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Happy Halloween - October 31, 2009
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Happy Mathoween!

Spiked Math Comic - Happy Halloween



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49 Comments

My clasmates won't agree with that.

Hey, nice jack-o-lanterns!

I'm afraid that I must kindly disagree. You did not at any point say that they where aloud to round up, or whether the 9 was indeed aloud to be repeating or if it had to be in fraction form (which would be 9/10 if I'm not mistaken. Shame on you Mike, SHAME I SAY, SHAME! SHAAAAAAAAAAAAAME!

Funny comic none-the-less, keep it up :)

surely maths geeks know how to spell?

you're an idiot. how would 9/10 be equal to .999999999 (repeating)?

Darkriku, you have overlooked the horizontal bar over the 9, and thus you are, in fact, deeply mistaken.

We're still presuming that they're both in base 10 though. But I guess that's a safe assumption in these times when other bases are nearly extinct. :-)

0.9 recurring isn't 1 though...it will never reach 1. Just because something trends towards something doesn't mean it equals that...

Actually, many mathematicians say that 0.999... = 1. http://en.wikipedia.org/wiki/0.999...#Generalizations
I personally believe they are different, but not everyone does.

"Many" mathematicians say? How about all mathematicians!

Seriously, it's a basic fact of mathematics. Just because you can't understand why 2 + 2 = 4, for instance, that doesn't make it any less true. What you believe doesn't matter. The aforementioned fact will continue to remain true, whether you choose to believe in it or not. Mathematics is a system of mutually consistent facts.

(Although it must therefore always be incomplete, due to Gödel.)

Um.. "Mathematics is a system of mutually consistent facts.
(Although it must therefore always be incomplete, due to Gödel.)"

I'm sorry, I lol'd.

Your misunderstanding of Godel's second is quite literally painful to see. Godel's second states that you need to be 1) sufficiently strong and 2) recursively axiomatizable. Last time I checked, the consistency of the theory wasn't one of the assumptions. By the way, I never understood why people cared about Godel's first: it's a very uninteresting result...

Um.. "Mathematics is a system of mutually consistent facts.
(Although it must therefore always be incomplete, due to Gödel.)"

I'm sorry, I lol'd.

Your misunderstanding of Godel's second is quite literally painful to see. Godel's second states that you need to be 1) sufficiently strong and 2) recursively axiomatizable. Last time I checked, the consistency of the theory wasn't one of the assumptions. By the way, I never understood why people cared about Godel's first: it's a very uninteresting result...

"I'm sorry, I lol'd.

Your misunderstanding of Godel's second is quite literally painful to see. Godel's second states that you need to be 1) sufficiently strong and 2) recursively axiomatizable. Last time I checked, the consistency of the theory wasn't one of the assumptions. By the way, I never understood why people cared about Godel's first: it's a very uninteresting result..."

Let's discuss the various things wrong with your response:

First of all, your misinterpretation of a reference to Godels first as a reference to Godel's second is "quite literally painful to me."

Actually, before going into the math part, was it really LITERALLY painful to you? Did you read it and then ACTUALLY contort in agony? I very much doubt it, but I'm sure you thought that "literally" is a nice word of emphasis to embellish your brilliant rebuttal. Literally.

So now that we've established that your understanding of the English language is lacking, let's address your understanding of math: Godel's first result states that:

"Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true,[1] but not provable in the theory (Kleene 1967, p. 250)" (thanks wikipedia).

As it happens, this result was hugely controversial at the time of it's discovery because many mathematicians were trying to formulate a
theory (literally) capable of expressing elementary arithmetic that was (literally) both consistent and complete, a task that had just been proven (literally) impossible by this result.

Also, your explanation of Godel's second has (literally) nothing to do with Godel's second, which states that

"For any formal effectively generated theory T including basic arithmetical truths and also certain truths about formal provability, T includes a statement of its own consistency if and only if T is inconsistent."

Last time I checked, the necessity of "you" being "sufficiently strong" and "recursively axiomatizable" wasn't one of the assumptions.

To review, you were just so many kinds of wrong that I (literally) lost count. In the future, if you plan to be smug and "lol" at other people's comments, please have the courtesy to be right, or at least a little bit less wrong. Thank you kindly.

1/3 = .333
3x 1/3 = .999
3x 1/3 = 1
go ahead, carry it out to .33333333, or further if you like. just copy this and add 3's to the end. it's still true.

you are correct!

But, even if you claim 0.99999.... is never equal to 1, the limit is certainly 1.0. And the only way to sum an infinite decimal is through limits...

they are different in the sense that they "look" different, but mathematically they have exactly the same value.

.9 repeating isn't a bloody sequence, it doesn't "approach" anything, it has no limit. It's a single number, and that number is 1.

Decimal representation is not unique. Which obviou just shown. Without the use of any limit. You only need to assume that 1/3 = 0.333333... (with an infinite number of 3:s)

This would have been funnier without the "finally! someone who understands."

x = 0.9999...
10x = 9.9999...
Subtract to get: 9x = 9
i.e. x = 1

Last proof is correct. Read some real analysis book and you will know :P

I don't see how they're equal. Seems to me that 0.999... would equal 1-(1/∞)

Scratch last comment. After thinking more about the above proof and the pattern I noticed below, I see how it works.

1/9 = 0.1111...
2/9 = 0.2222...
3/9 = 0.3333...
...
8/9 = 0.8888...
1 = 9/9 = 0.9999...

I refuse to beleive it. You all make the same mistake.

1/3 does not equal .3333333. Obviously if you say that 1/3 = .33333, then .33333 = 1/3 and .99999 = 1. But where do you get off saying 1/3 = .333333...? .3333 is a good approximation, I'll grant you, but it only approaches 1/3, it doesn't actually reach it.

0.3333 is an approximation while we are talking about the actual number, 0.3333333.... The same way you could say pi=3 which is also a good approximation :)

Oh dear. Obviously most people can't accept the fact that 0.999... = 1, because they tie so strongly the representation and the value.

The same kind of people will have difficulties to recognize patterns in (a^2 + 2a + 1), cos(x)^2 + (tan(x)*cos(x))^2, etc.

Also, most people are unable to gather a long enough attention span to read and understand a good proof. The 1/3*3 thing above is perfectly satisfying.

USE YOUR BRAINS DAMMIT!

It's also that most people cannot really understand infinity. They see a partial form, and go with that rather than realizing that it never ends!

Would another way to put it be this:
Let f(1)=0.9, f(2)=0.99, and so on. Clearly, f(x) -> 1 as x -> ∞.
Isn't writing 0.9 with a bar above the 9 more or less a shorthand for the above limit?

I think 0.9999.... of the entire mathematician think 0.9999...=1

That solves the struggle, :P

And still no one has finally convinced me, XD

Suppose that there existed a real number a with the following properties:

1.) a 2.) 1-10^-k

By (1) and (2) for k=0, we have 0

1-10^-m -10^-m 1-a

Since the last ones are between 0 and 1, inverting gives

10^m

and since by definition 1/(1-a)

10^m

This is a contradiction if you believe this inequality false for every natural number.

------------------------

Proving 10^k >= k by induction, notice that for k = 1 we have 10 >= 1 which is true. And if 10^k >= k, then

10^k * 10 >= 10k
10^(k+1) >= 10k = k + k + 8k > k + k >= k+1

so 10^(k+1) >= k+1.

------------------------

So there can't be a real number with the properties (1) and (2). If you insist on a

Can't see how to edit posts. Didn't read the warning sign with < signs.

Suppose that there existed a real number a with the following properties:

1.) a < 1
2.) 1-10^-k < a for all k in N. (i.e. .9999999....

By (1) and (2) for k=0, we have 0 < a < 1. So 1-a is a positive real number and its inverse 1/(1-a) is well-defined. Choose a natural number m larger than 1/(1-a). Applying (2) to m, we have

1-10^-m < a
-10^-m < a-1
1-a < 10^-m

Since the last ones are between 0 and 1, inverting gives

10^m < 1/(1-a)

and since by definition 1/(1-a)

10^m

This is a contradiction if you believe this inequality false for every natural number.

------------------------

Proving 10^k >= k by induction, notice that for k = 1 we have 10 >= 1 which is true. And if 10^k >= k, then

10^k * 10 >= 10k
10^(k+1) >= 10k = k + k + 8k > k + k >= k+1

so 10^(k+1) >= k+1.

------------------------

So there can't be a real number with the properties (1) and (2). If you insist on a <= 1, then either a does not exist or a = 1.

For two numbers to be different there must be a number in between them (an infinite amount of numbers, too).
There is no number in between 0.999... and 1. They are the same number.

The confusion is all a matter of definition. Whichever you believe is fine, and I personally believe that .999..... does not equal 1.

To help explain my viewpoint, I introduce a theoretical number I call the Nilbit. Since we are discussing infinity here, we've already entered the realm of theory.

Nilbit is what I've chosen to call the "smallest positive number".
Here, we have: .99999...+Nilbit=1;
because Nilbit is defined as being positive, it is not zero, and .999... therefore cannot equal 1. This is not a proof, as I introduced the Nilbit without proving it. It simply offers a possibility. However, I would like to think it is reasonable, and seems to hammer a wedge into every proof I've seen on the matter.

However, infinity is an impossibility. There are an infinite amount of numbers between 0 and 1, so there is no clearly defined smallest one. You can see this by observing that 1/3 = Nilbit+.333....
The Nilbit in that case would be 3 times smaller than the Nilbit I originally introduced, which should not be possible.

Anyway, when discussing theory it is important to have an open mind, as there are no facts. None of it can be measured. None of it can be calculated. Although you can know what infinity is and even use it in come situations, it cannot exist outside of the human mind.

Note how you can represent .999 repeating as an infinite geometric series.

0.999...=9/10+9/100+9/1000...

The convergent sum of an infinite geometric series is given as:

a/(1-r)

So the sum of the series is:

(9/10)/(1-1/10)
=(9/10)/(9/10)
=1

"The confusion is all a matter of definition." - No, it isn't.
"Whichever you believe is fine, ..." - False.
"...and I personally believe that .999..... does not equal 1." - Then you are simply wrong.

Let a = any single digit number {0, 1, ...,9}
0.aaaaa... = a/9.

Demonstration:
0.11111.... = 1/9
0.22222.... = 2/9
0.99999.... = 9/9 = 1.

That's really all there is to it.

The algebraic proofs are just as good.

Yeah, I figured that one out in elementary school from playing with a calculator. I didn't really believe it until I saw it confirmed in a math textbook though.

Or you could prove it this way:

Consider n=1-0.999...
For any epsilon (where epsilon is a real number), n < epsilon. If epsilon is 10^-10, for instance, that is greater than 1-.99999999999, which is greater than 1-.999... (I realize that doesn't look very rigorous, but I think the principal is sound.)

So 1-.999... is smaller than any positive number. If you'll allow me to assume 1-.999... is non-negative, then the only remaining possibility is 1-.999... =0 and .999...=1.

I used to have horrendous trouble with the concept of infinite repeating decimals, and limits. I used to say "but there'll always be a little piece missing!"

But at some point, I managed to understand limits. The limit of a function is not the value of the function at the point of the limit. It is the value the function approaches as the variables approach the point of the limit. That's what limits, repeating decimals, and infinite seem to all be about. Approach.

So .9=9/10!=1
.99=99/100!=1
.999=999/1000!=1
.9999=9999/10000!=1
...
and .999999999=999999999/1000000000!=1
But clearly as we add more digits, the value gets closer to 1.

So we can define 0.999... (0.9 repeating) as a limit (namely, the limit as n->infinite of 1-{10^[-n]}).

And limits have actual numerical values. The value in this case being precisely 1.

So 0.999...=1, if one accepts that repeating decimals are shorthand for limits. I think.

As everyone knows, 0.999...=1, and 1+1+1...=-1/2

say x = 0,99999...

10x = 9,99999...
-1x =-0,99999...
----------------
9x = 9,00000...

so x = 1

Actually, .99... isn't 1 - 1/∞, it's 1 - 1/(10^∞).

.9 = 1 - 1/10; .99 = 1 - 1/(10^2); .999 = 1 - 1/(10^3) and so on, therefore when you have a decimal point with ∞ 9's after it, you get .99... = 1 - 1/(10^∞), NOT simply 1 - 1/∞

Remember, in math, precision is important!

You prove yourself that your "nilbit" cannot exist (there is no smallest positive real number), thus it's use in your proof that 0.99... != 1 renders the proof invalid by contradiction (it makes an assumption which is provably false). In fact, the converse of this (absence of any number between 0.99... and 1) was precisely the thing you were trying to disprove, using this object you then demonstrate does not exist.

Bluh, refreshed the page and missed the reply tick box. That was addressed to John.

mhh well we have to differ between syntax and semantics here.
while those two numbers might be the very same number, there are apparently two ways of writing them down.
so semantically those are equal, whereas syntactically they ain't. therefore I must disagree. those two guys are not wearing the same custome..

just saying :P

arguing on the internet is like competing in the special olympics..

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