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

LunarsCry  Sep 29, 2017

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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