Surds

Surds are irrational \(n\)th roots.

What do you mean by that?

Well, consider \(\sqrt{4}\). This can be simplified down to an integer, \(2\), since \(2^2=4\).

Consider \(\sqrt{9}\), which simplifies to \(3\). What about \(\sqrt[3]{27}\)? That simplifies to \(3\) as well. Hence, all \(\sqrt{4}\), \(\sqrt{9}\) and \(\sqrt[3]{27}\) are rational \(n\)th roots.

But what about \(\sqrt{2}\)? Can it be boiled down into a rational number? Turns out that it can’t. There’s a nice proof too.

So, surds are \(n\)th roots that cannot be written as a fraction of two integers.

Example Please

Square roots of: \(2\), \(3\), \(5\), \(6\), \(101\) and anything that is not a square of a rational number (hence irrational). These are surds.

Cube roots of… well, of every integer that isn’t cubes. These are also surds.

And… so on.

Did you notice that if the square root of an integer is not an integer, the square root is irrational? Pretty cool, huh?