Topic awaiting preservation: The Greatest Above All? Or not? (Page 2 of 3)

Raptor! You're right there!

And yes you can divide a number by zero. But you'll get an error on a calculator. Its called an "imaginary number".

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"The Mystery of Life"
Vol. 841, Ch. 26
"All life begins with Nu and ends with Nu. This is the truth! This is my beleif! ...at least for now." - Chrono Trigger
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so are you saying that an imaginary number is the greatest number?

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and yes you can divide a number by zero. But you'll get an error on a calculator. Its called an "imaginary number".

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You do not get an imaginary number by dividing by zero. If you divide by zero on a calculator or computer you'll get a NaN as answer, which was stated above by Slime... The fraction a/b will go towards infinity as b approaches 0. This has nothing to do with imaginary numbers.

Imaginary numbers are those that are written a+ib, where the i is the same i that is beeing discussed above.

This thing with -1=1 is interresting. A professor showed us the 'proof' last semester and explained what is wrong with it, now I just have to figure out where my notes from that lection is...

btw, the greatest number is the Divine Proportion

_________________________
"There are 10 kinds of people; those who know binary, those who don't and those who start counting at zero"
- the Golden Ratio -

[This message has been edited by Veneficuz (edited 09-16-2003).]

so ruku, what is the greatest number now? the imaginary one?

You did not ask what the effect of multiplying a number by zero would be.. you asked what the biggest number was... this is a whole different thing, no hidden variables please.

That's like asking what is the biggest mountain in the world is? ...and I say Everest... and you say Everest times zero = nothing and so you conclude Everest does not exist anymore... mmm I don't have time for this foolery.



[This message has been edited by Xpirex (edited 09-16-2003).]

Well then just to keep everyone going, you are right about x/0 not being a real number. It's called an Imaginary Number. http://www.math.toronto.edu/mathnet/answers/imaginary.html

The answer is... 1/0, because 1/0*0=1 You can do this with any number really... 939248323.25248/0*0=939248323.25248. So therefore, this set of imaginary numbers are the largest number in the world . And sorry, I didn't pick up the x/0 part until I just thought now, Geez you can substitute any number for one and it'll come out the greatest... Crud!

---
"The Mystery of Life"
Vol. 841, Ch. 26
"All life begins with Nu and ends with Nu. This is the truth! This is my beleif! ...at least for now." - Chrono Trigger
><>

ok, once again:

~starts SCREAMING~
x/0 IS NOT AN IMAGINARY NUMBER !!!
IT AINT GOT NOTHING TO DO WITH IMAGINARY NUMBERS AT ALL !!!
IT AINT GOT NOTHING TO DO WITH IMAGINARY NUMBERS AT ALL !!!
IT AINT GOT NOTHING TO DO WITH IMAGINARY NUMBERS AT ALL !!!
IT AINT GOT NOTHING TO DO WITH IMAGINARY NUMBERS AT ALL !!!
IT AINT GOT NOTHING TO DO WITH IMAGINARY NUMBERS AT ALL !!!
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAARGH!!!
~runs off screaming and repeatedly banging his head on walls~

so.....every number that exists is the single "greatest" number aobve all.....

because if you do things to it that you can't *really* do you can make it the same as it was before you didn't do anything to it.....?

how cool.....





[This message has been edited by DL-44 (edited 09-17-2003).]

i is an imaginary number, like Emps guessed way early on, and other people mentioned earlier. i = square root of -1 or somesuch.

x/0 isn't any imaginary number, that's a good ol "undefined". (Some argue it's infinity, some undefined, same thing in my book... Though I do recall someone in our school got sent to ISP (in school suspension) for arguing with the math teacher that #/0 = infinity.. heh.)

imho, there is no "greatest number" besides the whole infinity+1 thing. (Actually, I prefer infinity^infinity. (infinity to an infinite power.)

-Xel


[This message has been edited by Xel (edited 09-17-2003).]

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Some argue it's infinity, some undefined

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I don't see how that's even arguable...

For any x, and given the equation x/n, you get two different possibilities for its limit as n approaches zero (as Slime has already pointed out for us). You get -infinity from as n approaches zero from the negative side, and +infinity as n approaches zero from the positive side. And simply put, +infinity != -infinity. I'd certainly call that "undefined."

*shrugs* I guess to me, arguing that n/0 is infinity is kind of like trying to say .9999999999999999999 (etc) isn't 1.



[This message has been edited by Raptor (edited 09-17-2003).]

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arguing that n/0 is infinity is kind of like trying to say .9999999999999999999 (etc) isn't 1.

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I've actually seen some very long arguements between math geeks, and people who refuse to believe that 0.999(rep) = 1.

Yet those same geeks would subscribe to the methods of rounding which would turn that 0.9(rep) into 1

Well 0.9999(rep) isn't 1 - you could get infinately close to it but you'd never reach it. On the other hand if you were covering a room with carpeting, I wouldn't say that I needed 0.999999(rep) yards, I'd say I just needed 1.

Hey its a controversial matter. Can you do anything in the real world with 1/0? Nope. Can you do it with math? Yep. Now you can go mouthing off to teachers saying that not everything multiplied by zero equals zero.

---
"The Mystery of Life"
Vol. 841, Ch. 26
"All life begins with Nu and ends with Nu. This is the truth! This is my beleif! ...at least for now." - Chrono Trigger
><>

Dan: people who don't believe are either
a) too blind to the truth to know when something is obviously spelled out for them or,
b) see a).

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Well 0.9999(rep) isn't 1 - you could get infinately close to it but you'd never reach it.

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Here we go...

eq1) .99999(rep) = 9/10 + 9/100 + 9/1000 + 9/10000 + ... + 9/10^n + 9/10^(n+1) + ... =

eq2) 9/10(1 + 1/10 + 1/10² + 1/10³ + ... + 1/10^n + 1/10^(n+1) + ... ) =

By definition of a geometric series; the sum of a*r^(n-1) from n = 1 to infinity =

eq3) a*r^0 + a*r + a*r² + a*r³ + ... + a*r^n + a*r^(n+1) + ...

Which, if it's convergent (it approaches a finite value - in this case, 1) =

eq4) a/(1-r) (edit: note that eq4 is only valid if r < 1)

Note that eq2 and eq3 are simliar. Note also that eq3 = eq4. Substituting eq2 into eq3 and converting it into eq4 we get the following:

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

Substitution yields eq1 = eq2 = 1; which also gives .9999(rep) = 1.



Now ask yourself the question: what's the smallest number greater than zero?



[This message has been edited by Raptor (edited 09-17-2003).]

Actually, I do remember a little trick one of our math teachers pulled on us kids back in 7th or 8th grade.. I'm absolutely positive there was something "wrong" with the way he did it, but we for the life of us couldn't figure out what it was.. but according to his equation wrangling trick, .9 equalled 1. Was funny. (Now I'll hafta go find this teacher again and see if he still remembers it.)

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Yet those same geeks would subscribe to the methods of rounding which would turn that 0.9(rep) into 1

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No rounding necessary, they're exactally the same number. (Any 2 real numbers which are different will always have an infinate number of real numbers between them)

1/9 = 0.111(rep) 2/9 = 0.222(rep) {3; 4; 5; 6; 7; 8; } /9, all the same. 45/99 3667/9999 - test em all out... observe the pattern. Now try 9/9.

...or you can prove it with algebra.


x = 0.99999999......

10x = 9.99999999......


9x = 10x - x = 9.999999.... - 0.9999999.....

9x = 9

x = 1



(i find it a little easier than induction)

reitsma

Bleh. My brain won't function this early. I'll just pretend I understand what you're talking about.

---
"The Mystery of Life"
Vol. 841, Ch. 26
"All life begins with Nu and ends with Nu. This is the truth! This is my beleif! ...at least for now." - Chrono Trigger
><>

Slime - Re: "I'm not actually sure if there's a problem with the conversion from (-1)^(2/6) to 6th-root((-1)^2). I think the problem is more subtle. I'm wondering if it has to do with the subtle difference between "square root" and the radical sign."

Yea, that was just what I pulled off the top of my head from the last time I had to convert between exponents and radicals (which has been awhile). But after searching a bit, it seems the two are equivalent (at least as long as you're working with the positive numbers): http://www.tpub.com/math1/8c.htm

I'll see if I can't spend a bit of time playing with that eq to find something maybe... ::shrugs::

l've looked some more at it and the problems with the proof is that the exponents rules aren't 'properly' defined for negative numbers. So error is spread among the following parts: (-1)^1/3 = (-1)^2/3=6th root((-1)^2), the 'biggest' error being in the first equal-sign. Talked to one of the math professor here at the Uni as well, and he said the same

_________________________
"There are 10 kinds of people; those who know binary, those who don't and those who start counting at zero"
- the Golden Ratio -

I think there are several differences in the way we are taught mathematics in France and in the USA.

- About dividing by 0 : your calculators will always return 'undef' because this expression doesn't exist. You can divide by expressions that tend towards 0 (when speaking about limits), but dividing by 0 is a mathematical nonsense. You can divide by 0, but it doesn't mean anything. Well, at least in France. And as MW finely said it, it is not an imaginary number. It simply doesn't exist.

- About the expression I formerly wrote : I asked a math teacher about it. What is wrong in this expression is writing roots of negative numbers. In fact, it is possible to write the roots of negative numbers, but we shouldn't, we should use complex numbers instead. For example, writing 'sqrt(-1)' is possible (although mathematically incorrect, as dividing by 0 is incorrect), but we should instead write 'i'.

But frankly, there are several differences about how everyone is taught this, and I won't rip my hair with you about that



[This message has been edited by Moon Shadow (edited 09-20-2003).]

quote:

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- About dividing by 0 : your calculators will always return 'undef' because this expression doesn't exist. You can divide by expressions that tend towards 0 (when speaking about limits), but dividing by 0 is a mathematical nonsense. You can divide by 0, but it doesn't mean anything. Well, at least in France. And as MW finely said it, it is not an imaginary number. It simply doesn't exist.

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No difference here in the US. It's mathematically undefined; some people are just mis-informed.

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- About the expression I formerly wrote : I asked a math teacher about it. What is wrong in this expression is writing roots of negative numbers. In fact, it is possible to write the roots of negative numbers, but we shouldn't, we should use complex numbers instead. For example, writing 'sqrt(-1)' is possible (although mathematically incorrect, as dividing by 0 is incorrect), but we should instead write 'i'.

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From what I've learnt about this one should optimally not write the root of negative numbers, but a proof can be valid even if it contains the root of a negative number. One just have to remeber to define what one means by that, since the normal root isn't defined for negative numbers. In this case we could have defined sqrt(-1) as isqrt(1) and that problem would have been 'solved'. This is legal as long as the new mathematical rules applying to normal roots still are valid in with the new definition.

This page from the Toronto Uni explains the same thing, but in a different way...

_________________________
"There are 10 kinds of people; those who know binary, those who don't and those who start counting at zero"
- the Golden Ratio -

Anyone with a computer knows that the greatest numbers is a 1 and a 0.


The programmer's national anthem is 'AAAAAAAAHHHHHHHH''

Wolfen's Sig Site

quote:

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x = 0.99999999......
10x = 9.99999999......
9x = 10x - x = 9.999999.... - 0.9999999.....
9x = 9
x = 1

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ACK! You guys are shattering my world view!

I'd like to see the proof that shows 10 x 0.999(rep) = 9.9999(rep)
If there's an infinate number of decimal places, can you really move the decimal over by one? I don't think so. (I'm sure in most cases it will work, but assuming it works in every case isn't very mathematical, and since 0.999(rep) can't be expressed as a fraction (that is, until after you prove that it equals 1), I don't think it's fair to say that you can determine the outcome of it being multiplied be any real number)

Raptor and Veneficuz : thanks to both of you for the little notifications. My vision of mathematics was being questioned, I feel better now

I'll leave again for a whole week soon, but it seems to me that this topic interested a lot of people. Therefore, I think that if there are some folks around who know other math riddles, I encourage them to post them.

Hey, who said once on this forum that our ability to program wasn't limited by our intelligence, but rather by our level at mathematics ?

[This message has been edited by Moon Shadow (edited 09-21-2003).]

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I'd like to see the proof that shows 10 x 0.999(rep) = 9.9999(rep)
If there's an infinate number of decimal places, can you really move the decimal over by one?

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Yes, IMO.
Because it´s obvious that 0.9 x 10 = 9 and 0.99 x 10 = 9.9 and from there you just keep going - it would be unlogical if there was a point where this suddenly wouldn´t be correct anymore.

This should be true for an infinite number of decimals, too, because Infinity = (Infinity + 1) .

Actually, if you want to get technical... 10 * .999(rep) = 1.

Since .999(rep) = 1 and all.

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Actually, if you want to get technical... 10 * .999(rep) = 1.
Since .999(rep) = 1 and all.

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I think you're mistaken. if .999(rep) = 1 then 10 * .999(rep) = 10 * 1 whcih equals 10 not 1

My poor head hurts...

dan - this is the technique for converting recurring decimals to fractions:

x = 0.34343434.....

100x = 34.343434.....
99x = 34

x = 34/99


see?


...and as i recall, this technique has also been supported bu induction, etc... plus it's fun.

reitsma

Oy this topic is huge. Anywho, yes that is a fun trick. ^^ So you can argue that any number over zero is the greatest number, and that 1=.999(rep) and that white is black and up is down... but I'd like to see the white/black up/down arguments... That'd be interesting.

---
"The Mystery of Life"
Vol. 841, Ch. 26
"All life begins with Nu and ends with Nu. This is the truth! This is my beleif! ...at least for now." - Chrono Trigger
><>

Now, what does any of this have to do with my future career as a professional basketball player?

(monday morning)

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I think you're mistaken. if .999(rep) = 1 then 10 * .999(rep) = 10 * 1 whcih equals 10 not 1

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...

note to self: don't post when drunk.

[This message has been edited by Raptor (edited 09-22-2003).]

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Now, what does any of this have to do with my future career as a professional basketball player?

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Just think how confused all the other players will be when you ask them these questions or start rambling about the proofs. And as they stand there dumb-struck you can easily walk over the field and score

_________________________
"There are 10 kinds of people; those who know binary, those who don't and those who start counting at zero"
- the Golden Ratio -

but isn't

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939248323.25248/0*0=939248323.25248

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is still equal to 0?

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but isn't
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939248323.25248/0*0=939248323.25248
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is still equal to 0?

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Divding by zero is an illegal operation, so I think we all agreed that Ruku's 'answer' was wrong. There is no greatest number according to his definition, that definition being that the number multiplied by zero should not equal zero.


_________________________
"There are 10 kinds of people; those who know binary, those who don't and those who start counting at zero"
- the Golden Ratio -

[This message has been edited by Veneficuz (edited 09-24-2003).]

quote:

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x = 0.99999999......

10x = 9.99999999......


9x = 10x - x = 9.999999.... - 0.9999999.....

9x = 9

x = 1

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Except that subtraction and addition supercede division and multiplication in the precedence of function. Your expression should read

9X = (10-1)X not 9X = 10X - X

Regardless of that. X is defined. Why redefine it? .9999.... != 1 it will always be .000000....1 short of 1. To believe otherwise is to believe that infinity doesn't exist.

*points up at using a geometric series to define .999(rep) = 1*

That proof is sound. .999(rep) = 1, and while I'm at it, .0000(rep)...1 doesn't exist, either. Same reason as .999(rep) doesn't exist. You can't have a "smallest positive real number," or a "greatest positive real number less than one."

More simply, how could you have a decimal, followed by an infinite number of zeros, ending with a one?