Does the set of irrational numbers form a group?

Question

Ethan Adkins · Accepted Answer

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

Since the sum or product of two irrational numbers can be a rational number and is not included in the set of irrational numbers, the set of irrational numbers does not form a group under addition or multiplication. Among the most basic examples would be: #sqrt(2) + (-sqrt(2)) = 0# #sqrt(2)*sqrt(2) = 2# #color(white)()# Footnote Addition, subtraction, multiplication, and division by non-zero numbers close some intriguing sets of numbers, which contain irrational 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.

Eva Adamson · Answer

undefined No, the set of irrational numbers does not form a group under addition or multiplication. While the irrational numbers are closed under both addition and multiplication, they do not satisfy the other group properties. Specifically, they lack an identity element and inverses.