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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