• What is the last digit of 1/3
    44 replies, posted
This is bothering me way too much If a third of something is 33.33%, then that would mean a one whole is 99.99i%, right? I realize that there aren't only two digits in the end, there are probably a whole lot (and not just 16 as my calculator suggests) Does anyone actually know what that last number is?
It doesn't really end, so there is no last number. But stop anywhere and you'll end at 3.
[QUOTE=paul simon;43770191]It doesn't really end, so there is no last number. But stop anywhere and you'll end at 3.[/QUOTE] But that can't be If 1/3 is 0.3 and 2/3 is 0.6 Then with what magic is 3/3 = 1?
[QUOTE=gokiyono;43770207]But that can't be If 1/3 is 0.3 and 2/3 is 0.6 Then with what magic is 3/3 = 1?[/QUOTE] Yes it actually can be, because we're dealing with infinity here. 99.999.... etc is (mostly) considered to be equal to 100.
[QUOTE=gokiyono;43770207]But that can't be If 1/3 is 0.3 and 2/3 is 0.6 Then with what magic is 3/3 = 1?[/QUOTE] That's called math magic. [editline]3rd February 2014[/editline] Either way, 3 goes into 3, once, so it would be 1 regardless of how you look at it. And 1/3 or 2/3 is just 0.3333 and 0.6666 repeating infinite, respectively.
[QUOTE=paul simon;43770214]Yes it actually can be, because we're dealing with infinity here. 99.999.... etc is (mostly) considered to be equal to 100.[/QUOTE] I think I could accept that, was it not because 5/3 is 1.666666666666667 That just confuses me even more
[QUOTE=gokiyono;43770281]I think I could accept that, was it not because 5/3 is 1.666666666666667 That just confuses me even more[/QUOTE] 5/3 is just 1 and 2/3's. So it's gonna end with a 0.666 repeating. At least until you stop and round up.
[QUOTE=gokiyono;43770281]I think I could accept that, was it not because 5/3 is 1.666666666666667 That just confuses me even more[/QUOTE] The 7 at the end is just the calculator rounding it up, it's actually 1.66666... infinitely.
If you want the true answer, drop the calculator and try the problem on paper. In a week, come back and tell us how many pages you used.
It can't have a last number by definition, 1/3 has infinite numbers after the point Infinite means that for every number, no matter how far from the point, there exists a number that's even further from the point
You'll enjoy this video OP [media]http://www.youtube.com/watch?v=TINfzxSnnIE[/media]
[QUOTE=paul simon;43770214]Yes it actually can be, because we're dealing with infinity here. 99.999.... etc is (mostly) considered to be equal to 100.[/QUOTE] Not mostly. It is exactly 1. 1/3 isn't 0.3, it's 0.33333... and that continues to infinity. 2/3 is 0.6666666... and 3/3 is 0.999999999... which in a mathematical sense is equal to one, since with every 9 you add, you get closer to 1. When you add an infinite amount of 9s, you get infinitely close to one, which is actually equal to being one. Also 3 divided by 3 is 1 (and 3/3 is just a division).
99.999999999999999999999999999999999999999999999999999999999999 is pretty much 100
Computerphile actually did a video on floating point that is really about basically the same stuff you're asking: [video=youtube;PZRI1IfStY0]http://www.youtube.com/watch?v=PZRI1IfStY0[/video] This is simply a "problem" with base 10, but since we humans understand repeating numbers, it's not a problem. The point is that 1/3 isn't equal to 0.3333.. (it's damn close, though), but 3/3 [I]is[/I] equal to 1. So when you're multiplying 0.333333.., you're not actually doing 3*1/3 at all. Kinda like you can express the square root of two (√2), but you can't actually write it in base 10 decimal.
[QUOTE=gokiyono;43770207]But that can't be If 1/3 is 0.3 and 2/3 is 0.6 Then with what magic is 3/3 = 1?[/QUOTE] 1/3 = 0.3333..... 0.3333..... x 3 = 0.9999..... 0.9999..... x 10 = 9.9999..... 9.9999..... - 1(0.3333..... x 3) = 9 9 / 9 = 1 Therefore 0.9999..... = 1 (The way I look at it at least)
[QUOTE=GoDong-DK;43771705]Computerphile actually did a video on floating point that is really about basically the same stuff you're asking: [video=youtube;PZRI1IfStY0]http://www.youtube.com/watch?v=PZRI1IfStY0[/video] This is simply a "problem" with base 10, but since we humans understand repeating numbers, it's not a problem. The point is that 1/3 isn't equal to 0.3333.. (it's damn close, though), but 3/3 [I]is[/I] equal to 1. So when you're multiplying 0.333333.., you're not actually doing 3*1/3 at all. Kinda like you can express the square root of two (√2), but you can't actually write it in base 10 decimal.[/QUOTE] It's not really the same thing. That video talks about representing numbers with a finite number of digits, here we're talking about infinte representations. 1/3 is actually equal to 0.333... and any rational number can be expressed in base 10 exactly, as long as you can use infinite digits. The problem here is that base 10 is a somewhat flawed way of expressing rational numbers, and in most cases it's much more practical to express them as fractions. Any fraction whose denominator has prime factors other than 2 or 5 doesn't have a finite representation in base 10. What's more is that even those numbers that do have a finite representation actually have another, infinite representation with recurring nines as people have already mentioned. The lesson here is: Don't bother too much with decimal representations of rational numbers, use fractions instead.
[QUOTE=pebkac;43771970]It's not really the same thing. That video talks about representing numbers with a finite number of digits, here we're talking about infinte representations. 1/3 is actually equal to 0.333... and any rational number can be expressed in base 10 exactly, as long as you can use infinite digits. The problem here is that base 10 is a somewhat flawed way of expressing rational numbers, and in most cases it's much more practical to express them as fractions. Any fraction whose denominator has prime factors other than 2 or 5 doesn't have a finite representation in base 10. What's more is that even those numbers that do have a finite representation actually have another, infinite representation with recurring nines as people have already mentioned. The lesson here is: Don't bother too much with decimal representations of rational numbers, use fractions instead.[/QUOTE] I do know that 0.3333.. to infinity equals 1/3, but I suspect OP wants something a bit more tangible (writing it out on paper, finite decimals). He talked about his calculator, and he kinda insinuated that there is a last digit. I simply (tried to) explained that while his calculator might spell it out 0.3333333333 that's not actually 1/3 (though multiplying it with 3, the calculator will normally give the result 1), and why we have this problem (same reason why base two has problems with rounding errors that seem very obvious to us).
[QUOTE=Antdawg;43771847]1/3 = 0.3333..... 0.3333..... x 3 = 0.9999..... 0.9999..... x 10 = 9.9999..... 9.9999..... - 1(0.3333..... x 3) = 9 9 / 9 = 1 Therefore 0.9999..... = 1 (The way I look at it at least)[/QUOTE] These "proofs" of that fact are always so bad and unconvincing, since they involve manipulation of infinite decimals in ways which you just kind of assume are true. "0.9999..... x 10 = 9.9999....." Says who? In fact, when you use that step, you're [I]implicitly assuming[/I] 0.999... = 1, because if you replace 0.999... by x, you get x*10 = x+9, and it's pretty obvious the only solution to that is 1. That would almost look like an alternate proof that 0.999... = 1, except you inserted that 0.9999..... x 10 = 9.9999..... step without any justification.
[QUOTE=JohnnyMo1;43772093]These "proofs" of that fact are always so bad and unconvincing, since they involve manipulation of infinite decimals in ways which you just kind of assume are true. "0.9999..... x 10 = 9.9999....." Says who? In fact, when you use that step, you're [I]implicitly assuming[/I] 0.999... = 1, because if you replace 0.999... by x, you get x*10 = x+9, and it's pretty obvious the only solution to that is 1. That would almost look like an alternate proof that 0.999... = 1, except you inserted that 0.9999..... x 10 = 9.9999..... step without any justification.[/QUOTE] The implicit assumption is basically that, if a line of 9's reaching to the right goes on for an infinity, and you pull it one digit to the left; there will still be an infinitely long line of 9's reaching to the right. The line will be no shorter than before even though it was pulled to the left. So when you multiply 0.999... by ten, you pull it to the left so it becomes 9.999... The proof only works by accepting that there's still the same number of nines after the decimal point. If there weren't, there wouldn't have been an infinite number of 9's to begin with, which has to be the case. It's just infinity being weird, see [URL=http://www.youtube.com/watch?v=faQBrAQ87l4]Hilbert's Infinite Hotel paradox.[/URL] Also just leaving this here: in base twelve, one half is 0.6, a quarter is 0.3 and a third is 0.4. We should convert.
In 0.3, it's 3. In 0.6, it's 7 when you decide to end it, since 6 rounds up and 3 rounds down. So 0.3 + 0.3 = 0.67. Because reasons.
[QUOTE=Sherow_Xx;43772751]The implicit assumption is basically that, if a line of 9's reaching to the right goes on for an infinity, and you pull it one digit to the left; there will still be an infinitely long line of 9's reaching to the right. The line will be no shorter than before even though it was pulled to the left. So when you multiply 0.999... by ten, you pull it to the left so it becomes 9.999... The proof only works by accepting that there's still the same number of nines after the decimal point. If there weren't, there wouldn't have been an infinite number of 9's to begin with, which has to be the case. It's just infinity being weird, see [URL=http://www.youtube.com/watch?v=faQBrAQ87l4]Hilbert's Infinite Hotel paradox.[/URL] Also just leaving this here: in base twelve, one half is 0.6, a quarter is 0.3 and a third is 0.4. We should convert.[/QUOTE] If that's the case, then the proof can literally be shortened to just that line, because I showed that it's implies 0.999... = 1. You can just say, "If you accept that 0.999... * 10 = 9.999... then 0.999... = 1."
I guess there's just two things to accept: That when you multiply any number by ten, you can just move the decimal point one digit to the right, i.e. 0.5 becomes 5, 3.14 becomes 31.4, 0.123 becomes 1.23, 0.999 becomes 9.99. That when you do that to an infinitely repeating number, there is no end to the numbers in the line and thus when you move the decimal point to the right, it doesn't affect the number of repeating digits after the decimal point. Normally you'd think that now there's one less since we pulled the first one over to the other side of the decimal point - but we have to throw that idea away when were talking about infinity. Also, go to [URL="http://youtu.be/TINfzxSnnIE?t=5m"]05:00 in the video posted earlier[/URL] (reason #7), maybe that one is more convincing...
You don't need to convince me, I know 0.999... = 1. I'm just criticizing the common proofs I see. The point is that you're working with assumptions about the "decimal in the infinitieth place" which is dangerously informal, and if they believe you that that's how it works, they're already convinced that 0.999... = 1, they just don't know it yet. You're only one step away from showing it and the rest of the proof is a waste of space. [editline]3rd February 2014[/editline] Throwing that step in is basically asking, "Do you accept that 0.999... = 1?" in disguise.
33.3 1/3d?
[QUOTE=JohnnyMo1;43772991]You don't need to convince me, I know 0.999... = 1. I'm just criticizing the common proofs I see. The point is that you're working with assumptions about the "decimal in the infinitieth place" which is dangerously informal, and if they believe you that that's how it works, they're already convinced that 0.999... = 1, they just don't know it yet. You're only one step away from showing it and the rest of the proof is a waste of space. [editline]3rd February 2014[/editline] Throwing that step in is basically asking, "Do you accept that 0.999... = 1?" in disguise.[/QUOTE] Hmm, yeah you're right. Do you know of any proof or example that [I]is[/I] convincing, or in other words one that avoids circular reasoning? Infinity is one hell of a concept sometimes. The seventh reason in the video does it fairly well if you ask me, do you agree? I mean, showing that you can't come up with a single number between 0.999... and 1 means that they have to be the same.
[QUOTE=DrDevil;43770813]Not mostly. It is exactly 1. 1/3 isn't 0.3, it's 0.33333... and that continues to infinity. 2/3 is 0.6666666... and 3/3 is 0.999999999... which in a mathematical sense is equal to one, since with every 9 you add, you get closer to 1. When you add an infinite amount of 9s, you get infinitely close to one, which is actually equal to being one. Also 3 divided by 3 is 1 (and 3/3 is just a division).[/QUOTE] Notice where I said "considered". I wrote that because I'm fairly sure there will be mathematicians out there that disagree with the concept.
[QUOTE=Sherow_Xx;43773083]Hmm, yeah you're right. Do you know of any proof or example that [I]is[/I] convincing, or in other words one that avoids circular reasoning? Infinity is one hell of a concept sometimes.[/QUOTE] The one in that video that you mentioned in the post just above is pretty good, the fact that there is always another number between two real numbers. If they can't name the number in between, they must be the same. If they're not convinced of [I]that[/I] fact, the proof of it is maybe a little long for most people, but elementary, requiring nothing more than basic algebra. My favorite one though is using the geometric series summation formula to add up 0.9 + 0.09 + 0.009... etc. It's pretty easy to convince someone that that has infinitely many terms, or else it's obviously a different number than one but not the "closest" number to one, which many people are convinced 0.999... is. Then you can equate 0.999... = sum from 0 to infinity of (9/10)(1/10)^n, derive the geometric series formula (which is quite elegant and convincing and requires only algebra and taking one limit) and apply it to show that it's just 1. If they [I]really[/I] have problems, and don't accept that the limit in the proof of the geometric series formula works, then you can do it directly with a limit, showing (or just declaring, as it's pretty intuitive) that the sum of an infinite series (e.g. 0.9 + 0.09 + ...) is the limit of the sequence of its [I]partial[/I] sums, i.e. {0.9, 0.99, 0.999, ...}. If they have misgivings about limits (a lot of people see it as a kind of process, not simply a value), it's not hard to apply the limit definition [I]directly[/I] to the sequence to show that its limit is 1. [editline]3rd February 2014[/editline] [QUOTE=paul simon;43773159]I wrote that because I'm fairly sure there will be mathematicians out there that disagree with the concept.[/QUOTE] There are absolutely no mathematicians that disagree with "0.999... = 1," except maybe N. J. Wildberger, but he is a crank. I don't know that he disagrees, but he disagrees with the foundations of set theory on which all modern mathematics is based and nobody really takes him seriously. Actually now that I think about it, Wildberger and other radical finitists probably reject the question entirely, since infinite decimals don't have any meaning to them.
The way I look at it, it 0.9... it goes on for infinity, so the number that is "missing" (the number you would add to make it 1) is infinitely small, due to the way infinity would work, infinitely small is the equivalent to 0, so to make 0.9... 1, you would have to add 0. But we know adding or taking away 0 never makes a difference in what the number is, you can automatically know that 0.9... = 1.
[QUOTE=DrDevil;43770813] 3/3 is 0.999999999... which in a mathematical sense is equal to one, since with every 9 you add, you get closer to 1. When you add an infinite amount of 9s, you get infinitely close to one, which is actually equal to being one.[/QUOTE] Might be a but late to say, but 3/3 is not 0.999 repeating. It's just 1. 3 can go into 3 once. With no remainders.
[QUOTE=Sgt. Khorn;43779738]Might be a but late to say, but 3/3 is not 0.999 repeating. It's just 1. 3 can go into 3 once. With no remainders.[/QUOTE] 3/3 [I]is[/I] 0.999 repeating because 0.999 repeating is 1.
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