned

I think you meant hyberbolas and that can't be represented by an elementary function in the way that I think you are thinking, it can be represented as a vector valued function or as the set of zeros of a polynomial. If we adjoin the "line at infinity" and pass to projective coordinates, then it does have a value at infinity, namely 0, and is really just an elipse that intersects the line at infinity in two places. The fact that it doesn't have a value at infinity is really just a defect of the non-projective coordinates. A parabala, in that case also has a value at infinity and is just an ellipse that intersects the line at infinity twice at the same place in the same way that (x - 0)^2 = x^2 intersects the x axis twice at 0.

Ok, all is well enough, but, help me out with this particular problem. Let's say that I have,

1/(1-|x|)

I always imagine having different answers for x=1 and x=.999~
But the REASON I think that way is because we are only always able to portray the infinite string of 9's as a limited string (say ten, a hundred, or a thousand decimal places) in any equation, so I've had the bad habit of viewing .9999999999999999999999999999999999999999999999999 99999~ as a number that approaches 1 but is never at 1.

I can humbly admit that I was terribly, terribly wrong. But not without understandable cause!

Except parabolas don't even have any asymptotes.

Philosopherstoned already corrected me there, but there are no second place prizes. And I understand that the example isn't compatible with the problem. I'd imagine that one with a mind such as yours, Xei, would opt towards helping me without the negative attitude.

*gets popcorn and scratches head with confused look*