Date: 09/09/98 at 17:04:51 From: stephanie Subject: Contest question How can I show that the rationals are countable and the irrationals are uncountable? It seems like so many numbers. Help!
Date: 09/09/98 at 17:56:41 From: Doctor Floor Subject: Re: Contest question Hi Stephanie, To show that rationals are countable, you have to put them in a square like this: 1/1 2/1 3/1 4/1 5/1 ... 1/2 2/2 3/2 4/2 5/2 ... 1/3 2/3 3/3 4/3 5/3 ... 1/4 2/4 4/4 4/4 5/4 ... ........ There are double counts. That's no problem, because if the number of these is countable, then it certainly is when you have even less. Now you can count these in the following way: 1 2 6 7 15 .... 3 5 8 14 4 9 13 10 12 11 So you can count all rationals. Now for the real numbers. Suppose the number of real numbers is countable. Let's just write down the real numbers between 0 and 1. You can write them down in decimal progression, and in a row, because they are countably many. It could start something like this: 0.23498514750193845019847509845104975... 0.54394857852398509283450928374509827... 0.09182560197650921874310298650198473... 0.98798452908435435114757183591761251... .... We'll create a new number, digit-for-digit. Now if you take the first digit after the dot not being 2, then you have a number unequal to the first in the row. If you take the second unequal to 4 then the number is also equal to the second number. The third not equal to 1, the fourth not equal to 9, etc. If you go on to infinity, you create a real number between 0 and 1 and not listed. So you have a contradiction, because apparantly the list is not complete - and the number of real numbers is not countable. If you have a math question again, please send it to Dr. Math. Best regards, - Doctor Floor, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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