Question

Mmm good question...

Squared numbers must end in  1,4,9,6,5 or 0

So if your whole number ends in 2,3,7 or 8 then the square root will have to be irrational.

I have not answered your question. I have only looked at the most obvious exclusions.  

Perhaps if you give some specific examples I could tell you what other logic I would use. ://

Maybe someone else would like weigh in?

A whole number's square root is rational if and only if the number is the square of another whole number.

Example- 4 is a whole number, and also the square of 2 (2²=4). 4's square root is 2.

Proof: no need to prove that the square root of a number that is a square of a whole number is rational (if a=b² and b is whole then the square root of a is b and b is whole therefore also rational).

Proving that if a whole number's square root is rational then it is also whole is a bit harder:

suppose a=(q/r)²=(q²/r²) where r and q are relatively prime:

q²/r²=a. suppose r is not 1 (meaning q/r is not a whole number). now i will prove that if q and r are relatively prime then q² and r² are also relatively prime: we can present r as some powers of prime numbers multiplied together:r=a₁^b₁*a₂^b₂.......*a_n^b_n where a₁.......a_n are  prime numbers and b₁.........b_n are natural numbers. that means:

r²=a₁^b₁*2*a₂^b₂*2.......*a_n^b_n*2. q=c₁^d₁*c₂^d₂.......*c_k^d_k and q²=c₁^d₁*2*c₂^d₂*2.......*c_k^d_k*2 similarly. because r and q are relatively prime, the primes a₁,a₂.......a_n and the primes c₁, ......, c_k are different (meaning there aren't any primes a_j and c_i so that a_j=c_i). the fundamental theorem of arithmetics (https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic) states that there is exactly one way to present a number as a multiplication of powers of primes.

Suppose r² and q² are not relatively prime. that means there exists a divisor d so that d*x₁=r² and d*x₂=q² . d can be presented as a multiplication of powers of prime numbers therefore we can change the equation:

(multiplication of powers of primes that results in d)*x₁=r² and (multiplication of powers of primes that results in d)*x₂=q² 

IN CONTRARY TO THE FUNDAMENTAL THEOREM OF ARITHMETICS (because that equation means there is a way to write q² and r² as a multiplication of powers of primes in a way that THERE EXISTS A_I AND C_J SO THAT A_I=C_J IN CONTRARY TO THE FACT THAT Q AND R ARE RELATIVELY PRIME)

And what did we get from that exhausting proof? that q² and r² are relatively prime. We know that q²/r²=a meaning they are not relatively prime (unless r=1 but we already mentioned that r is not 1). we assumed that r is not 1 and we got a contradiction meaning that r=1.

this means that if the square of a rational number is whole then that rational number is also whole.

~blarney master~