Does point 999 repeating equal 1?

[Mohammed_Rots_In_Hell]

I have seen several proofs of this and want to see a good debate by the mathematicians of the group.

 

Does .9999(repeating) truely equal 1 (one).

 

I believe so... see the following

 

1) any rational number is defined as number that can be represented as a fraction with a whole number as the numerator and denominator. Any repeating decimal value can be represented as a fraction (is a rational number) by placing the repeating digits as the numerator and the equivolent number of 9's as the digits in the denominator. For instance .333(repeating) can be expressed as 3/9 which is equal to 1/3. So by the same formula .999(repeating) can be expressed as 9/9 which is equal to 1.

 

2) In long division, it is possible to substitute a rule for another as long that rule is consistant and does not violate any mathematical principles. for example: if I want to divide 10 by 5, I could use long division and substitute the rule that every time I come across 5 into 10 I use only the first increment (in this case 1) so the first iteration is 1, I multiply 1 times 5 and subtract from 10 leaving 5. Now I divide 5 into 5 and get 1, now adding the 1 back to the original 1 I get 2 and multimply the 5 by 1 get 5 subtract and it leaves 0. I am done and the answer is two as expected. This will work for any two different numbers (try it). Now try it with two numbers that are the same. take 5 and 5 and change the rule that every time you get 5 divided by 5 you factor out a value of .9. -ok- the first iteration is 5 / 5 you get .9 multiply .9 times the 5 you get .45 subtract from the 5 in the numerator you get .5 move the decimal and start over ... you get .09 add back to .9 and you get .99 multimply the .09 times 5 you get .045 subtract from the .5 you get .005 move the decimal lace and start over you get .009 add is back and you get .999 this will continue for ever giving you .999(repeating).

 

What do you think?

[phreakwars]

Hmm... gets out scientific calculator....

 

switches to abacus

 

searches google

 

makes own hypothesis.

 

I agree for the following reason.

 

1 is the sum of a whole, since .999 is infinite, it should be treated as a whole

 

Besides, you ever try getting your dime bag the right size ?? That .999 could mean alot to the picky.

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[cool_dude]

According to significant figures rules it does.

[sixes]

This is a trick question, and a old one at that. No it cant eqal one.

[Mohammed_Rots_In_Hell]

This is a trick question, and a old one at that. No it cant eqal one.

Yes, the question is over 2000 years old, and it stumps mathematicians to this day. It is hardly a "trick" question, please offer an explanation either way.

 

Another way to look at this problem is on a number line. There is a rule in geometry that states for any 2 distinct points on a line there must exist another point between them, so if .999(repeating) is a distinct point, what point exists between 1 and .999(repeating)? I don't think there is one, which either makes .999(repeating) and 1 equal, or either .999(repeating) or 1 is not a real number.

 

There are more, but I want to see some MATH, please.. not just your personal conjecture.

 

Hint: Smoke some really good herb prior to contemplation of this question. It is very fun!

[eisanbt]

Since an infinite number is well, infinite, would it not be possible then to place it on a graph since there is no possible way for us to pin-point its definate... (Inhales)... position . Positions on say a t-graph are all finite which is why we are able to give a increasing and decreasing set of numbers with definate, known positions as opposed to just slapping shit around. I suppose you COULD have a t-grapth with only infinite numbers ( 0.333, 0.666, 0.999 etc...) with a range of 1 and -1 or something to that effects. It certainly is a melon scratcher though...

[cool_dude]

what point exists between 1 and .999(repeating)? I don't think there is one, which either makes .999(repeating) and 1 equal, or either .999(repeating) or 1 is not a real number.

 

actually they are not equal since the difference between 1 and infinite .99999 is infinite .11111

[ToriAllen]

Yes, they are equal. You will learn this in College Algebra along with the proofs, if you can stay awake.

[TheJenn88]

We were talking about asymptotes in math class the other day, and half the class could not understand for the love of god that the line on the graph would NEVER touch the asymptote, it would infinitely get closer, but never touch.

 

That's the same principle as 0.9999 will never actually equal one. It can get infinitely closer, but it will never touch. It will never be.

 

0.99999999999999999999999999999999 not equal to 1.

 

They even look different

[TheJenn88]

Yes, they are equal. You will learn this in College Algebra along with the proofs, if you can stay awake.

 

Just curious, but did they present this as an opinion, or as something that's cold hard fact?

[sixes]

Yes, the question is over 2000 years old, and it stumps mathematicians to this day. It is hardly a "trick" question, please offer an explanation either way.

 

Another way to look at this problem is on a number line. There is a rule in geometry that states for any 2 distinct points on a line there must exist another point between them, so if .999(repeating) is a distinct point, what point exists between 1 and .999(repeating)? I don't think there is one, which either makes .999(repeating) and 1 equal, or either .999(repeating) or 1 is not a real number.

 

There are more, but I want to see some MATH, please.. not just your personal conjecture.

 

Hint: Smoke some really good herb prior to contemplation of this question. It is very fun!

 

Ok, I smoked a fat joint the other night and went over this. I had to go to the ER assclown! lol Thanks alot.

[Mohammed_Rots_In_Hell]

We were talking about asymptotes in math class the other day, and half the class could not understand for the love of god that the line on the graph would NEVER touch the asymptote, it would infinitely get closer, but never touch.

Ah yes, infinity ... hits on bong what a concept!

 

That's the same principle as 0.9999 will never actually equal one. It can get infinitely closer, but it will never touch. It will never be.

Not quite you see asymptotes are part of a line, not a point but neither a segment it is the term used to refer to the part of graph of a function that can never be on a point (one point) but can exist on other points around it. The NUMBER .999(repeating) is one point, just like the number .333(repeating) which is equal to 1/3 (or .666(repeating) which is 2/3).

 

another proof (but I use the term "proof" loosely because I am not sure if math is defined for a repeating decimal.)

 

take x = .999(repeating) and multiply by 10. You get 9.999(repeating)

10x = 9.999(repeating)... now subtract x from both sides

9x = 9

x = 1

 

interresting is it not?

 

0.99999999999999999999999999999999 not equal to 1.

 

They even look different

Yes, they do look different. Just like 5/5 looks diferent than 1 or 9876 - 9873 + 2 looks different even still.

 

I'm going back to my bong now.

[Anna Perenna]

I have seen several proofs of this and want to see a good debate by the mathematicians of the group.

 

Does .9999(repeating) truely equal 1 (one).

 

I believe so... see the following

 

1) any rational number is defined as number that can be represented as a fraction with a whole number as the numerator and denominator. Any repeating decimal value can be represented as a fraction (is a rational number) by placing the repeating digits as the numerator and the equivolent number of 9's as the digits in the denominator. For instance .333(repeating) can be expressed as 3/9 which is equal to 1/3. So by the same formula .999(repeating) can be expressed as 9/9 which is equal to 1.

 

2) In long division, it is possible to substitute a rule for another as long that rule is consistant and does not violate any mathematical principles. for example: if I want to divide 10 by 5, I could use long division and substitute the rule that every time I come across 5 into 10 I use only the first increment (in this case 1) so the first iteration is 1, I multiply 1 times 5 and subtract from 10 leaving 5. Now I divide 5 into 5 and get 1, now adding the 1 back to the original 1 I get 2 and multimply the 5 by 1 get 5 subtract and it leaves 0. I am done and the answer is two as expected. This will work for any two different numbers (try it). Now try it with two numbers that are the same. take 5 and 5 and change the rule that every time you get 5 divided by 5 you factor out a value of .9. -ok- the first iteration is 5 / 5 you get .9 multiply .9 times the 5 you get .45 subtract from the 5 in the numerator you get .5 move the decimal and start over ... you get .09 add back to .9 and you get .99 multimply the .09 times 5 you get .045 subtract from the .5 you get .005 move the decimal lace and start over you get .009 add is back and you get .999 this will continue for ever giving you .999(repeating).

 

What do you think?

 

Look, it all equals 42 in the end.

[Mohammed_Rots_In_Hell]

actually they are not equal since the difference between 1 and infinite .99999 is infinite .11111

What you mean to say is that the difference is .000(repeating) with some digit somewhere that is equal to one, but that digit is infinitely small (or in the infinitelth position, which of course does not exist)

 

Try this: The law of tricotomy states that for any set of figures one of the following (and only one) must be true:

 

A > B or

A < B or

A = B

 

set A = .999(repeating)

B = 1

 

A > B is false because the most significant digit of A must be either greater than or equal to the most significant digit of B

 

So lets view A < B. reverse it through the associative property of addition and you get B > A subtract A from both sides

0 < B - A therefore there must exist a digit in B - A that is positive. Look at this infinite series.

 

1 - 9/10 - 9/100 - 9/1000 - ... - 9/(10 ^ n) so the digits (the infinite series) becomes... 1/10 - 1/(10 ^ 2) - 1/(10 ^ 3) - ... - 1/(10 ^ n) ok great, lets write the digit so that it d >= 1/(10 ^ n) (because the digit is a positive integer between 1 and 9) is that positive number positive (where n => infinity) Now take A and truncate all the remaining 9's after the nth digit you get A > A' (A' is A after we trucate all the digits after 1/(10 ^ n)) from this we get 1 - A < 1 - A', but wait, isn't d = 1 - A, so how can the digit be both less than 1/(10 ^ n) and greater than or equal to 1(10 ^ n)... It can't by the law of tricotomy! So the only answer left is... you guessed it

 

A = B!

[phreakwars]

Well DUH !! everybody knows that...

 

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[builder]

0.9999999999999999999999999999999999999999999999999999999999

 

needs 0.0000000000000000000000000000000000000000000000000000000001

added to it to make 1.

[phreakwars]

not if the 9 is infinite

 

PAY ATTENTION !!!

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[builder]

How many nines and zeroes do you want to see?

 

It still don't make 1. If it was 1, it would not be 0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999................................

 

It would be..............1.

[phreakwars]

Only proves you REALLY DO need to be high to figure it out.

 

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[RoyalOrleans]

I believe the angle of the dangle is directly proportionate to the heat of the beat of the meat.

[builder]

I believe the angle of the dangle is directly proportionate to the heat of the beat of the meat.

 

Now that (1) I can relate to. Welcome back, cracker.

[Mohammed_Rots_In_Hell]

0.9999999999999999999999999999999999999999999999999999999999

 

needs 0.0000000000000000000000000000000000000000000000000000000001

added to it to make 1.

builder, I see where you are going with this but since the preceding 0's are infinite something really weird happens to the first significant digit look two posts above http://Off Topic Forum.com/showpost.php?p=421998&postcount=14 Do you see? The first significant digit after infinity zeros is both negative and positive at the same time which violates the law of tricotomy (it means the digit does not (can not) exist)

[phreakwars]

You could actually apply this science into one of the PRANKS I have pulled off on members by having them click one of my links, which subsequently crashes their machine. I merely lead to an image that was written with this interger in mind, THE CPU has a brain fart trying to figure it all out, and the PC reboots itself..Well, if it's not an Athlon 64, it will probably crash. I call it the Pentium killer

 

 

I'm funny like that ya know .

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[phreakwars]

For a better explanation, CLICK HERE .

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[Mohammed_Rots_In_Hell]

For a better explanation, CLICK HERE .

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I wouldn't click that link for all the tea in China!