square root of a irrational square

Answer 1

The square root of an irrational square will still be irrational.

To understand this, let's consider the definition of an irrational number. An irrational number is a real number that cannot be expressed as a fraction of two integers.

If we take the square of an irrational number x, we get an irrational square, which means it cannot be expressed as a fraction of two integers.

Now, let's assume that there exists a rational number y, such that y is the square root of the irrational square (y = √x^2).

In this case, we can square both sides of the equation:

y^2 = (√x^2)^2
y^2 = x^2

Now, if y is rational, then it can be expressed as a fraction of two integers. Let's say y = a/b, where a and b are integers with no common factors.

Then, we have:

(a/b)^2 = x^2
a^2/b^2 = x^2
a^2 = b^2 * x^2

Since b^2 and x^2 are both integers, a^2 must also be an integer. This implies that a is an integer.

Now, let's consider the prime factorization of a. Since a is an integer, it can be expressed as a product of prime factors:

a = p1^m1 * p2^m2 * ... * pn^mn

where p1, p2, ..., pn are prime numbers, and m1, m2, ..., mn are positive integers.

Now, substituting the prime factorization of a into our equation:

(p1^m1 * p2^m2 * ... * pn^mn)^2 = b^2 * x^2
(p1^(2m1) * p2^(2m2) * ... * pn^(2mn)) = b^2 * x^2

This implies that each prime factor on the left side of the equation must also appear in the prime factorization of b^2 * x^2.

Now, let's consider the prime factorization of b^2 * x^2. If we assume that b and x are irrational, then each of their prime factorizations would only contain primes to the power of 0 or 1. Therefore, their square would still have primes to the power of 0, 1, 2, or 3, but no higher power.

This means that the right side of the equation b^2 * x^2 can only contain primes to the power of 0, 1, 2, or 3, while the left side contains primes to the power of 2m1, 2m2, ..., 2mn.

Since the left side and the right side of the equation represent the same integer, we have a contradiction. Therefore, our assumption that y is rational must be false.

Hence, the square root of an irrational square is always irrational.