Re: It is irrational to put rational and irrational numbers on the same line.

On Sep 16, 6:59 am, "Jesse F. Hughes" <je...@xxxxxxxxxxxxx> wrote:
lwal...@xxxxxxxxx writes:
On Sep 14, 12:39 am, adamle...@xxxxxxxxxxxx wrote:
Imaginary numbers have their own axis.

Notice that the OP goes by the name "finite guy,"
which strongly suggests that he's a finitist.

I really think that you're being far too charitable here. Finitists
do not tend to be quite so incoherent in their questions.

I see what you mean. I thought that it didn't take
a great leap to get from the username "finite guy"
to finitism, but you argue otherwise.

[You should be] answering questions as literally
as possible---and doing so using classical mathematics

Here's an answer to the OP's question in classical
mathematics -- and ironically, it turns out that
"finite guy" was right after all!

In the thread "relation of irrationals to rationals,"
Christopher Heckman and Gerry Myerson tell us that
contrary to popular belief, the proper definition of
the set of irrationals is not R\Q but C\Q! In other
words, the set of irrationals contains every complex
number with nonzero imaginary part. (Once again,
please refer to that thread for their explanation as
to why the set of irrational numbers is C\Q.)

OP argues "it is irrational to put rational [i.e., Q]
and irrational [i.e., C\Q] numbers on the same line."

On what line do the rational numbers lie? The real
line obviously. The line on which the rationals must
lie is R.

Now do the irrational numbers C\Q belong on R? Of
course not -- in particular, the irrational numbers
with nonzero imaginary part don't. Indeed, it is
irrational to put irrational numbers like i on the
same line as the rational numbers! And notice that
the OP even mentions complex number in his post:

"Imaginary numbers have their own axis."

So using the right (per Heckman and Myerson)
definition of the irrationals, we see that the OP
is actually correct -- even more correct than the
OP himself even realizes. The set of irrational
numbers is not a subset of the line on which
the rationals belong, the real line.

.

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