Thread: Does 0.9 repeated = 1?

Right.

All rational numbers can be written as whole number fractions. For example 4/2 = 2. Pi is irrational, and cannot be written as a fraction.

Originally Posted by xXSomeGuyXx

Okay, okay, okay.

I believe .999~ is .999~.

I believe 1 is 1.

It doesn't matter what you believe. .99... = 1 and this can be (and has been) proved. Whether or not you believe it doesn't change reality.

I'm not a troll. I'm having a debate. Plus, I actually am a nice guy. Trolls suck.

Originally Posted by xXSomeGuyXx

I'm not a troll.

You are spouting random crap about a topic you're obviously ignorant about. If that's not trolling that's just plain immaturity.

What?

2 is a rational number.

This is 2 in decimal form: 2.

It terminates. Thus, it is rational.

The definition of a rational number is that it is a fraction of two whole numbers. Decimals have no place in pure maths.

Your definition includes things like 0.12112211122211112222... which is not rational. In fact it's not even irrational.

You can also see the proof in certain graphs. 1/x-1 for example cannot equal 1, but its value rises or drops indefinitely as it approaches the asymptote. And no matter how close the value gets to 1 (in other words, how many times 9 is repeated), it is never equal to one.

Asking if .9999999~ repeated is equal to a rational number is like asking the same of infinity. The answer is that it's not a number to begin with, but an irrational system: a term we use to stand for something our minds couldn't normally contend with. Sort of like how we use i to stand for the square root of a -1 (which can't exist, but it's still an essential component of mathematics).

Originally Posted by Xei

The definition of a rational number is that it is a fraction of two whole numbers. Decimals have no place in pure maths.

Your definition includes things like 0.12112211122211112222... which is not rational. In fact it's not even irrational.

No, no. I mean repeating. Not patterns. Sorry, wrong word.

0.12112211122211112222 is irrational, because it doesn't repeat, and it doesn't terminate.

And yeah, I was just stating the definition I knew. You said "Fractions of whole numbers", which threw me off.

Originally Posted by hungrymanz

You are spouting random crap about a topic you're obviously ignorant about. If that's not trolling that's just plain immaturity.

Okay, I know absolutely nothing about maths. And I'm not immature. Trolling? Because I believe something else than you do, and am trying to prove it, you think I'm trolling? Yeah, okay. Remind me never to have a debate with you.

Originally Posted by Tamias.Squirrel

Asking if .9999999~ repeated is equal to a rational number is like asking the same of infinity.

NO.

The answer is that it's not a number to begin with, but an irrational system:

ABSOLUTELY NOT.

1/3 is clearly a real, rational number. Look at its decimal representation:

.333....

IT IS A NUMBER. Irrational system? It is one single number. It doesn't do anything. It is as solid as 2 or 3 or 1/5 or 3/7.

You can't take a third of a pie and then say that pie's calculating itself out to infinity. You just have a third of a pie. And if you have three thirds, you have one whole, equal to .999... of the pie.

Remind me never to have a debate with you.

Fine by me.

Originally Posted by xXSomeGuyXx

0.999~ doesn't terminate.

Rational numbers either terminate or have repeated decimals.

Originally Posted by xXSomeGuyXx

That isn't even close to algebra. That is a numerical expression, not an algebraic expression. ERGO, I do not suck a algebra. Test me. Give me a loong ass algebraic equation and I'll find x for ya.

Here's a couple that cropped up in my second year astronomy. Good. Fucking. Luck.

5[exp(x)-1] = x*exp(x)

and

(G*M*m) /((r–R)^2) – (G*m*n) /(R^2) = (m*v^2)/(r–R)
where your goal is to solve for big R as a function of G, M, m, n, v, and r, which can be taken as constants.

O_O See ya in a few weeks...lol. Those actually look fun.

Originally Posted by hungrymanz

NO.

You can't take a third of a pie and then say that pie's calculating itself out to infinity. You just have a third of a pie. And if you have three thirds, you have one whole, equal to .999... of the pie.

That's actually a good point xD Hmm... This is a lot more complicated than I first thought. Maybe we should all just agree that nobody really knows?

Lol, sure.

Originally Posted by xXSomeGuyXx

O_O See ya in a few weeks...lol. Those actually look fun.

Hint: There's no exact answer for either of them. Kinda blows your mind if you think about that fact.

EDIT: One thing. The first equation has a trivial solution x=0. I want the non-trivial solution.

Originally Posted by Tamias.Squirrel

That's actually a good point xD Hmm... This is a lot more complicated than I first thought. Maybe we should all just agree that nobody really knows?

No, we do know. It has been proven dozens of different ways, including real number construction, the be-all end-all proof of numbers, that 0.9~ DOES EQUAL 1.

Well, yeah., Ima go use a table of solutions.

What about this?

There are no numbers between 1.000...1 and 1, right?

So they are the same, because adding the one to the end is redundant.

So basically, .999... is just 1 - 0.000...1, which is zero, making .999... 1.

Originally Posted by drewmandan

No, we do know. It has been proven dozens of different ways, including real number construction, the be-all end-all proof of numbers, that 0.9~ DOES EQUAL 1.

He is right. Maybe you don't understand it, Tamias, but there are people who do. I encourage you to keep trying to understand it.

Question, exp(x) is to the xth power, correct?

Originally Posted by xXSomeGuyXx

Question, exp(x) is to the xth power, correct?

exp(x) is e^x, where e is the usual e

Originally Posted by drewmandan

No, we do know. It has been proven dozens of different ways, including real number construction, the be-all end-all proof of numbers, that 0.9~ DOES EQUAL 1.

But it doesn't: Like I said before, the graph 1/x-1 gets infinitely close to 1, but never, ever touches it. The y-values just keep going up or down. So no matter how close to one you 1, you are never exactly on 1.



I guess I don't get it.

Do you mean 5e^(x+1) = x.e^x ?

Or 5e^x + 5 = x.e^x?

Neither of these has solution x=0.

Originally Posted by Xei

Do you mean 5e^(x+1) = x.e^x ?

Or 5e^x + 5 = x.e^x?

Neither of these has solution x=0.

Learn2distribute minus sign

5[exp(x)-1] = x*exp(x)

Meaning 5*e^x - 5 = x*e^x, which has a solution x = 0.

Originally Posted by Tamias.Squirrel

But it doesn't: Like I said before, the graph 1/x-1 gets infinitely close to 1, but never, ever touches it. The y-values just keep going up or down. So no matter how close to one you 1, you are never exactly on 1.



I guess I don't get it.

What is this? Proof by screen shot? Wtf? By the way, 1/(1-x) has nothing to do with the constant number 0.9~. At all.

Originally Posted by Tamias.Squirrel

But it doesn't: Like I said before, the graph 1/x-1 gets infinitely close to 1, but never, ever touches it.

You have to understand that .999... is a number, not a function or a graph.

I understand that, hungrymanz.
What I don't understand is why this thread continues and people still view the issue as debatable despite being shown the 'proof' that they are the same. It is... fact. It's really that simple. Whether or not it makes sense is irrelevant.

Your logic on the issue is irrevelant.

Human logic means nothing next to numbers!

You are all powerless next to the awsome might of calculus! Understand?