Mathematical Cranks and Fallacies

[kingjoshi]

Pi is neither on the interval [0, 1] nor rational...

this method covers every real number, including irrationals. one just has to extend to include all such subsets {[1,2] [2,3] ...}

 

[pboily]

Re: the spoilered hint-

That was my first thought, but if n continues to infinity without a problem, repeating decimals will be included eventually.

Your should always trust your first thought, then, even if you're an engineer.

See, I can tell you where 1/4 will fall, it will be in the second list, since 1/4=0.25 (two digits).

1/200 will be in the third list since 1/200=0.005 (three digits).

In what list would 1/3 fall? (that's really the crux of the problem: can you take any real number between 0 and 1 and find the list in which it falls? If so, then [0,1] would have to be countable, because it would be a countable union (indexed by N) of countable lists.)

In fact, you can't, as Cantor has shown [0,1] to be uncountable.

 

[warpus]

You're only counting rational numbers; you should be counting irrational numbers (1/3, pi) as well. The set of rational numbers is countable, while the set of irrational numbers is uncountable.

The set of real numbers contains both rational and irrational numbers. The technique you propose to count all the real numbers is not including any irrational numbers, which is why the set you end up with is countable.

So you might ask, why is my technique not counting any irrational numbers?

ALL the numbers you list can be written as fractions. Your technique is recursive - you could easily use induction to show that every single number produced by your technique can be expressed as a fraction.

I leave that proof as an exercise to you, the OP!

 

[Mise]

1/3 is rational but I know what you mean.

 

[Truronian]

Is this (the 'proof' in the OP) the basis for proving that the set of rational numbers is countable (repeating using each integer as a base rather than just ten)?

 

[warpus]

Haha 1/3 is irrational

Whoops!!!!

I shouldn't have hit that bong during the Arsenal game

 

[pboily]

You're only counting rational numbers

And not even all of them (see the 1/3 kerfaffle). That's another good answer.

 

[pboily]

Is this (the 'proof' in the OP) the basis for proving that the set of rational numbers is countable (repeating using each integer as a base rather than just ten)?

Short answer: yes. Long answer: not quite, there are some annoying details.

list all the rationals that have 1 as a denominator (there are countably many of them.)

1/1 2/1 3/1 4/1 5/1 ...

list all the rationals that have 2 as a denominator (there are countably many of them.)

1/2 2/2 3/2 4/2 5/2 ...

there are also countably many of them, although some are repeated from the first list: 1/1=2/2, etc...

...

list all the rationals that have n as a denominator (there are again countably many of them.)

1/n 2/n 3/n 4/n 5/n ...

(again, with repetitions: n/n=1/1, etc...)

...

For each n, the list above is countable. The set of all positive rationals is exactly the (countable) union of all these (countable) lists, minus the repetitions, so the positive rationals are countable. The positive rationals are in 1-to-1 correspondance with the negative rationals, so their union is also countable. Add the set {0} in there, and you've got all Q, and it's rational.

EDIT: one of the many bijections from Q to N is

[U]n[/U]   <-----> n + [U](m+n-1)(m+n-2)[/U]
m                     2

 

[pboily]

Now, shouldn't the Transactions of the Wisconsin Academy of Sciences, Arts and Letters have published some sort of retraction? I know this was published in 1974 (seems like an eternity ago) but how seriously can that journal be taken after?

EDIT: I read that there was a huge commotion about this. Dilworth sued U. Dudley, who exposed his proof as crankiness incarnate. Dudley won in the end, but some of the court transcripts...

Circuit Opinion for Dilworth vs. Dudley:
http://www.law.emory.edu/7circuit/jan96/95-2282.html

 

[sysyphus]

Bunch of geeks.

Of course, as an engineer I favour applied math, but delving into pure math on occasion is quite interesting. I had a prof in uni who demonstrated that the hypoteneuse of a right triangle with sides both of length 1 was 2 (rather than sqrt(2)).

The proof relied on modelling the hypoteneuse as an infinite series of vertical and horizontal lines.

 

[warpus]

Assumption: Each element of a set C of all real numbers with x # of digits can be expressed as a fraction, if we use the method from the OP to obtain set C.

Base case 1: x=0
----------------
0 = 0/1
1 = 1/1

Base case 2: x=1
----------------
0.0 = 0/10
0.1 = 1/10
0.2 = 2/10
...
0.9 = 9/10

Theorem holds for both base cases.

Inductive hypothesis: If each element in the set A(n) of all real numbers with n # of digits can be expressed as a fraction, then each element in the set B(n+1) of all real numbers with n+1 # of digits can be expressed as a fraction.

Inductive step:

We know that we can express any element from A(n) as a fraction (this was assumed). We denote an element from A(n) as:

e(i) = i / [ 10^(n-1) ]

where i goes from 1 to 10^(n-1). So, for example, for n=2, we have elements e(1) = 1/ 10^1, e(2) = 2/10^2, e(3) = 3/10^3 ... e(9) = 9/9^3 which is another way of writing out the set 0.1, 0.2, 0.3, ... 0.9

Now, note that any element in B(n+1) can be written as:

e(i) = i / [10^(x-1)], where i goes from 1 to 10^(x-1).

or as e(i) = i / [10^n], where i goes from 1 to 10^n, and where x=(n+1)

Now, we know that i / [10^(n-1)] is a fraction and we have to show that i / [10^n] is a fraction too. If we can do that, the proof is done.

If i / [10^(n-1)] is a fraction, then i has to be a rational number. Therefore i is a rational number.

If i is a rational number, then i / [10^n] is a rational number as well.

Therefore our initial assumption must be correct.

Therefore each element of a set C of all real numbers with x # of digits can be expressed as a fraction, if we use the method from the OP to obtain set C.

Therefore the method used to count real numbers in the OP is ONLY including fractions; it's excluding all rational numbers.

Therefore it is not a correct method that can be used to list all real numbers, since there exist irrational real numbers.

By assuming that the method is correct we arrive at an incorrect conclusion - namely that the set of real numbers is countable. Therein lies the paradox.

QED

 

[pboily]

...

much overkill... suffices to notice that the Dilworth list contains only the real numbers between [0,1] with a finite decimal representation, and so cannot be the complete list of real numbers between [0,1].

Of course, we also need to show that the real numbers with unending decimal representation are uncountable, but Cantor's diagonal argument will do the trick.

Another common fallacy is to use the fact that R is uncountable to show that R is uncountable...

 

[NewbieHere]

This is unrelated to the previous spook "proof" but it is quite interesting as well.

I will now prove that 1 = -1

1 *** -1
--- = ---
-1*** 1

* __ ***__
\/ 1***\/-1
----- = ----
*__ *** __
\/-1***\/ 1


1 *** i
--- = ---
i *** 1

** Ignore the asterisks, they're there to make sure it looks good.

Now we cross-multiply

1 = i^2
1 = -1

 

[Irish Caesar]

...Except that 1/i is -i...

 

[Mise]

When you take the square root of something, you get two solutions, a positive and a negative one. You have assumed the positive solution, which, since you got the wrong answer, was the wrong solution to pick.

 

[NewbieHere]

When you take the square root of something, you get two solutions, a positive and a negative one. You have assumed the positive solution, which, since you got the wrong answer, was the wrong solution to pick.

Very nice. I'm glad someone picked this up so soon

Now I shall go along with you. Indeed, when you take the square root, you get two roots, one positive, one negative, so here goes:

* __ ***__
\/ 1***\/-1
----- = ----
*__ *** __
\/-1***\/ 1


+-1 *** +-i
---- = ----
+-i *** +-1

We still cross multiply:

(+-1)(+-1) = (+-i)(+-i)

Since both terms on both sides of the equations are the same, we can rewrite them as:

(+-1)^2 = (+-i)^2
1 = -1



Irish Caesar:

Yes, 1/i is -1 (assuming both is positive, which is not a reasonable assumption as mentioned by Mise). But if we look at this step:

1 *** i
--- = ---
i *** 1

And now simply it down (without cross multiplying)

1/i = -i
i/1 = i

-i = i

 

[Irish Caesar]

Merely pointing out more reason that it is wrong.

 

[warpus]

much overkill... suffices to notice that the Dilworth list contains only the real numbers between [0,1] with a finite decimal representation, and so cannot be the complete list of real numbers between [0,1].

Of course, we also need to show that the real numbers with unending decimal representation are uncountable, but Cantor's diagonal argument will do the trick.

Another common fallacy is to use the fact that R is uncountable to show that R is uncountable...

Yeah, I know, I was bored

 

[RedWolf]

I hate math its my worst subject. Too bad everything in the computer field requires lots of math.

I'm gainfully employed in the computer industry yet yours is the onl post in this thread I understood. Seriously.

It's good to realize every once in a while that there are a whole lotta people smarter then yourself. Keeps you humble.

 

[Tenochtitlan]

I find math to be a waste of time, money and resources (for me atleast). All you need to know is a few applicable concepts and know how to apply them. They are all what you will need. Why do I have to learn in school all the unnecessary math when I do not even enjoy it?
I will never need to use 95% of it anyway. We have computers that can do the math for people like me (who are also the majority).