Without caluculating, how can you tell whether the square root of the whole number is rational or irrational?

LunarsCry  Sep 29, 2017

3+0 Answers


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.  smiley


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


Maybe someone else would like weigh in?

Melody  Sep 29, 2017

Thank You Melody ^^

LunarsCry  Sep 29, 2017

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=a1b1*a2b2.......*anbn where a1.......an are  prime numbers and b1.........bn are natural numbers. that means:

r2=a1b1*2*a2b2*2.......*anbn*2. q=c1d1*c2d2.......*ckdk and q2=c1d1*2*c2d2*2.......*ckdk*2 similarly. because r and q are relatively prime, the primes a1,a2.......an and the primes c1, ......, ck are different (meaning there aren't any primes aj and ci so that aj=ci). 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 AI AND CJ SO THAT AI=CJ 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~

Guest Sep 29, 2017
edited by Guest  Sep 29, 2017
edited by Guest  Sep 29, 2017

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