Obscure & Useless

[Resha Caner]

Question 1: Do you have an obscure science or math problem that you find fascinating - even though it's most likely useless?

Mine would be the question: Does 0.999 repeating = 1?

Question 2: What do you think is the most underappreciated aspect of science or math? In other words, little known but very important.

Mine would be the assumptions behind space-time measures.

 

[AV1611VET]

Question 1: Do you have an obscure science or math problem that you find fascinating - even though it's most likely useless?

I'm intrigued that 0/0 ≠ 1.

 

[Blayz]

Question 1: Do you have an obscure science or math problem that you find fascinating - even though it's most likely useless?

Mine would be the question: Does 0.999 repeating = 1?

let x = 0.9 recurring
=> 10x = 9.9 recurring

10x - x = 9.9 recurring - 0.9 recurring
9x = 9
x = 1

I see no problem.

 

[Tinker Grey]

let x = 0.9 recurring
=> 10x = 9.9 recurring

10x - x = 9.9 recurring - 0.9 recurring
9x = 9
x = 1

I see no problem.

To be fair, he didn't say 'problem'; he said fascinating.

 

[Nick665]

Yes, thats a fascinating problem .

 

[Gracchus]

I'm intrigued that 0/0 ≠ 1.

Think of it this way. Let (x, y) be a vector in two-space. That is: it is the vector (0 + x, 0 + y) Then y/x would be the slope of the vector. So every vector has a slope. But (0, 0) is a point, and has no slope.

 

[Blayz]

To be fair, he didn't say 'problem'; he said fascinating.

I don't get the fascinating either, it's just 1 of many examples of a convergent series.

lim (x -> inf) sum (0.9 x 10^-x)

 

[Tinker Grey]

I don't get the fascinating either, it's just 1 of many examples of a convergent series.

lim (x -> inf) sum (0.9 x 10^-x)

I don't disagree with your assessment ... but many message boards have threads that involve pages upon pages discussing it.

'Fascinating' is subjective.

 

[Exiledoomsayer]

let x = 0.9 recurring
=> 10x = 9.9 recurring

10x - x = 9.9 recurring - 0.9 recurring
9x = 9
x = 1

I see no problem.

I'm utterly unfamilour with this whole thing so forgive my ignorance.
Could somebody explain to me why "10x - 0.9recurring = 9x" instead of 9.1x ? Looks like assuming the conclusion?
Again im a complete noob and just honestly interested in how that works.

 

[SithDoughnut]

Could somebody explain to me why "10x - 0.9recurring = 9x" instead of 9.1x ? Looks like assuming the conclusion?

If you replace x with 0.9 recurring (remember that x was defined as 0.9 recurring at the start), you get:

10(0.9 recurring) - 0.9 recurring = 9(0.9 recurring)

To make it even easier to visualise, you can write it like this, as x is the same thing as 1x (if you multiply anything by 1, it does not change, after all).

10(0.9 recurring) - 1(0.9 recurring) = 9(0.9 recurring)

I'm not sure how you arrived at 9.1, but does the above way of writing the equation make more sense? You have 10 things (in this case 0.9 recurring), you take away 1 thing and you're left with 9 things.

 

[Nostromo]

Could somebody explain to me why "10x - 0.9recurring = 9x" instead of 9.1x ? Looks like assuming the conclusion?
Again im a complete noob and just honestly interested in how that works.

Because with 0.99999 recurring, the 9s never end.

x = 0.99999999 recurring
10x = 9.99999999 recurring

In 10x, the number of 9s after the decimal place is the same as in x, i.e. infinity. So when you subtracting x from 10x, everything after the decimal place cancels out, leaving 9.

 

[Exiledoomsayer]

Ah yes I understand now. My thanks to both of you.
I completely overlooked that x=.9recurring which was the whole point heh.

 

[Exiledoomsayer]

Woops, double post.

 

[Resha Caner]

I'm intrigued that 0/0 ≠ 1.

There are some slight variations on this where, by L'Hopital's rule, one can get a variety of answers (depending upon how the original problem is structured). The rate of convergence becomes important.

Playing with infinity (and infinitesimals) can yield some odd results. Another one is taking any number to the zero power.

 

[Resha Caner]

let x = 0.9 recurring
=> 10x = 9.9 recurring

10x - x = 9.9 recurring - 0.9 recurring
9x = 9
x = 1

I see no problem.

Yes, that's the common trick taught to beginning algebra students. However, if one goes on to advanced math, one learns that the trick is only an approximation. So, it works most of the time, but not all the time.

As shown by Richard Dedekind in the 19th century, the proper way to deal with infinitesimals is through what became known as the "Dedekind cut". Using that mathematical technique, one can show that 0.9... is not equal to 1.

Richman, Fred. "Is 0.999 ... = 1?" Mathematics Magazine 72(5), Dec 1999, 396-400.

Intuitively I would have said they were not equal, so I guess I like that Richman confirmed my intuition. I think of it this way. Though numbers are symbols, they are not variables. So, 0 cannot represent anything other than zero, 1 cannot be anything other than 1, etc. Therefore, if a sequence containing only numerical symbols is not the same, the numbers are not the same. 0.9... is not the same as 1.0..., therefore the numbers are not the same.

It is a similar argument to what Godel used in his incompleteness theorum.

 

[SithDoughnut]

Ah yes I understand now. My thanks to both of you.
I completely overlooked that x=.9recurring which was the whole point heh.

Meh, algebra. I stopped at calculus and never looked back.

 

[SithDoughnut]

As shown by Richard Dedekind in the 19th century, the proper way to deal with infinitesimals is through what became known as the "Dedekind cut". Using that mathematical technique, one can show that 0.9... is not equal to 1.

At the risk of not remotely understanding the answer, how? Surely there is no difference between 0.9... and 1, as there is no rational number between them?

 

[Jazer]

At the risk of not remotely understanding the answer, how? Surely there is no difference between 0.9... and 1, as there is no rational number between them?

Yes there is a difference. A guy who calls himself a master carpenter told me that when he measures he looks to see if the measurement is light or heavy. Under or over. Then he makes his cut accordingly. I try to find the most ultrafine pens I can. I had pens that were .2 but now the best I can do is find a pen that is .5 That seems very thick to me.

 

[Resha Caner]

At the risk of not remotely understanding the answer, how? Surely there is no difference between 0.9... and 1, as there is no rational number between them?

Hmm. The Dedekind cut is one of those things that makes my head hurt. I have to think about just the definition for an hour before I can even start solving a problem. So, I don't think I would be able to be brief. My best explanation was the one I gave in my previous post - simply that because the two numbers are symbolically different, their value is different.

It doesn't matter that you can't find a rational number between them. A Dedekind cut focuses on the "gap" between two numbers. It defines a set "A" that is less than "B", and yet does not specify the greatest value for "A." Therefore, "A" is unbounded and yet less than "B" (I'm not sure I said that quite right). Anyway, the two sets (I think they're called "rings") "A" and "B" have no intersection. Therefore, it is impossible for any of their members to be equal. Richman then shows that a Dedekind cut of the number line puts 0.9... and 1.0... in two different sets. So, they are not equal.

If it helps, think of this. Blayz arbitrarily multiplied x by 10. You don't have to do that. What if you multiplied x by 9. The trick should still work, but it doesn't. How do you do the multiplication? Maybe someone would argue that you get:

-x = -0.9...
9x = 8.9...
----------
8x = 8

x = 1

But that's not right. Think of a truncated version:

-x = -0.999
9x = 8.991
-----------
8x = 7.992

x = 0.999

Since multiplication requires starting at the right end of the number, you're actually approximating the infinite series in that algebra trick in order to bootstrap the process. You're also depending on an artifact of a 10-based system to make the trick work. If we worked in a hexadecimal system (like computers do), the trick wouldn't work because you would be using the wrong approximation. In other words, you would have to use the hexadecimal equivalent to make the trick work, and you would end up proving a different result (even though they are very similar).

Did that actually help?

 

[Tinker Grey]

The problem with the above is that you cannot start at the right of an infinite number.