quote:

Originally posted by PersianMafia Yeah, it's those far more complex things that the college boards actually bother to test you on... :runs off screaming thinking of his ap calc exam this may:

i will also be taking that test...


OMG TImE to STUDAY!!!

quote:

1=1? How can one number be equal to another? I could never see that.

they are the same number..

1+1 = 2 2 of one thing. maybe thats how different combinations of numbers can equal the same thing.

DID YOU KNOW - .99999999999 = 1 it is equal to one. think about it. what is 3/3 in decimal form. 1/3 = .33333333 2/6 = .6666666666 3/3 = .999999999999 but we all know something like 3/3 or 5/5 = 1. so all .9999999 = 1

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quote:

Originally posted by ::TranceVanDyk:: we're doing quadratic equations in algebra 2 now. they suck but they're do'able.

HAHAHHA


quadratic... ah, kids

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quote:

Originally posted by ::TranceVanDyk:: they are the same number.. 1+1 = 2 2 of one thing. maybe thats how different combinations of numbers can equal the same thing. DID YOU KNOW - .99999999999 = 1 it is equal to one. think about it. what is 3/3 in decimal form. 1/3 = .33333333 2/6 = .6666666666 3/3 = .999999999999 but we all know something like 3/3 or 5/5 = 1. so all .9999999 = 1

That is incorrect, if you round it then yes, ".99999999999 = 1" but as it is, what you wrote is 99999999999/100000000000, which is in fact not equal to 1, but 1/100000000000 away from equaling 1.

First I will postulate that a number divided by the exact same number one time only will equal 1.

3/3 does in fact equal 1, because you have the exact same number in the numerator as you do in the denominator. The same goes for 5/5. However, in your example, your numerator is not exactly the same as your denominator, and is therefore not 1.

And for reference, 2/6 does not equal .6666666666. If you reduce the fraction, you will see that 2/6 is the same as 1/3, which is the repeating decimal .3

quote:

Originally posted by Zenchowdah HAHAHHA quadratic... ah, kids

lol what an asshole

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Originally posted by Orbital32 i got a cool math trick: start/run/calc Wahlaaa!

OMG don't give away that secret!!

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Originally posted by Tranc3 And for reference, 2/6 does not equal .6666666666. If you reduce the fraction, you will see that 2/6 is the same as 1/3, which is the repeating decimal .3

that was a little mistake.

i meant 2/3 = .66666666666666666.

i will research and show u that .99999999 really does equal one. a mathematician showed me once and i will find out how u write it out.

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Why does 0.9999... = 1 ?
This answer is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright (c) 1994 Hans de Vreught ([email protected]).
The first thing to realize about the system of notation that we use (decimal notation) is that things like the number 357.9 really mean "3*100 + 5*10 + 7*1 + 9/10". So whenever you write a number in decimal notation and it has more than one digit, you're really implying a sum.

So in modern mathematics, the string of symbols 0.9999... = 1 is understood to mean "the infinite sum 9/10 + 9/100 + 9/1000 + ...". This in turn is shorthand for "the limit of the sequence of numbers

9/10,
9/10 + 9/100,
9/10 + 9/100 + 9/1000,
...."


One can show that this limit is 9/10 + 9/100 + 9/1000 ... using Analysis, and a proof really isn't all that hard (we all believe it intuitively anyway); a reference can be found in any of the Analysis texts referenced at the end of this message. Then all we have left to do is show that this sum really does equal 1:

Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)

= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)

= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1


Not formal enough? In that case you need to go back to the construction of the number system. After you have constructed the reals (Cauchy sequences are well suited for this case, see [Shapiro75]), you can indeed verify that the preceding proof correctly shows

lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1
0.9999... = 1

Thus x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.


Another informal argument is to notice that all periodic numbers such as 0.9999... = 9/9 = 1 are equal to the digits in the period divided by as many nines as there are in the period. Applying the same argument to 0.46464646... gives us = 46/99.

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quote:

Originally posted by ::TranceVanDyk:: Why does 0.9999... = 1 ? This answer is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright (c) 1994 Hans de Vreught ([email protected]). The first thing to realize about the system of notation that we use (decimal notation) is that things like the number 357.9 really mean "3*100 + 5*10 + 7*1 + 9/10". So whenever you write a number in decimal notation and it has more than one digit, you're really implying a sum. So in modern mathematics, the string of symbols 0.9999... = 1 is understood to mean "the infinite sum 9/10 + 9/100 + 9/1000 + ...". This in turn is shorthand for "the limit of the sequence of numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, ...." One can show that this limit is 9/10 + 9/100 + 9/1000 ... using Analysis, and a proof really isn't all that hard (we all believe it intuitively anyway); a reference can be found in any of the Analysis texts referenced at the end of this message. Then all we have left to do is show that this sum really does equal 1: Proof: 0.9999... = Sum 9/10^n (n=1 -> Infinity) = lim sum 9/10^n (m -> Infinity) (n=1 -> m) = lim .9(1-10^-(m+1))/(1-1/10) (m -> Infinity) = lim .9(1-10^-(m+1))/(9/10) (m -> Infinity) = .9/(9/10) = 1 Not formal enough? In that case you need to go back to the construction of the number system. After you have constructed the reals (Cauchy sequences are well suited for this case, see [Shapiro75]), you can indeed verify that the preceding proof correctly shows lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1 0.9999... = 1 Thus x = 0.9999... 10x = 9.9999... 10x - x = 9.9999... - 0.9999... 9x = 9 x = 1. Another informal argument is to notice that all periodic numbers such as 0.9999... = 9/9 = 1 are equal to the digits in the period divided by as many nines as there are in the period. Applying the same argument to 0.46464646... gives us = 46/99.

Well certainly, the limit as N approaches infinity of the sum(i) from i=1 to n will equal infinity....if it's, say, a 9 behind a decimal, then yes the repeating number does indicate that it approaches 1. However, the only correct way to interpret your notation would be with significant figures....that is, you were measuring out a number to the nth significant place (and therefore simply truncating the remainder, changing the value of the sum, as it no longer approached infinity but rather a finite number).

The formal argument makes perfect sense and is absolutely correct logically, but you didn't follow the conditions necessary for the argument to support your conclusion.

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Originally posted by Sunsnail I went over to a friend's house, and his mother teaches math.

go up to his mom and be like "i'm a math genious like you! You + Me - Clothes / your legs, and hope we don't multiply, biatch!!"

Anyone else here a math major? This is all fun stuff to me, or at least it should be since I'm probably gonna end up doing it for a living in a few years...

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