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POLL: Math time!
 
Neo7  posted on Oct 10, 2011 6:51:56 PM - Report post

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Just a quick review:

Rational numbers are real numbers that can be expressed in the form a/b where b is not zero.

Irrational numbers are real numbers that cannot be simplified into a/b and exhibit a non-terminating nor repeating decimal.

Let us call the "both are the same" option (3rd one) as the default that does not require proof. If you do pick the other two, let's see if you can come up with a reason why you picked your answer (because mathematics is all about trying to prove why a statement is true).

[Edited by Neo7, 10/10/2011 6:54:40 PM]

 
benduhova  posted on Oct 10, 2011 7:03:15 PM - Report post

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before i give an answer what is your definition of a nonreal or false number please
 
Neo7  posted on Oct 10, 2011 7:14:02 PM - Report post

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A "non-real" number (formally known as a Complex Number) is a number that has the square root of -1 in it's expression.

Actually a "non-real number" is known as an imaginary number and revolves around the square root of -1 and it being raised to various powers.

A Complex Number is a number containing a real component and an imaginary component. It is expressed as the form a+bi where a is any real number and bi is a scalar (b) times the square root of -1 (i).

[Edited by Neo7, 10/10/2011 7:18:25 PM]

 
lamile  posted on Oct 10, 2011 8:24:30 PM - Report post

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Cheated a little for explanation.

There is a proof, called Cantor's diagonalization argument, which shows that the set of irrational numbers (non-terminating, non-repeating decimals) is uncountably infinite (while rational numbers are countably infinite).

 
Neo7  posted on Oct 10, 2011 8:47:02 PM - Report post

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And there goes the entire topic. Was kind of hoping it would last longer.
 
benduhova  posted on Oct 11, 2011 9:18:55 AM - Report post

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if you follow the view of constructive mathematics then a collection of numbers is subcountable if there is a partial surjection from the natural numbers onto it(or that a collection is no bigger than the counting numbers) so the answer could be c depending on your stance
 
Neo7  posted on Oct 11, 2011 11:11:59 AM - Report post

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quote:
originally posted by benduhova

if you follow the view of constructive mathematics then a collection of numbers is subcountable if there is a partial surjection from the natural numbers onto it(or that a collection is no bigger than the counting numbers) so the answer could be c depending on your stance

I think in this case, you need the bijection function (not the surjection function). You can represent an infinite amount of irrational numbers between 0 and 1 by using infinite strings of just 1s and 0s while you can represent the natural numbers as 1,2,3,~

Using Cantor's diagonalization, you will generate a particular number that will not be able to pair up with another rational number so long as they go out infinitely long (which makes creating a bijection impossible)

Day[9] probably explained it much better here: Link

[Edited by Neo7, 10/11/2011 11:14:13 AM]

 
benduhova  posted on Oct 11, 2011 11:31:30 AM - Report post

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The interpretation of Cantor's result will depend upon your view of mathematics. cantors argument shows no more than that there is no bijection between the natural numbers and T( be a set consisting of all infinite sequences of 0s and 1s). It does not rule out the possibility that the latter are subcountable
Cantor's argument contradicts epsilon-delta proof.
According to the epsilon-delta proof, width/length of the list converges to zero.
When we increase digits of the list, width/length monotonically decreases.
So, we can easily accept the result.

However, Cantor's argument is as follows.
When the number of digits reaches the actual infinity, suddenly width/length becomes 1.

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