<p>whoops! I see his first one now, but his second one still doesn't make much sense...</p>
<p>Once again, problem is you are multiplying an infinite term by something that is not infinite. It just creates problems.</p>
<p>You cant say 2 * inf = 2inf. It's a concept, not a value you can apply mathematical computations to.</p>
<p>right I'll tell him that. I'm sure he'll care ;) But you do have a few interesting points.</p>
<p>There is no such thing as one short of infinity. I agree with both proofs so far. .999999.... is indeed 1. Like asking if .3333..... is 1/3. </p>
<p>Here's another way to look at it:</p>
<p>1/3 = .3333333... = 3/10 + 3/100 + 3/1000 + ....</p>
<p>Now multiply both sides by 3:</p>
<p>3/3 = 9/10 + 9/100 + 9/1000 + ....... = .9999.........</p>
<p>So 3/3=1 is .999.......</p>
<p>Ands its not an infinite term, just an infinite series. We deal with these all the time. When you take higher math courses you'll see how we can easily deal with them.</p>
<p>Look at randomperson's proof. Most (if not all; I skipped over the last few posts, but feel confident it's all) of the proofs offered so far rely on convergent series. However, randomperson offers an example where the convergent series methodology is unsatisfactory.</p>
<p>The multiply by 10 proof is valid. It does not add an extra term because there're an infinate number of 9s. So basically,</p>
<p>10*.9999~ has the same decimal places as .9999~ </p>
<p>3/3 = .99999~ = 1</p>
<p>
[quote]
But, if you meant that that was the "final answer" to the thread, then I'm just going to ignore you seeing that I don't believe you have the authority to "close" or end a thread just through your wise words
[/quote]
</p>
<p>Actually, the fact that .99999 repeating = 1 is universally accepted by mathematicians. If you get into Harvard, you can ask the professors there.</p>
<p>Mathwiz, you can't do that, I may not have put it eloquently, but still, your "proof" is severly flawed. And, you too, Aurelius are wrong.</p>
<p>...harvard's beyond me...i didn't even bother with applying</p>
<p>I'm starting to understand how .999~ = 1</p>
<p>So while everyone jostles for correctness, I'll pose a new problem:</p>
<p>Prove that 0 = 0.</p>
<p>And an even better one:</p>
<p>Given x, prove the universe is infinite.</p>
<p>Oh, I'm irreverent :-D</p>
<p>Guys, all these "proofs" involving multiplying .333~ by 3 are silly, not to mention more than a bit short of rigorous.</p>
<p>The issue here is that we can't represent some numbers with a finite number of decimal digits. Instead, we represent them with infinite series of decimal terms, series which converge to the desired results. In a sense, "converges to" does mean "equals." The problem is in how we treat these series. For instance, if we're taking a function of our infinite sum (say, a function f(x) that is discontinuous at x=1), we have to decide "when" we're taking the limit - before we evaluate the function or after it, because we'll get different results each way.</p>
<p>If we're taking a function of an infinite series, we should take the limit before we evaluate the function, so that f(.999~)=f(1), despite the fact that the function is discontinuous at 1. This means that, more or less, .999~=1. But you have to be careful, since the first is an infinite series and the second is a number, and my point is that doing things like this sloppily may get you into trouble. You need to know what and when you're evaluating. In any case, there's definitely more here than the handwaving that's going on.</p>
<p>It does, and the proof is below. '¬' refers to the squiggly line, but my keyboard is screwy.</p>
<p>Let x = .111¬, and .999¬ = 9x
.999¬ = 1
Divide both sides by 9, and you get:
.111¬ = 1/9
By substitution,
x = 1/9
Multiply both sides by 9, and you get:
9x = 1
From what we denoted above, 9x = .999¬
Therefore, .999¬ = 1</p>
<p>
[quote]
Actually, the fact that .99999 repeating = 1 is universally accepted by mathematicians. If you get into Harvard, you can ask the professors there.
[/quote]
</p>
<p>or ask your hs math teacher, heh :P</p>
<p>Logic is fundamentally flawed. Best not to bother really.</p>
<p>Somebody wants to write an essay on the non-uniqueness of the decimal representation of real numbers?</p>
<p>
[quote]
1/3 = .33333333333333
2/3 = .66666666666666
3/3 = .99999999999999 or 3/3 = 1
[/quote]
</p>
<p>Are you kidding me????</p>
<p>That's one of the worst "proofs" I've ever seen O__O
If one were to accept that 1/3=.333333 then that would be the same as accepting that 1 = .99999999</p>
<p>EDIT: Come to think of it, every proof on here is fundamentally flawed. I suppose it's just a matter of definition.</p>
<p>^^because .999999... does = 1. I was going to post the proof, but gospy got to it first. :)</p>
<p>If any of you are still unsure, just ask your math teacher.</p>
<p>Infinite series don't equal what they are solved down to, they are extremely close but not equal. </p>
<p>Again, all of these proofs take into account something which is infinite and slightly approximates.</p>
<p>BTW, I never said that .999~ doesn't equal 1.</p>
<p>Wooow. I hope this isn't a precursor for the world of academia. I suppose I've been naive and thought that people generally get along...</p>
<p>All of you have criticized one another's proofs without giving a solid proof of your own. Please, this isn't how you rebutt another's claims.</p>
<p>It's amazing how arrogant some people can be while spewing out garbage. If you want to see a conclusive proof go to <a href="http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1%5B/url%5D">http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1</a> . It is definitive, and correct.</p>