For a so-called intelligent person, taking the administrators to court would be an incredibly dumb move.BlacklightVirus said:
Poll: 0.999... = 1
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We don't link to it because they either won't read it, or they'll read it but still insist that they are different numbers because .99... can't possibly equal 1 because they are different numbers!Jaime_Wolf said:
Plus, it's fun to argue.
I must say, I read your wikipedia link. ANd it says one of the reasons the kids accept it is because someone of authority did. hwich is just intimidiation, and not reasoning.Jaime_Wolf said:
Do you accept that 2 = 2?emeraldrafael said:
Isn't that only true because someone of authority said it is? That's just intimidation, you should use your reason.
Could have sworn this post was done many moons ago. 0.9999 can equal 1 as long as it bumps my grade from a B to an A.
I probably wouldn't win, and I accept that but I would still do it on principal, that principal being that fact is imperatively true regardless of how many people believe it and fact must not be silenced due to opposition.smithy_2045 said:
M'kay. The number 0.999... is equal to an infinite series 0.9 + 0.09 + 0.009 + 0.0009 and so on. If you know anything about slightly advanced maths, you'll know that the sum of an infinite geometric series is equal to a/(1-r) when |r| < 1 (explained below for those who aren't so good at maths)
In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).
So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.
Explanation of maths involved:
A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].
The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]
Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]
Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]
This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])
And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)
Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero.
Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1
Satisfied now?
In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).
So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.
Explanation of maths involved:
A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].
The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]
Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]
Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]
This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])
And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)
Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero.
Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1
Satisfied now?
So what you're saying is that you read the summary AND completely ignored the context of the sentence?emeraldrafael said:
I certainly can't see why you don't understand the reasoning behind the original claim.
The point there is that mathematics educators have a strong interest in understanding why learners accept or reject the claim given that it's demonstrably true. One of the reasons that students accept it is on appeal to authority. That has nothing to do with the actual truth or falsity of the claim, only why students typically accept or reject it.
They silenced it because it was disrupting the ability for other discussions to commence. At the time they outlawed it, there had been something like _hundreds_ of threads created and thousands of posts in the main thread discussing it.BlacklightVirus said:
More of a "Damn...we have created a monster." rather than a "Stop having discussion!"
That and the "discussion" (like we see so often on the intertubez) devolved into each side slamming eachother because the other was just _wrong_ and an idiot for it.
M'kay. The number 0.999... is equal to an infinite series 0.9 + 0.09 + 0.009 + 0.0009 and so on. If you know anything about slightly advanced maths, you'll know that the sum of an infinite geometric series is equal to a/(1-r) when |r| < 1 (explained below for those who aren't so good at maths)
In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).
So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.
Explanation of maths involved:
A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].
The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]
Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]
Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]
This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])
And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)
Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero. (If |r| = 1, we end up with 0/0, and I don't wanna go there (it's not 1)).
Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1
Satisfied now?
In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).
So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.
Explanation of maths involved:
A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].
The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]
Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]
Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]
This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])
And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)
Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero. (If |r| = 1, we end up with 0/0, and I don't wanna go there (it's not 1)).
Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1
Satisfied now?
Infinity doesn't scare us at all, which is why we do math with it on a regular basis.SomethingAmazing said:
And it actually equals both .999... and 1, but that's semantics.
1/3 = .333...
.333... * 3 = .999...
1/3 * 3 = 3/3
3/3 = 1
.999... = 1
I'm not saying that there arent other good ideas backing it up to be "true." I'm just saying that this is the same logic that explains how the holocaust started. Someone of authority said something was right, and the people followed.Rabid Toilet said:
and dont tell me it has nothing to do with this, because it has everything to do with the reasoning behind as to why some believe its right and others dont.
Also:
I'm glad that we can make fun of others in the thread.Jaime_Wolf said:
Unfortunately, we've used that proof at least twice, to no avail. Good effort though!Redingold said:
And yet someone in authority said that 2 + 2 = 4, and the people followed.emeraldrafael said:
Why does 2 + 2 = 4? Because they say so.
Using the very laws of mathematics that were invented so long ago by those people, .99... and 1 are the same number.
Oh, well that's completely different. I thought you were referring to a situation like this one.IMakeIce said:
You're right that .3333 does not equal 1/3.SomethingAmazing said:
However, .333... (note the repeating, there's an infinite number of threes) exactly equals 1/3.
I like the fact that most people don't seem to know that this is true only because it is defined to be so within the most commonly used numbering system. You can define systems where 0.(9) =/= 1 you know. You just have to define numbers such as 0.000...0001 (an `infinitesimal'). Of course, then many of the things we take for granted (like three thirds making up a whole) don't work any more.