Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: LDStudios on December 10, 2013, 09:16:43 pm
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Someone please help explain this to me because I cant decide whether or not .9 repeating = 1
Here is my reasoning for why it should:
1) Fraction Approach
.3 repeating = 1/3
1/3*3=1
.3 repeating*3=.9 repeating
2) Algebraic Approach
I will write .999 to represent .9 repeating to simplify things
.999=.999
x=x (substitute x for .999)
10x=10x
10x-x=9x
10(.999)-.999=9x
9.999-.999=9x
9=9x
1=x
.999=1
Here is my reasoning why it does not equal 1:
1) It does not make any sense! It should not equal 1! :crazy:
2) You could argue that 1 as a value for x in the algebraic approach is extraneous because when plugged back in, it comes out to .999=1
So thats my reasoning so far. Can someone please help to explain to me whether or not it does equal 1? Thank you :)
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I went to wikipedia for this one: http://en.wikipedia.org/wiki/0.999...
It equals 1.
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It approaches 1 as the length goes to infinity. For any value \delta more than 0 you can choose a natural number N so that the number .99999... with N digits or more will be within \delta of 1.
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It does not aporoach 1. It IS 1. There is no way you could say that multiplying fractions with infinite digits can result in a number with finite digits otherwise.
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1) It does not make any sense! It should not equal 1! :crazy:
Do you have a specific argument against it?
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@harold
The explanation I have heard is that it approaches one without reaching it, like an asymptote on a graph, it will get infinitely close, but cannot equal the value of the asymptote. Im not saying that I completely believe it does not equal 1, but it does hurt my head to think it does :P
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I agree with the asymptote thing... It gets infinitely close to but never equals 1. That's just how it works in my head though.
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I'm with Kenoi on this one. It equals 1.
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This is how I think of it:
0.9 = 1 - 1/10 so with 1 digit behind the comma, it is 1/10th less than 1
0.99 = 1 - 1/100 so with 2 digits behind the comma, it's 1/100th less than 1
0.99 ... 9 (n digits behind the comma) = 1 - 1/(10^n)
and for infinity:
0.99... (∞ digits behind the comma) = 1 - 1/∞, and anything divided by infinity (except for infinity itself) is 0, so it equals 1-0=1.
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I've had this argument with a friend. To me, a decimal number is still just that, a decimal. Nothing other than 1 can equal 1. If anything, the proof only suggests to me not that .99999 repeating equals 1, but that instead that the decimal system is a flawed way to represent partial numbers.
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If you were in a limit, 0.999... = 1-. But since we're not in a limit, it would, for all intents and purposes, equal 1.
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To me, it equals 1.
Simply accept that as long as 41.9999...99958=42, I'm right.
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The explanation I have heard is that it approaches one without reaching it, like an asymptote on a graph, it will get infinitely close, but cannot equal the value of the asymptote.
I agree.
My math teacher has this picture on a shirt :P
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@harold
The explanation I have heard is that it approaches one without reaching it, like an asymptote on a graph, it will get infinitely close, but cannot equal the value of the asymptote. Im not saying that I completely believe it does not equal 1, but it does hurt my head to think it does :P
Yeah I agree too. If you take a sequence u with first term u1=0 and u(n)=u(n-1)/10+0.9 then it makes sence that u(n) converges to 1 without reaching it. But the convention is that 0.999... = 1
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I've had this argument with a friend. To me, a decimal number is still just that, a decimal. Nothing other than 1 can equal 1. If anything, the proof only suggests to me not that .99999 repeating equals 1, but that instead that the decimal system is a flawed way to represent partial numbers.
have you ever taken calculus? that these two things ARE equal is essential to the fundamental theorem, and, thus, most modern mathematics.
think of it this way:
.9 is close to 1, but does not equal it, .99 is closer, and, every time you add another 9, the gap between the two grows even smaller. if you add infinitely many 9s on the end, then, the gap between the two numbers will be infinitely small. an infinitely small gap is not a gap at all, and, thus, the two are equal. take a look at the concepts of limits and convergence.
EDIT: i think where people are getting tripped up is the concept of infinity.
The explanation I have heard is that it approaches one without reaching it, like an asymptote on a graph, it will get infinitely close, but cannot equal the value of the asymptote.
I agree.
My math teacher has this picture on a shirt :P
"infinity" is not a number; it's a concept. if something is "infinitely x", then it is as x as something can possibly be. if two things are "infinitely similar to one another", they are as similar as it is possible to be. the most similar two things can be to one another is to match one another in every single regard, so two things that are infinitely similar are the same thing. likewise, if something is "infinitely close to another thing", the two occupy the same space. a decimal place followed by infinitely many 9s, thus, equals one, e(-infinity) = 0, and so on.
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an infinitely small gap is not a gap at all
Well, it is. It's just infinitely small.
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@Shmibs: No, I haven't taken calculus. A decimal can extend as far as it wants in one direction, but that still will not make it equal to one from what I can see.
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Art_of_Camelot:
If you want to learn calculus: I'm making a series of lessons here:
http://ourl.ca/20293
It does not cover convergence yet, but it has covered limits (to an extent) :D
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To me, it equals 1.
Simply accept that as long as 41.9999...99958=42, I'm right.
That "notation" is heavily misleading and, somewhat unfortunately, crops up whenever this debate does. What does the use of the ellipsis denote? An infinite string of characters extending to the right. The tricky part here is the word "infinite". An infinite string has very different behavior from a finite one. In particular, the notion of an "end" does not exist. You can't append to the end of an infinite string because there isn't an end to append to. By placing numbers after the ellipsis, you're attempting to denote this. It's not a valid operation and hence what you get is wrong. 41.999...=42 would be the only way to write that statement and thankfully, it's true.
To see why, let's take a quick look at what the reals are:
Reals intuitively consist of all the numbers along a number line. This can be formalized with something called a dedekind cut, which is essentially as follows: For every point X along a line, define sets A and B such that A consists of all the points below X and B consists of all points not in A. We can define the reals as every point that we can do this for. Note that each A has the property that while there is an upper bound on the set (X), there is no greatest member of A. In the discussion 0.9999... = 1, what we're really asking is whether there's a a distinct real number from 1 such that there is an infinitesimal difference between them. In other words, is there a greatest member in set A? As I stated earlier, this is a property of the way the set is defined that there is not. Therefore, if 0.9999... is distinct from 1, it cannot be in A. It must therefore be a member of B, as the smallest member. Unfortunately, this is exactly where 1 is located. Thus 0.999... = 1.
If this confuses you, that's fine. It's simply a more intuitive explanation of the formalization of reals. Note that, unlike what shmibs, says, I am an uninteresting person (err, i mean you can have systems where the equivalence is not true), but it's not possible with the *reals*.
EDIT: Thanks to shmibs for edits and *ahem* punctuation.
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Dedekind cuts probably aren't going to convince the deniers, though. Let's be honest, they don't really make intuitive sense, and the biggest reason the deniers have is "but it doesn't make sense"..
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@Shmibs: No, I haven't taken calculus. A decimal can extend as far as it wants in one direction, but that still will not make it equal to one from what I can see.
that's true. no matter how many 9s you care to add to the end, you will still have a finite number of them. adding 9s to the end can never help you to reach infinitely many 9s any more than it can help you to reach 1. thus, both 1 and .9 repeating are greater than any number you can reach by simply adding more and more 9s. as .9 repeating is also, obviously, not greater than 1, the two must be equal.
>inb4 that's not exactly true from fish :P
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/me acknowledges the wisdom of the shmibs and that it's a much better way of saying what he tried to get across with dedekind cuts
/me also acknowledges that he is in fact a banana
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0.9999... is just 9 * (1/10 + (1/10)^2 +(1/10)^3 +...) = 9 * (1/10) / (1- 1/10) = 1. To do this formally, you'll need calculus and infinite series.
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if you have infinite 9s after the comma then it is equal to 1, if you have a finite number, it will only approche 1 but never equal it.
Also:
0.1111...... = 1/9
0.2222...... = 2/9
0.3333...... = 3/9
0.4444...... = 4/9
0.5555...... = 5/9
0.6666...... = 6/9
0.7777...... = 7/9
0.8888...... = 8/9
0.9999...... = 9/9
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/\ That solves it for me.
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To me, it equals 1.
Simply accept that as long as 41.9999...99958=42, I'm right.
That "notation" is heavily misleading and, somewhat unfortunately, crops up whenever this debate does. What does the use of the ellipsis denote? An infinite string of characters extending to the right. The tricky part here is the word "infinite". An infinite string has very different behavior from a finite one. In particular, the notion of an "end" does not exist. You can't append to the end of an infinite string because there isn't an end to append to. By placing numbers after the ellipsis, you're attempting to denote this. It's not a valid operation and hence what you get is wrong. 41.999...=42 would be the only way to write that statement and thankfully, it's true.
To see why, let's take a quick look at what the reals are:
Reals intuitively consist of all the numbers along a number line. This can be formalized with something called a dedekind cut, which is essentially as follows: For every point X along a line, define sets A and B such that A consists of all the points below X and B consists of all points not in A. We can define the reals as every point that we can do this for. Note that each A has the property that while there is an upper bound on the set (X), there is no greatest member of A. In the discussion 0.9999... = 1, what we're really asking is whether there's a a distinct real number from 1 such that there is an infinitesimal difference between them. In other words, is there a greatest member in set A? As I stated earlier, this is a property of the way the set is defined that there is not. Therefore, if 0.9999... is distinct from 1, it cannot be in A. It must therefore be a member of B, as the smallest member. Unfortunately, this is exactly where 1 is located. Thus 0.999... = 1.
If this confuses you, that's fine. It's simply a more intuitive explanation of the formalization of reals. Note that, unlike what shmibs, says, I am an uninteresting person (err, i mean you can have systems where the equivalence is not true), but it's not possible with the *reals*.
EDIT: Thanks to shmibs for edits and *ahem* punctuation.
I don't really grasp what you're saying here... It seems you assume calculus knowledge which most deniers do not have. I for myself can accept the convention that 0.99... is 1 but I don't agree with it, mainly because only 1 is 1.
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I love that youtube channel!
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If your having math problems, I feel bad for you son, I've got 98.999... problems but .999... is one! Vi's the best!
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In ninth grade my friends and I got so mad about this. We would argue for hours about this, no joke. :D I was always in the camp that said it was equal, but whatever.
As said above, in summation notation it could be expressed as the summation of 9/(10^x) as x goes from 1 to infinity, which of course would be 1.
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She also posted this video :P
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It is most definitely 1, and that is a cool feature of how the Reals work. Think of it this way:
Try to find a number between .9999... and 1. You can't and that is why they are equal!
I had to take a proof-writing course with a pretty neat professor and he once asked me something like this for homework:
Let f(x) be the function that returns the value of the first digit after the decimal point in base 10. Is this a function? (Meaning that f(x)=f(x))
On initial inspection, I thought it was indeed a function and I wrote the proof for it. That was until I realised the morning it was due that .999... =1 and f(.99...) returns 9 and f(1) returns 0, yet the inputs are identical! Crazy shtuff, that :D
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I've had three math teachers prove it to me in three different ways (algebraically, with calculus, and then with real analysis), but I've personally always accepted it as fact without any proof. It might have something to do with using calculators a lot—type .99999999999 on your graphing calculator and see what it comes up with. Calculators never lie :P
EDIT: For the record, I don't think calculators count as valid proof :P
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Standard analysis they are equal
Non-standard analysis [the hyper-reals], they are not equal
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Relevant xkcd (http://forums.xkcd.com/viewtopic.php?t=288)
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The following is an argument that makes sense to me: two real numbers are different if and only if their difference (the result of subtracting them) is not 0. If we look at the result of subtracting .9repeating from 1, it is certainly nonnegative, but it is also less than .1, .01, .001, and so on. So whatever this number is, if it is positive, we can find a (negative) power of 10 that is smaller than it and that it must be smaller than. Since two numbers x and y cannot have both x>y and x<y, we have arrived at a contradiction. Therefore, this number cannot be positive, and must be 0.