IntegerQ[x] gives false. Just leave it as it is

Question

Since x is an alphabet not an integer,

Clear[x];IntegerQ[x]

gives

False

But sometimes I want to leave it as it is

I want to construct function IntegerQ2 :

Clear[x];{IntegerQ2[22],IntegerQ2[4.915],IntegerQ2[x]}

would give

{True,False,IntegerQ2[x]}

Can you help me? This problem is not just confined to IntegerQ. It applies to almost every predicate function.

small note: the kind of thing x is is a Symbol, not an alphabet! (just to help you search documentation) :) — thorimur, Commented Sep 11, 2021 at 19:08
Predicate functions are supposed to behave that way (maybe you knew this but stating it in case not). — Daniel Lichtblau, Commented Sep 12, 2021 at 4:26

kglr · Accepted Answer · 2021-09-11 13:05:30Z

12

ClearAll[IntegerQ2]
IntegerQ2[x_?NumericQ] := IntegerQ[x]

IntegerQ2 /@ {2, 4.95, x}

{True, False, IntegerQ2[x]}

answered Sep 11, 2021 at 13:05

kglr

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$\begingroup$ Thank you! General solution for BlahblahQ is related to my another question. Constructing DeterminedQ : mathematica.stackexchange.com/questions/242142/… $\endgroup$
– imida k
Commented Sep 11, 2021 at 13:14
1

$\begingroup$ I think replacing NumericQ with DeterminedQ would improve the code. Note that the question with URL in the comment is not perfectly solved. $\endgroup$
– imida k
Commented Sep 11, 2021 at 13:18

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