Question

Prove that the set of real numbers is a field with respect to addition and multiplication. ​

Answer 1

Answer:

Show that the following set A of real numbers under addition and multipication is a field:

A=a+b2–√:a,b rational

I am not sure if I am right but here is what I have thus far:

Closure: Let a1+b12–√, a2+b22–√∈R. Then (a1+b12–√)+(a2+b22–√)=(a1+a2)+(b1+b2)2–√∈A.

Now, (a1+b12–√)×(a2+b22–√)=(a1×a2+2×b1×b2)+(a1×b2+b1×a2)2–√∈A So both addition and multipication are closed on A

Associativity: As addition and multipication are associative on R it follows from Restriction of Operation Associativity that they are also associative on the set A.

Commutativity: As addition and multipication are commutative on R it follows from Restriction of Operation Commutativity that they are also commutative on the set A.

Identity: We have (a+b2–√)+(0+02–√)=(a+0)+(b+0)2–√=a+b2–√ and similarly for (0+02–√)+(a+b2–√). So, (0+02–√) is the identity for addition on A. Now, for multipication we have (a+b2–√)(1+02–√)=(a×1+2×b×0)+(b×1+a×0)2–√=a+b2–√ and similarly for (1+02–√)(a+b2–√). So, (1+02–√) is the identity for multipication on A.

Inverses: (a+b2–√)+(−a+(−b)2–√)=(a−a)+(b−b)2–√=0+02–√ and similarly for (−a+(−b)2–√)+(a+b2–√). So, (−a+(−b)2–√) is the inverse of (a+b2–√) for addition on A.\ For product inverse consider the difference of squares: (a+b2–√)(a−b2–√)=a2−2b2 which leads to (a+2–√)(a−b2√a2−2b2)=1=1+02–√. So, demonstrating that the product inverse of (a+b2–√) is aa2−2b2−b2√a2−2b2 as a,b are rational, it follows that so are aa2−2b2 and ba2−2b2. So, the product inverse (a+b2–√)∈A