Show that the set of all real numbers is an abelian group with respect to addition?

Answer 1

Sure thing, honey. The set of all real numbers is indeed an abelian group under addition. It's closed because adding two real numbers gives you another real number. It's associative because math plays nice like that. The identity element is 0, and every real number has an inverse (just slap a negative sign in front of it). Plus, addition is commutative, so you can add those numbers in any order and still get the same result. Voilà, you've got yourself an abelian group!

Answer 2

To show that the set of all real numbers is an abelian group with respect to addition, we need to verify the group properties:

1. Closure: For any two real numbers a and b, their sum a + b is also a real number.
2. Associativity: Addition of real numbers is associative, meaning (a + b) + c = a + (b + c) for all real numbers a, b, and c.
3. Identity element: The real number 0 serves as the identity element since a + 0 = a for all real numbers a.
4. Inverse element: For every real number a, its additive inverse -a exists such that a + (-a) = 0.
5. Commutativity: Addition of real numbers is commutative, meaning a + b = b + a for all real numbers a and b.

Since the set of real numbers satisfies all these properties, it is indeed an abelian group with respect to addition.

Answer 3

Let R represent the set of real numbers. Then

Closure

For all x and y in R, x+y belongs to R.

Associativity

For all x, y and z in R, (x + y) + z = x + (y + z).

Identity element

There exists an element in R, denoted by 0, such that for every x in R, x + 0 = x = 0 + x.

Inverse element

For each x in R, there exists an element y in Rsuch that x + y = 0 = y + x where 0 is the identity element (defined above). y is denoted by -x.

The above proves that R is a group.

Commutativity

For any x and y in R, x + y = y + x.

The group is, therefore, Abelian.