Does 0.9999(repeating) = 1?

Yes 705 54.95%
No 578 45.05%
Voters: 1283. You may not vote on this poll

Does 0.9999(repeating) = 1?

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WhattayaBrian

Engineer

01-23-2013
29 of 32 Riot Posts

Quote:
Originally Posted by Kaolla View Post
except that it makes logical sense, unlike whatever you people use.

Attachment 598680

and yes i know the subtraction in that pic is totally wack. it's late and i don't feel like fixing it

the pic just illustrates a point

[edit] nm better edit before **** ppls cry
Pattern recognition is super useful, but it is not the same as a proof, and it will sometimes lead you astray, especially when it comes to infinity.

Imagine a square with side length 1. The perimeter is 4.
Now remove the top right portion of the square, removing 25% of its total area. The perimeter is still 4.
For each of the two points you have, remove a similarly proportioned square. The perimeter is still 4.
You can continue on for as long as you like, and the perimeter of the polygon never changes. It is still 4.

However, as soon as you cut it infinitely many times, you are left with simply an isosceles triangle with sides equal to 1, which as we know has a hypotenuse of length sqrt(2), which means the perimeter is 2 + sqrt(2), which is most certainly not 4.

I have attached below a reason why I am an engineer and not an artist.

The black lines are the original square.
The red box is what you remove for the first iteration.
The green boxes are what you remove for the second iteration.
The blue boxes are the third, and so forth...
Until you reach your state at infinity, which is the purple line.

Had I drawn this perfectly to scale, none of the box's edges would have crossed the purple line (in fact their bottom left corners would all be exactly touching it).

The .9... question is fun because the question is very simple for people to understand. But it actually requires quite a hefty amount of math to really believe. You need to understand...

1. That numbers with an infinite number of decimal places are just as real and valid as any natural number.
2. That you can perform all the same arithmetic operations on these numbers, including such strange-looking operations as .9... * 10 = 9.9... and 9.9... - .9... = 9. These are true, but they are admittedly odd, and you are right to question their legitimacy.
3. That patterns (more formally "invariants") do not hold at infinity sometimes, and that these are on a case-by-case basis.
4. That infinite series are not a process. There are no "steps" to infinite summations, the entire statement is exactly equivalent to its end result at all times. There is not a time when .9... is .9, and a time when it is .99, and then .999, and .9999, and so on. The "process" is how we think about it sometimes, but that doesn't mean the number itself is one.

...and probably more that I'm not thinking of. It's a complicated problem even though its proofs are relatively short. It's why it sparks so much debate--because everyone can pitch in.

And that, to both trolls, and to people who love math, is really pretty fun.

Edit: Fixed my silliness.


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Graffers

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Senior Member

01-23-2013

X=1.000...1
10X=10.000...1
9X=9
X=1
It would appear that any two numbers that are .000(repeating)1 apart are equal to each other. Call it Graffers' Law or whatever you want, but I think my point is made. It's almost like an infinite amount of nothing is equal to nothing. What's so special about .999(repeating) if there are 17 other equally impossible numbers that are equivalent to 1?


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ploki122

Senior Member

01-23-2013

Quote:
Originally Posted by WhattayaBrian View Post
Pattern recognition is super useful, but it is not the same as a proof, and it will sometimes lead you astray, especially when it comes to infinity.

Imagine a square with side length 1. The perimeter is 4.
Now remove the top right portion of the square, removing 25% of its total area. The perimeter is still 4.
For each of the two points you have, remove a similarly proportioned square. The perimeter is still 4.
You can continue on for as long as you like, and the perimeter of the polygon never changes. It is still 4.

However, as soon as you cut it infinitely many times, you are left with simply an isosceles triangle with sides equal to 1, which as we know has a hypotenuse of length 2 * sqrt(2), which means the perimeter is 2 + 2 * sqrt(2), which is most certainly not 4.

I have attached below a reason why I am an engineer and not an artist.

The black lines are the original square.
The red box is what you remove for the first iteration.
The green boxes are what you remove for the second iteration.
The blue boxes are the third, and so forth...
Until you reach your state at infinity, which is the purple line.

Had I drawn this perfectly to scale, none of the box's edges would have crossed the purple line (in fact their bottom left corners would all be exactly touching it).

The .9... question is fun because the question is very simple for people to understand. But it actually requires quite a hefty amount of math to really believe. You need to understand...

1. That numbers with an infinite number of decimal places are just as real and valid as any natural number.
2. That you can perform all the same arithmetic operations on these numbers, including such strange-looking operations as .9... * 10 = 9.9... and 9.9... - .9... = 9. These are true, but they are admittedly odd, and you are right to question their legitimacy.
3. That patterns (more formally "invariants") do not hold at infinity sometimes, and that these are on a case-by-case basis.
4. That infinite series are not a process. There are no "steps" to infinite summations, the entire statement is exactly equivalent to its end result at all times. There is not a time when .9... is .9, and a time when it is .99, and then .999, and .9999, and so on. The "process" is how we think about it sometimes, but that doesn't mean the number itself is one.

...and probably more that I'm not thinking of. It's a complicated problem even though its proofs are relatively short. It's why it sparks so much debate--because everyone can pitch in.

And that, to both trolls, and to people who love math, is really pretty fun.

You might want to fix you hypothenuse to sqrt(2) instead of 2*sqrt(2)... I'm not quite sure you can have an hypothenuse (~2.82) longer than the sum of both other sides (1+1). Otherwise it was a nice read, especially coming back from a whole season of Extra Credits... Knowledge is always entertaining to discover...

Still, I don't get how we can explain that while the result should be a P of 4, we end up with a P of roughly 3.41, while maintaining an equal P from steps to step... It's even more appalling to me if you add in the A, which, with A(n) being the Area after n cuts and A(0) = 1, A(1) = .75, you end up with A(n) = A(n+1) - (A(n+2)-A(n+1))/2 or A(n) = 1.5A(n+1) - .5A(n+2)... which basically means that every steps you remove half of what you removed the last time, or that you remove half of what's left of the upper triangle... Which means that you will eventually end up with A(inf) = .5+.000(repeating)1 or .5.

So, you have a normal triangle who has a Perimeter of ~3.41 and an Area of .5, and our weirdly shaped Triangle who has an hypothenuse made out of microscopic dents that makes the triangle actually bigger, have a Perimeter of 4 while having an Area of 5.000(repeating)1. So we have a shape with a perimeter of ~3.41 and a negligible Area. Or basically a 1/inf Area split into an infinity of equal triangles...

I love things that are easy to picture... I hate this problem you gave us...



Onto the main topic, 0.999 there is no tangible way to represent a period (other than the bar over the numbers), that I know of at the very least. And so, to be written, the number has to be shoved into a "container" or basically a digit limit... which in the ends will round up to 1, explaining why mathematicians usually consider them of equivalent value (however, you shouldn't ever hear them say they are equal, since they aren't...).


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Crowcide

Senior Member

01-23-2013

Quote:
Originally Posted by WhattayaBrian View Post
Pattern recognition is super useful, but it is not the same as a proof, and it will sometimes lead you astray, especially when it comes to infinity.

Imagine a square with side length 1. The perimeter is 4.
Now remove the top right portion of the square, removing 25% of its total area. The perimeter is still 4.
For each of the two points you have, remove a similarly proportioned square. The perimeter is still 4.
You can continue on for as long as you like, and the perimeter of the polygon never changes. It is still 4.

However, as soon as you cut it infinitely many times, you are left with simply an isosceles triangle with sides equal to 1, which as we know has a hypotenuse of length 2 * sqrt(2), which means the perimeter is 2 + 2 * sqrt(2), which is most certainly not 4.

I have attached below a reason why I am an engineer and not an artist.

The black lines are the original square.
The red box is what you remove for the first iteration.
The green boxes are what you remove for the second iteration.
The blue boxes are the third, and so forth...
Until you reach your state at infinity, which is the purple line.

Had I drawn this perfectly to scale, none of the box's edges would have crossed the purple line (in fact their bottom left corners would all be exactly touching it).

The .9... question is fun because the question is very simple for people to understand. But it actually requires quite a hefty amount of math to really believe. You need to understand...

1. That numbers with an infinite number of decimal places are just as real and valid as any natural number.
2. That you can perform all the same arithmetic operations on these numbers, including such strange-looking operations as .9... * 10 = 9.9... and 9.9... - .9... = 9. These are true, but they are admittedly odd, and you are right to question their legitimacy.
3. That patterns (more formally "invariants") do not hold at infinity sometimes, and that these are on a case-by-case basis.
4. That infinite series are not a process. There are no "steps" to infinite summations, the entire statement is exactly equivalent to its end result at all times. There is not a time when .9... is .9, and a time when it is .99, and then .999, and .9999, and so on. The "process" is how we think about it sometimes, but that doesn't mean the number itself is one.

...and probably more that I'm not thinking of. It's a complicated problem even though its proofs are relatively short. It's why it sparks so much debate--because everyone can pitch in.

And that, to both trolls, and to people who love math, is really pretty fun.
This is wrong on so many levels. With arguments like this you can show 1=2. The reason .999~ = 1 is because how repeating decimals are defined, .999~ is just short had notation for a particular geometric series which happens to equal 1.


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Crowcide

Senior Member

01-23-2013

Quote:
Originally Posted by Graffers View Post
X=1.000...1
10X=10.000...1
9X=9
X=1
It would appear that any two numbers that are .000(repeating)1 apart are equal to each other. Call it Graffers' Law or whatever you want, but I think my point is made. It's almost like an infinite amount of nothing is equal to nothing. What's so special about .999(repeating) if there are 17 other equally impossible numbers that are equivalent to 1?
While this is a slick trick it'd essentially a circular argument. .999~ is 1 cause of the reasons I stated in my previous post.


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WhattayaBrian

Engineer

01-23-2013
30 of 32 Riot Posts

Quote:
Originally Posted by ploki122 View Post
You might want to fix you hypothenuse to sqrt(2) instead of 2*sqrt(2)... I'm not quite sure you can have an hypothenuse (~2.82) longer than the sum of both other sides (1+1).
The derp is strong with me tonight. :P Fixing...

Quote:
Originally Posted by Crowcide View Post
This is wrong on so many levels. With arguments like this you can show 1=2. The reason .999~ = 1 is because how repeating decimals are defined, .999~ is just short had notation for a particular geometric series which happens to equal 1.
Could you tell me how I'm wrong? I'm pretty sure I gave you no tools to prove that 1=2.


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Zerglinator

Senior Member

01-23-2013

Would I be wrong if I said that .999... ad infinitum was equal to 1 minus 1 times ten raised to the negative infinity?


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ploki122

Senior Member

01-23-2013

Quote:
Originally Posted by Crowcide View Post
This is wrong on so many levels. With arguments like this you can show 1=2. The reason .999~ = 1 is because how repeating decimals are defined, .999~ is just short had notation for a particular geometric series which happens to equal 1.
hmm... isn't saying that 0.9p = 1 another way of saying that 1=2? More than Brian's explanation?

if s (smallest positive number) is 1-0.9p, then an infinite number of s, or inf*s, = 1 since basically, the smallest possible number is 1 divided an infinite amount of times...

so we have :
0.9p+s = 1, add s to both sides, and we have
0.9p +2s = 1+s, mutiply an infinite amount of times and we have
inf + 2 = inf + 1, substract the infinite number on both sides and you get?
2 = 1

Now, the only way yo ucan disprove what I just showed you is if you acknowledge that inf*0,9p isn't equal to inf*1, and thus that 0.9p doesn't equal 1...

P.s. there is actually a fallacy in there, but it's grounded deeply in a way that can'T be easily seen :P


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ploki122

Senior Member

01-23-2013

Quote:
Originally Posted by Zerglinator View Post
Would I be wrong if I said that .999... ad infinitum was equal to 1 minus 1 times ten raised to the negative infinity?
That's 100% true... but it raises exactly the same problem... the fact that infinity isn't logical...


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Crowcide

Senior Member

01-23-2013

Quote:
Originally Posted by WhattayaBrian View Post
The derp is strong with me tonight. :P Fixing...



Could you tell me how I'm wrong? I'm pretty sure I gave you no tools to prove that 1=2.
I just wrote a very lengthy post about metrics and how you were equivocating the taxicab metric with the standard metric but then I reread your post and saw you typed you CAN'T carry the process out "to infinity" and expect the finite case to speak to the limit. I originally read CAN instead of CANT my bad, appologizes.