Radical

I am not yet sure how to define radicals generally. The natural concept to define here in standard mathematics would be algebraic numbers, but it's not clear what those mean in real life.

Square roots

The square root of an area $x$ , denoted ${\sqrt {x}}$ , is the side length a square with area $x$ .

Loosely speaking, standard mathematics defines the square root of, say, 2, to be the positive number $r$ such that $r^{2}=2$ . It is well known that no rational number satisfies that equation.^{[note 1]} Solutions to that equation can at best be approximated by a sequence of fractions. This means that standard math's definition of ${\sqrt {2}}$ comes hand in hand with a requirement for an infinite amount of precision.

Objective Mathematics does not require an infinite amount of precision in its definition of square root. The amount of precision required is determined by one's context. This means that if one desires to convert a square root into a fraction, then which fraction one chooses depends on the level of precision at which one wishes to measure the real, physical square that is under consideration. [TODO I'm not completely sure about this.]

Cube roots

The cube root of a number $x$ , denoted ${\sqrt[{3}]{x}}$ , is the side length of a cube with volume $x$ .

Imaginary numbers

We can extend the concept of square root to signed areas, beyond just areas. A signed area is a signed difference of areas (see the article integers for a detailed explanation), meaning that it reduces to some ordered pair of areas.

A complex integer reduces to an ordered pair of integers (which reduces to an ordered 4-tuple of natural numbers). These things satisfy some special multiplication law. The point of the multiplication law is that, if $x$ is any signed area, then we want the multiplication law to be defined in such a way that ${\sqrt {x}}\cdot {\sqrt {x}}=x$ , just like was the case for the square root.

Notes

↑ Proof by contradiction: Suppose $(p/q)^{2}=2$ , where $p,q$ are integers sharing no common factors. Then $p^{2}=2q^{2}$ , so 2 divides $p^{2}$ . It follows that 2 divides $p$ , so write $p=2k$ . Then $4k^{2}=2q^{2}$ , so $2k^{2}=q^{2}$ . But that means that 2 divides $q$ as well. We have reached a contradiction.

[1] Proof by contradiction: Suppose $(p/q)^{2}=2$ , where $p,q$ are integers sharing no common factors. Then $p^{2}=2q^{2}$ , so 2 divides $p^{2}$ . It follows that 2 divides $p$ , so write $p=2k$ . Then $4k^{2}=2q^{2}$ , so $2k^{2}=q^{2}$ . But that means that 2 divides $q$ as well. We have reached a contradiction.

[note 1]

Radical

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Square roots

Cube roots

Imaginary numbers

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Radical

Square roots

Cube roots

Imaginary numbers

Notes

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