.9~=1

College Confidential Community
1

<p>Proof 1:</p>

<p>n=.9~
10n=9.9~
10n-n=9.9~-.9~
9n=n
n=1</p>

<p>(This one is disputed because it is believed that multiplying non terminating decimals is an illegal operation. However, I don't see the problem with multiplying .9~ by 10. Its a simple movement of the decimal place</p>

<p>Proof 2:</p>

<p>1/3=0.333...
2/3=0.666...</p>

<p>0.333...+0.666...=0.999...
1/3+2/3=3/3</p>

<p>The argument against this proof is really weak. The people against the concept of .9~=1 will say that decimals are approximations of fractions, and not exact, and that this "proof" is merely an exploit of the 10 base numbering system</p>

<p>Proof 3:</p>

<p>The first law of calculus states that in between any 2 real numbers is an infinite set of other real numbers. Not only is there not an infinite set of real numbers between .9~ and 1, there's not a single number between them.</p>

<p>There's no real argument to this proof, its detractors usually just recycle their points that .9~ always gets closer, but never reaches. </p>

<p>Everyone I've spoken to with actual credentials that has studied higher level math agress that .9~=1. Are the people who think they aren't equal just being stubborn, or do they have a case?</p>

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<p>I thought everyone knew this.. .9~ (repeating to infinity) is obviously 1.</p>

3

<p>what's ur point. .99999~ is an infinite geometric sequence and converges to 1. So it is one. Also there's a simpler version of proof.
.99999~=x
9.9999~=10x
subtract from the top
9x=9
x=1 so .9999999~=1</p>

4

<p>9n=n</p>

<p>you wanna correct that to 9n=9, btw. :)</p>

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<p>imiracle.. isn't that the exact same proof he just did? (except with n instead of x.. lol)</p>

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<p>I wish the people who thought up the college grade point system thought like you guys. <em>sigh</em> I would kill for ability to turn a 3.99999 back into a 4.0. :-)</p>

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<p>n=.9~
10n=9.9~
10n-n=9.9~-.9~
9n=n
n=1</p>

<p>The only part that bugs me is step 3. 10n-n=9.9~-.9~
You subtract n from one side. This means you subtract it from the other side. However you are already setting it up by saying that it is the same as subtracting .9~. So you're saying n is .9~ here just so the variables can be eliminated to reveal that n can be 1 later. It seems like a cheap "proof" to me. Although I do agree that .9~ is 1, I don't think this is the best proof.</p>

<p>The only reason this works is because you are able to multiple n by ten and the other side by ten while being able to manipulate the equation by allowing for the placement of a .9... although technically you can say that on that same note, you can add the placement of a .0 on the 10-side, which allows it to work.</p>

<p>I don't know what I am talking about. Ignore me.</p>

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<p>oh he did? didn't c that lol</p>

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<p>well u could think of the infinite geometric sequence proof. mind me for being redundant.</p>

<p>.99999~=.9+.09+.009+.0009~=sigma from 1 to n 9(.1)^k=.9/1-.1=1</p>

10

<p>.9 != 1</p>

<p>Just a flaw with the denary (base-10) system</p>

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<p>Pff... all the fancy math in the world won't change the fact that .9~ is .0~1 less than 1.</p>

<p>...</p>

12

<p>.0~1 can't exist though, that's the problem.</p>

13

<p>It can exist the same way .9~ exists. Can't it?</p>

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<p>No. Where do you put the one after an infinite series of 0's? Where ever you put the 1, you've broken the infinite set of 0's.</p>

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<p>Actually, .000~1 is 0 if you look at it mathematically. What is .9999 etc. etc. In terms of series it is the sum from n = 1 to infinity of (9^n divided by 10.) We see that the partial sums converge to 1. In the same manner, the sum from n=1 to infinity of (1^n) over ten is equal to .000000~1 or if we decide to find the limit of the partial sums, 0.
There is a more mathematically rigorous theories of infintesimals (more so than the epsilon delta calculus) Google: Nonstandard Analysis if you are interested.</p>