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Author | Topic: A question of numbers (one for the maths fans) | |||||||||||||||||||||||||||
kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
you subtract infinity from inifinity and get zero.
You were asking what ∞ - 1 was. And that's just it. ∞ - 1. Same holds ture for ∞ + 1. It equals inifinity. All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
10/3 can be represented by decimals. 10/3 is just more conrete.
.99999999999999999999 does equal 3/3. .33333333333333333333 = 1/3. .33333333333333333333 * 3 = .999999999999999999999.33333333333333333333 * 3 = 3/3 = 1. assume all the long decimals strecth for infinity. this means that 3/3 = .999999999999999999999999, which means that .99999999999999999999 = 1. if A = B, and B = C, then A = c.if (A).99999999999999999999999999 = (B)3/3, and (B)3/3 = (C)1, then (A).999999999999999999999 = (C)1. Edited by kuresu, : fixed logic proof, so that A = C, instead of A =B at the end. All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
I'll amswer both questions.
what times 2 = .99999999999999999999999999?.5 * 2 = .999999999999999999999999999, because .999999999999999999999 = 1 .99999999999999999999999 * 2 = ?.99999999999999999999= 2, because .9999999999999999999 = 1. Anywho, these calculators have been programmed by mathemiticians and are notorious for getting the right answer. So I'll trust them to get the answer to whatever I put in them. I just have to put in the right data. All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
actually, Mod's equation makes perfect sense.
This is going to asssume you have no basic algebra skills, just for simplicity.
x=0.999999... 10x=9.999999... 10x-x = 9.999999... - 0.999999... 9x = 9 Therefore x=1 What you do on one side of the equation, you must do to the other side of the equation. We establish that the variable, x, is equal to .999 . . .If we multiply .999 . . . by 10, we get 9.999 . . . If we subtract .999 . . . from 9.999 . . ., we get 9. On the other side of the equation,if we multiply x by 10, we get 10x. Subtract x from 10x, and you will get 9x. then, 9x = 9. Now then, all that's left to do is find out what x is.To do that, we divide the 9 (that's attached to the x) into the other 9, which gives us x = 9/9, or x = 1. So, 1 = .999 . . .
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
+ 1 - = 0 I do believe that that should equal 1, not zero, but you might be right because adding one to infinity would give you infinity. Well, let's take a look at your .333 . . . problem. x = .333 . . .10x = 3.333 . . . 10-x = 3.333 . . . - .333 . . . 9x = 3 x = 3/9 = 1/3. 1/3 = .333 . . . wheras in the other equation, 1 = .999 . . . All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
10x is just easier to use. But I'll go ahead with 2x.
x = .999 . . .2x = 2 x = 2/2, or 1. x = .333 . . .2x = .666 . . . x = .666 . . ./2, or .333 . . ., or, 1/3. .333 . . .does resolve, as a fraction. if you're looking to get it to resolve as a decimal, it will keep going for infinity. All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
THe original equation, where you multiply by 10 proves that .999 . . . is 1.
.999 . . . * 2 does not equal 1.999 . . .. It would make less sense if this were so. Now, .999 * 2 is 1.998, because you are dealing with a non-recurring decimal.
x = .999... 2x = 1.999... 1x = 1 and you still end up with 1 = .999 . . . in that equation.if x = .999 . . ., and if 1.999 . . . divided by 2 is 1, then x = 1 and .999 . . .. All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
From a quick glance, those equations appear to be slightly wrong.
x=3 .5x=1.5 -.5x= -1.5 you divide the 1.5 by .5, not multiply by -1, to find out what x is.x = 3, in that equation at the end. x=.999... .5x=.4999 (5?)... or maybe just .4999... let's use .4999... -.5x = -.499 x=1 Here again, you do not multiply by -1. You would divide .499 . . . by .5 to get 1. And you should use .5, instead of .499 . . .. reason being, is that .499 . . . is not half of .999 . . ..(number in line above have recurring decimals) .499 is not even half of .999. .4995 is half of .999 (numbers in line above are finite) x=.333... .5x=.166... -0.5x=.166 x=?
okay, this equation needs to be fixed. x = .333 . . ..5x = .166 . . . x = .333 . . . You have to divide the .166 . . . by .5 in order to find out what x is.
Shouldn't the 10x in mod's equation be 9.999...0 - .999... be 8.999...1 This is going to sound wierd.No. first, you would have to subtract by 1, but you still wouldn't get 8.999 . . .. That's because .999 . . . is 1. So you are in effect subtracting 1 from 10, without rounding the numbers. Now then, if the number was 9.999 and you subtract .999 you will get 9. If you subtract 1 from 9.999 you will get 8.999. That's because all the numbers in this segment have decimals that don't repeat. All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
I wasn't saying that yuor answer's were wrong--well, not yet anyways. It was your setup that was wrong.
There's no reason to multiply both sides of the equation by -1, and that's what you're doing. Also, without that last step of isolating the x by division, we've no clue what x is (unless we do it on our won, like I did).
why can't .333... be .4 In order for .333 . . . to be .4, it would have to .3999 . . ., but .3999 . . . is not .333 . . ..
x should always = 1 if the formula is a valid way of proving .999... = 1 and it is, if the equation you are referring to is the one we've been using the entire time. It doesn't matter what you multiply x by, you will always get 1 = .999 . . .. You say it doesn't--come up with a problem where 1 does not equal .999 . . . using that equation. All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
we could at any point in the equation replace .999... with 1, correct? That is correct. Just keep in mind, when replacing .999 . . . with 1, you have to add that 1 to the number that originally had .999 . . .. Otherwise, you would have:8.999 . . . (replace .999 . . . with 1) 8.1, which doesn't work out. SO just add the 1 to 8, and you'll get 9. All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
x=.999... .5x=.4999 (5?)... or maybe just .4999... let's use .4999... -.5x = -.499 x=1.0020040080160320641282565130261 You said you changed it, so I took another look--you now have the wrong value for x. Your equation is right up until you've added the negative (-) signs to both sides. First off, the (-) isn't necessary. Second, why is -.499 not -.4999 . . .? There's a big difference. Third, you divide -.499 by -.5, which will give you .998 (refer to second criticism). What you did was to divide -.5 by -.499, which would make your equation look like this: -.5x/-.499 = 0. It should look like this:x = -.499/-.5 All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
It's a little wierd, I'll admit that.
It's a matter of precision--and I like precision. Besides, that you you avoid making the mistake of writing .499 instead of .499 . . .. Technically speaking, .499 . . . does not equal .999 . . ., because.499 . . . is actually .5. And you'll never get .999 . . . to be .499 . . . when dividing by 2. It will always be .5. Just a matter of precision All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
It's partially my fault.
If you divide .999... by 2, you will get .5 or .499....But since .499... = .5, it's easier to write .5 than .499.... And my calculator will never give me .499... as the answer to x = .999.../2. Just .5. I like precision, but if .5 is = .499..., then why not write .5?Besides, it helps prevent you from screwing up and writing .499 instead of .499...--and that can be a major difference. So technically, you are right. It's just better to write .5 in my opinion. All a man's knowledge comes from his experiences
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kuresu Member (Idle past 2540 days) Posts: 2544 From: boulder, colorado Joined: |
No, you missed what I said.
When I was evaluating one of your equations, you used .499... at one point, and .499 at another. The difference is important. Because if the number is .499... (or .5), then it is half of .999... But if the number is .499, then it is NOT half of .999.... Somewhere along the line, you wondered if .499 was half of .999. And I said no--.4995 is half of .999. The important fact here is that these decimals are nonrecurring. They stop. Which changes things completely. The reason I make the jump from .999... to 1 is because of Modulous's equation that proves that .999... = 1. The same equation can be used to prove that .499... =.5. Keep in mind, my calculator, a Ti-86, is programmed by professional mathematicians--and if they can't do the actual programming, then they explain the operations to the programmer. Either way, my caculator is right, because it follows the very same rules that Modulous used to prove that .999... IS 1. All a man's knowledge comes from his experiences
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