It looks to me as if the fact that 0 is being compared is significant.

   0 (~:!.0) 1e_45
1
   0 (~:) 1e_45
1

I would expect the intolerant ~:!.0 to return 1, but the tolerant ~: to return 
0, except that I think that 0 may be multiplied by the default tolerance to 
produce the intolerant fit of 0 

You could also do a definition using match (-:) instead of not equal (~:)

   matchnotreal =. -.@-: +
   #I.0=(iscomplex=matchnotreal)1 j. -:^:a:1
1

J can be so surprising.

Cheers, bob

On Nov 29, 2014, at 10:09 PM, Henry Rich <[email protected]> wrote:

> Yes, I believe it would.
> 
> Whether that's desirable is a different question.
> 
> Henry Rich

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