Why is the square root of every non-perfect square integer irrational?

Answer 1

If n is your number, use the rational root test on the polynomial
x^2 -n. The only possible rational roots are the factors of n, which are integers. But if one of those is a root, n is a perfect square.

This proof also works to show that the cube root of any non-cube is irrational.

Answer 2

Basically because you can break any square root of non-perfect square into the square root of prime factors.

sqrt(10) = sqrt(5) * sqrt(2)
sqrt(5.3) = sqrt(53) / (sqrt(2) * sqrt(5))

The square root of any prime number is always irrational.

Here is the proof. You can play around with the proof by using any root of a prime multiplied or divided by any other root of a prime and get the same basic result.

Let a and b be any integers, and p is a prime number
If there in an a and b that works in the equation a / b = sqrt(p)
Then sqrt(p) is rational.

a / b = sqrt(p)
a^2 / b^2 = p
Factor a and b into primes
(c^2)(d^2)(e^2) .... / ((f^2)(g^2)(h^2)...) = p
Now reduce the fraction on the left by illiminating all common factors from the numerator and denominator. Since it has been broken down to primes, for every pair elimiated in the denominator, a pair must be elimiated from the denominator.

The left side of the equation cannot be reduced further, since it contains no common factors.

The denominator must equal 1 since all reduced fractions are decimal numbers and p cannot contain a decimal.

(j^2)(k^2)(l^2) = p

The left side must only contain pairs of prime numbers, multiplied by other pairs of prime numbers. Since the left side of the equation must contain at least a single pair of prime numbers, and no two prime numbers can be multiplied together to result in a prime number, there is no a and b where a / b = sqrt(p). Therefore sqrt(p) is not rational.

Answer 3

There is a well known proof that shows that the square root of 2 is not a rational number.

The same proof works for arbitrary n where n isn't a perfect square.

You assume a/b is rational number written in lowest terms, i.e. a and b have no common factors and (a/b)^2 = n and show that a and b must have a common factor p, which is a contradiction.

Answer 4

I think your question is only relevant to the first part of this answer, but I wanted to clear up the second part, since Michael, although giving a very nice answer, was not thorough enough for me (he assumed that for two distinct primes p and q, if √p and √q are irrational, then √pq is irrational. this of course is true, but I don't think you can assume it in this type of proof.

This first part should answer your question:
A perfect square integer is an integer that has an integer square root. Your question is to prove that for any integer that has a rational square root, it has an integer square root:

Proof:

Lets a be an integer and √a=m/n where m and n are relatively prime (this means that no number other than 1 divides both; all rational numbers can be written of this form) integers.. then a=m^2/n^2. Since a is an integer m^2/n^2 is an integer, thus n^2 divides m^2. Since m and n are relatively prime, this means that n=1 and m/n=m is an integer.

This is just a more thorough (and probably a little more confusing) proof of what Michael M did.

Secondly, by the Fundamental Theorem of Arithmetic any number larger that 1 can be written as a product of powers of distinct primes (this product can be only one prime): Thus a=(p1)^(n1) • (p2)^(n2) • (p3)^(n3) • … • (ps)^(ns).

Consider √a= √[(p1)^(n1) • (p2)^(n2) • (p3)^(n3) • … • (ps)^(ns)]. First note that if (ni) is even, then √[(pi)^(ni)]= (pi)^((ni)/2) is an integer if (ni) is odd then √[(pi)^(ni)]= (pi)^(((ni)-1)/2)•√(pi). Label all of the primes that have odd powers distinctly by (q1), (q2), (q3), … (qt).

Thus √a is irrational if and only if √[(q1) • (q2) • (q3) • … • (qt)] is irrational.

Let's assume that √a is irrational, then √[(q1) • (q2) • (q3) • … • (qt)] is irrational and √[(q1) • (q2) • (q3) • … • (qt)] = m for some integer m (using the above arguments, since (q1) • (q2) • (q3) • … • (qt) is an integer).

Thus [(q1) • (q2) • (q3) • … • (qt)] = (√[(q1) • (q2) • (q3) • … • (qt)]) = m^2. Since (q1)•(q2)•(q3)•…•(qt)=m^2, (q1) divides m^2. By the property of primes this means (q1) divides m and thus (q1)^2 divides m^2. Therefore (q1)^2 divides (q1)•(q2)•(q3)•…•(qt) and (q1) divides (q2)•(q3)•…•(qt). But this is a contradiction since (q1) is distinct from (q2), (q3), … , (qt) and (q2), (q3), … , (qt) are all primes. Thus we have reached a contradiction is assuming that √a is rational.

Answer 5

A square root is the "root" of a square number. In other words the number that was squared to make the number that you want rooted, it's the reverse of squaring. for example the square root of 49 is 7 because 7 squared is 49. As for the definition of a square number "the result of any whole number multiplied by itself?"

Answer 6

This is a circular issue. In order for the square root to be "rational", it must be the square root of a perfect square.

Answer 7

Because non-perfect square roots cannot be represented as a normal fraction.

For instance, sq root of 3 is not definite. Can't be stated as a normal fraction.

Answer 8

try to do one by hand with no calculator.
ex. square root of 2=approx. 1.4142135623730950488016887242097
but if you do it by hand,
1.4*1.4=1.96
1.41*1.41=1.9881
1.414*1.414=1.999396
see as the numbers get closer to 2 but they are never going to end. a great example is that none of the double digits that are multiples of 10 is a perfect square. so that's the reason why they are irrational.
if one of double digits were a multiple of ten then none of the non-perfect squares would be rational.

Answer 9

Well, if you multiply a rational non-integer by itself, you get another rational non-integer.