off topic but please forgive me me and answer

Steven D'Aprano steve at REMOVE-THIS-cybersource.com.au
Sat Apr 3 22:24:28 EDT 2010


On Sat, 03 Apr 2010 10:56:37 -0700, Patrick Maupin wrote:

>> The square root of 2 is irrational, but if you multiply it by itself
>> then the result isn't irrational, so not all operations involving
>> irrational numbers will result in an irrational result (unless that's
>> what you mean by "closely related irrational numbers").
> 
> Yes, I think I am closely related to myself.  But in addition to that
> particular disclaimer, I qualified the statement with "most" and I also
> mentioned that zero is special.  I stand by the assertion that if you
> take a random assortment of non-zero numbers, some irrational, some
> rational, and a random assortment of numeric operators, that most
> operations involving an irrational number will have an irrational
> result.


There are an infinite number of rational numbers. There are an infinite 
number of irrational numbers. But the infinity of the rationals is 
countable (1, 2, 3, 4, ... or aleph-0) while the infinity of the 
irrationals is uncountable (c or aleph-1), so there are infinitely more 
irrationals than rationals.

To put it another way, even though there are an infinite number of 
rationals, they are vanishingly rare compared to the irrationals. If you 
could choose a random number from the real number line, it almost 
certainly would be irrational.

(This is not to be confused with floats, which of course are all rational 
numbers.)


-- 
Steven



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