The square root of 0 = 0.
Rational is defined as a value that can be expressed as a quotient of integers.
For example, 0/1 = 0 and 0/2 = 0. Although 0/0 is undefined, the former examples are still true. Therefore, the square root of 0 is rational.
For your knowledge, the square root of any number is rational, except for negative numbers, which are irrational. This is because two identical integers cannot be multiplied to produce a negative integer.
For instance, the square root of 25 is 5 because 5x5 = 25.
However, the square root of -25 has no real roots because neither -5x-5 nor 5x5 = -25.
No. The cube root of 3, for example is irrational. But the square of cubert(3) is 3 to the power 2/3, which is irrational. Another example, pi2 is irrational (in fact so is pi to any non-zero power).
Rational zero test cannot be used to find irrational roots as well as rational roots.
The same as you would a rational number. Its distance from zero will represent the number, whether it is rational or irrational.
Zero (0) is a rational number, because it is a whole number and an integer.
An irrational number is a number that can't be expressed by a fraction having integers in both its numerator and denominator. A rational number can be. 2 is rational. The square root of 2 is irrational.
It is irrational. * The square root of any positive integer, except of a perfect square, is irrational. * The product of an irrational number and a rational number (except zero) is irrational.
Neither. It is not defined.
sqrt(2) is irrational. 3 is rational. The product of an irrational and a non-zero rational is irrational. A more fundamental proof would follow the lines of the proof that sqrt(2) is irrational.
No: Let r be some irrational number; as such it cannot be represented as s/t where s and t are both non-zero integers. Assume the square root of this irrational number r was rational. Then it can be represented in the form of p/q where p and q are both non-zero integers, ie √r = p/q As p is an integer, p² = p×p is also an integer, let y = p² And as q is an integer, q² = q×q is also an integer, let x = q² The number is the square of its square root, thus: (√r)² = (p/q)² = p²/q² = y/x but (√r)² = r, thus r = y/x and is a rational number. But r was chosen to be an irrational number, which is a contradiction (r cannot be both rational and irrational at the same time, so it cannot exist). Thus the square root of an irrational number cannot be rational. However, the square root of a rational number can be irrational, eg for the rational number ½ its square root (√½) is not rational.
The square root of 50 is an irrational number. Irrational numbers are real numbers that cannot be expressed as a fraction a/b where a and b are integers and b is non-zero. Rational numbers are numbers that can be expressed as a fraction a/b where a and b are integers and b is not zero. The square root of 50 is approximately 7.071067812, which cannot be expressed as a fraction of integers. For example, the square root of 50 is between 7 and 71/100, but even (707110/100000)2 is approximately 50.00045521, which is still not quite 50.
The square root of 50 is an irrational number. Irrational Numbers are real numbers that cannot be expressed as a fraction a/b where a and b are integers and b is non-zero. Rational numbers are numbers that can be expressed as a fraction a/b where a and b are integers and b is not zero. The square root of 50 is approximately 7.071067812, which cannot be expressed as a fraction of integers. For example, the square root of 50 is between 7 and 71/100, but even (707110/100000)2 is approximately 50.00045521, which is still not quite 50.
Rational numbers can be expressed in the form p/q (where q is not equal to zero). Irrational numbers cannot be expressed in this form. For example, the square root of 2 cannot be expressed as p/q.
Well, honey, the square root of zero is actually just zero. And guess what? Zero is as rational as it gets in the number world. So, yes, the square root of zero is a rational number. But hey, don't sweat it if math isn't your thing, we can't all be numbers wizards.
When a rational numbers is divided by an irrational number, the answer is irrational for every non-zero rational number.
No. The cube root of 3, for example is irrational. But the square of cubert(3) is 3 to the power 2/3, which is irrational. Another example, pi2 is irrational (in fact so is pi to any non-zero power).
Sure. For example, the sum of:root(2) and: - root(2) is zero, which is rational.
Such a product is always irrational - unless the rational number happens to be zero.