# Does the set of irrational numbers form a group?

##### 1 Answer
Apr 3, 2017

No - Irrational numbers are not closed under addition or multiplication.

#### Explanation:

The set of irrational numbers does not form a group under addition or multiplication, since the sum or product of two irrational numbers can be a rational number and therefore not part of the set of irrational numbers.

About the simplest examples might be:

$\sqrt{2} + \left(- \sqrt{2}\right) = 0$

$\sqrt{2} \cdot \sqrt{2} = 2$

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Footnote

Some interesting sets of numbers that include irrational numbers are closed under addition, subtraction, multiplication and division by non-zero numbers.

For example, the set of numbers of the form $a + b \sqrt{2}$ where $a , b$ are rational is closed under these arithmetical operations.

If you try the same with cube roots, you find that you need to consider numbers like: $a + b \sqrt{2} + c \sqrt{4}$, with $a , b , c$ rational.

More generally, if $\alpha$ is a zero of a polynomial of degree $n$ with rational coefficients, then the set of numbers of the form ${a}_{0} + {a}_{1} \alpha + {a}_{2} {\alpha}^{2} + \ldots + {a}_{n - 1} {\alpha}^{n - 1}$ with ${a}_{i}$ rational is closed under these arithmetic operations. That is, the set of such numbers forms a field.