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

LunarsCry
Sep 29, 2017

#1**+1 **

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?**

Melody
Sep 29, 2017

#3**+1 **

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 (2^{2}=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=b^{2} 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}=(q^{2}/r^{2}) where r and q are relatively prime:

q^{2}/r^{2}=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 q^{2} and r^{2} are also relatively prime: we can present r as some powers of prime numbers multiplied together:r=a_{1}^{b1}*a_{2}^{b2}.......*a_{n}^{bn} where a_{1}.......a_{n} are prime numbers and b_{1}.........b_{n} are natural numbers. that means:

r^{2}=a_{1}^{b1*2}*a_{2}^{b2*2}.......*a_{n}^{bn*2}. q=c_{1}^{d}^{1}*c_{2}^{d}^{2}.......*c_{k}^{d}^{k} and q^{2}=c_{1}^{d1*2}*c_{2}^{d2*2}.......*c_{k}^{d}^{k*2 }similarly. because r and q are relatively prime, the primes a_{1},a_{2}.......a_{n} and the primes c_{1}, ......, 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 r^{2} and q^{2} are not relatively prime. that means there exists a divisor d so that d*x_{1}=r^{2} and d*x_{2}=q^{2} . 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)*x_{1}=r^{2} and (multiplication of powers of primes that results in d)*x_{2}=q^{2}

IN CONTRARY TO THE FUNDAMENTAL THEOREM OF ARITHMETICS (because that equation means there is a way to write q^{2} and r^{2} 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 q^{2} and r^{2} are relatively prime. We know that q^{2}/r^{2}=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

edited by Guest Sep 29, 2017