The New Sudoku Players' Forum

[Cec]

Thanks rep'nA for explaining that 10^{-15} = .000000000000001

I have to confess after wracking my brains on this teaser that whilst I'm still sticking with my belief that 0.99999 (repeating) simply gets closer to equalling 1 but never reaches this number I can't discount rep'nA's argument using Zeno's paradox in failing to get from point A to point B.

"For your argument to be correct, you are only allowed to continue the exercise a finite number of times, placing a finite number of 9's after the decimal point. As soon as you allow an infinite number of 9's, all bets are off.."

It seems my argument is logical if continued for a finite number of times but not logical when taken to an infinite number. My understanding is that an "infinite" number of times can never be reached which means 0.99999 (repeating) would never reach 1.

Cec

[udosuk]

It seems my argument is logical if continued for a finite number of times but not logical when taken to an infinite number. My understanding is that an "infinite" number of times can never be reached which means 0.99999 (repeating) would never reach 1.

Physically/practically it can never be reached... But in mathematics we're talking about theoretical stuffs... You only need to imagine it can be repeated an infinite number of times, although in reality you can never see such a number written out in full...

My final attempt to explain this concept...

[MCC]

The infinity sign "∞" is called the "lemniscate".


MCC

[re'born]

The infinity sign "∞" is called the "lemniscate".


MCC

Or for those of you with a sense of humor (and a background that includes a little bit of complex numbers), ∞ = e^{πi/2} ⋅ 8

[MCC]

"⋅" is this a mathematical sign
Or is it Chinese and my computer doesn't recognise it

Perhaps [∞ = e^{πi/2} ⋅ 8] doesn't exist and you only imagined it


MCC

[re'born]

"⋅" is this a mathematical sign
Or is it Chinese and my computer doesn't recognise it

Perhaps [∞ = e^{πi/2} ⋅ 8] doesn't exist and you only imagined it


MCC

The "⋅" (tongue placed firmly in cheek) represents multiplication.

[Bigtone53]

Dont forget the most beautiful formula of them all

e^ (i x pi) + 1 = 0

where,

e is the base of natural logarithms 2.718....
i is the square root of -1
pi is pi

It looks better visually, but I have not worked out how to do it.

Benjamin Peirce, a noted nineteenth century mathematician and Harvard professor, said, "It is absolutely paradoxical; we cannot understand it, and we don't know what it means, but we have proved it, and therefore we know it must be the truth.

[underquark]

And of course you'd need to calculate pi. How about ArcTan(1).4 [or ATAN(1)*4 in Excel]? I'm having chicken pie for dinner, BTW.

[udosuk]

I was so desperate to work out the meaning of that symbol, I went into the source code and extracted this line:

Or for those of you with a sense of humor (and a background that includes a little bit of complex numbers),
<span style="font-style: italic">&#8734; = e^{&#960;i/2} &#8901; 8</span>

So the symbol I cannot see is of code 8901, or 22C5 in hexadecimal base...

Naturally I went into this page to find some info about the symbol:

http://www.fileformat.info/info/unicode/char/22c5/index.htm

So, turns out it is a "dot operator"... Furthermore, I followed this link to find out what font I can use to display it:

http://www.fileformat.info/info/unicode/char/22c5/fontsupport.htm

I don't have many in that list, but I do have Lucida Sans Regular & Lucida Sans Unicode... So I copy & paste the whole equation to Notepad & change the font to Lucida Sans... Finally!

I did say I was desperate...

Now that I have the tools, Bigtone53's equation can be easily typed out with "e^{&#960;i} + 1 = 0", or

e^{πi} + 1 = 0

Or in better visual effect:

    πi
e   + 1 = 0

[re'born]

Perhaps I should explain my 'joke' a little.

First, a little background about complex numbers and the complex plane. Any complex number is of the form a+bi where a and b are real numbers. These numbers can be represented graphically on the complex plane. This is like the typical Cartesian plane, but the x-axis is now the real numbers and the y-axis is the imaginary line. A complex number a+bi then shows up at the point (a,b) in the plane. Often we draw a vector from (0,0) to (a,b)

Now, let θ be an angle (given in radians). Then one can show that multiplying a+bi by e^{θi} will rotate the vector counterclockwise by θ radians. In the case of Euler's formula (provided above by Bigtone53), we see that multiplying by e^{πi} is the same as rotating by π radians (or 180 degrees, if you like). Take the complex number 1 = 1+0i. Rotating the vector π radians will take your vector to -1=-1+0i. Therefore, e^{πi} = e^{πi} ⋅ 1 = -1 and hence e^{πi} + 1 = 0.

In my case, multiplying by e^{πi/2} rotates by π/2 radians (or 90 degrees). The joke is then that multiplying 8 by e^{πi/2} will rotate 8 by 90 degrees, which is the symbol ∞. Hah, hah wasn't that funny? (Man, I need to get a life.)

[udosuk]

In my case, multiplying by e^{πi/2} rotates by π/2 radians (or 90 degrees). The joke is then that multiplying 8 by e^{πi/2} will rotate 8 by 90 degrees, which is the symbol ∞. Hah, hah wasn't that funny? (Man, I need to get a life.)

So in value, 8*e^{πi/2}=8i? Well, this "i" could stand for "infinity", if not "imaginary number"...

Just came across this interesting page:

http://www.maths.nottingham.ac.uk/personal/anw/Research/Hack

It used the game of "Hackenstrings" to work out problems such as the "0.999...=1?" problem... Quite surprising to me...

According to it, upon systems other than real numbers, such as surreal numbers (including infinity and infinitesimal numbers), "0.999..." is not equal to 1, with their difference being the infinitesimal number "ε"...

So perhaps Cec, MCC & Luna etc can justify their views... They just need to do it surreally...

[Cec]

It seems my argument is logical if continued for a finite number of times but not logical when taken to an infinite number. My understanding is that an "infinite" number of times can never be reached which means 0.99999 (repeating) will never equal 1.

Physically/practically it can never be reached... But in mathematics we're talking about theoretical stuffs... You only need to imagine it can be repeated an infinite number of times, although in reality you can never see such a number written out in full...

My final attempt to explain this concept...

"...So perhaps Cec, MCC & Luna etc. can justify their views..."

Oh dear.. just when I assumed this matter had been put to rest Anyway here's my justification that 0.999999 (repeating) never equals 1.

Chessmaster originally asked whether 0.99999(repeating) ever reached exactly 1. My view was, and still is, that the longer the exercise is repeated the answer becomes progressively closer to 1 but never reaches 1. As udosuk himself has quoted above that "physically/practically it can never be reached" (presumably "it" means 1) then it seems udosuk is agreeing with me after all.

Cec

[udosuk]

Cec,

It all depends on how you interpret "0.99999(repeating)".

If you regard it as a real number with the representation "0." followed by an infinite number of 9s, then the answer is yes (it equals to 1).

If you regard it as a continuous attempt to add 9s after "0.", or accept that there are surreal numbers such as the "infinitesimal number" ε, then the answer is no.

Let's say I agree that it is not a well-defined question...

[Cec]

Chessmaster originally asked whether 0.99999 (repeating) ever reached exactly 1.

I've just noticed Chessmaster actually quoted .99999(repeating) and not 0.99999(repeating) which I have been incorrectly quoting. I don't think my mis-quoting makes any difference to my previous answer but thought I should mention this.

udosuk, in arriving at my answer I chose your above second option as to what "0.99999(repeating)" means which is to continually add 9's after "0." and so never reach 1. As to your first option being to regard 0.99999(repeating) as "a real number with the representation "0." followed by an infinite number of 9s", I don't mind confessing I simply don't understand what this means even after reading about real numbers . My brain has just about cooked trying to know what the "correct" answer is to all this and I wonder what Chessmaster now thinks the answer is

Cec

[re'born]

Chessmaster originally asked whether 0.99999 (repeating) ever reached exactly 1.

I've just noticed Chessmaster actually quoted .99999(repeating) and not 0.99999(repeating) which I have been incorrectly quoting. I don't think my mis-quoting makes any difference to my previous answer but thought I should mention this.

You are correct, it makes no difference.

dosuk, in arriving at my answer I chose your above second option as to what "0.99999(repeating)" means which is to continually add 9's after "0." and so never reach 1.

Then you are answering a different question. There are two distint questions:

1) Does .999... = 1?

and

2) Does any term in the sequence .9, .99, .999, .9999,... equal 1?

The answer to the first question is yes. The answer to the second question is, of course, no.

As to your first option being to regard 0.99999(repeating) as "a real number with the representation "0." followed by an infinite number of 9s", I don't mind confessing I simply don't understand what this means even after reading about real numbers . My brain has just about cooked trying to know what the "correct" answer is to all this and I wonder what Chessmaster now thinks the answer is

You have found the heart of the matter right here. This is not a simple question and does not have a simple answer. It took some very smart mathematicians a good deal of time to decide just what should be meant by 'real number' (and I'm sure their brains cooked a bit in the process).