Does the set of irrational numbers form a group?

1 Answer
Apr 3, 2017

Answer:

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) + (-sqrt(2)) = 0#

#sqrt(2)*sqrt(2) = 2#

#color(white)()#
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+bsqrt(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+broot(3)(2)+croot(3)(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+...+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.