# Prove that F satisfies all field axioms by method of direct verification

by GOsuchessplayer
Tags: axioms, field, method, prove, satisfies, verification
 P: 11 1. The problem statement, all variables and given/known data Consider the collection F of all real numbers of the form x+y√2, where x and y are rational numbers. Prove (by direct verification) that F satisfies all the field axioms (just like R) under the usual addition and multiplication. 2. Relevant equations Field axioms: There exist two binary operations, called addition + and multiplication ∗, such that the following hold: 1) commutativity x + y = y + x, xy = yx 2) associativity x + (y + z) = (x + y) + z, x(yz) = (xy)z 3) distributivity x(y + z) = xy + xz 4) Existence of 0; 1 such that x + 0 = x, 1 · x = x, 5) Existence of negatives: For every x there exists y such that x + y = 0. 6) Existence of reciprocals: For every x ̸= 0 there exists y such that xy = 1. 3. The attempt at a solution I just want to make sure I did this right. If you were to prove that a collection F ( a collection of all real #'s of the form x+y$$\sqrt{2}$$ where x & y are rational numbers ) satisfies all of the field axioms by direct verification, would you just do something like suppose m,n,o belong to F. then m=x1+y1$$\sqrt{2}$$, etc. and then you just say m+(n+o) = x1+y1$$\sqrt{2}$$ + ..... until you return to m+n+o = (m+n)+o ??? And then proceed to do so for all the axioms mentioned above? It's supposed to be really trivial right? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
 Sci Advisor HW Helper PF Gold P: 2,606 You're on the right track. To be completely rigorous, I'd add the steps to verify that the set F is actually closed under addition and multiplication. That is, for any $$x,y\in F$$ 1. $$x+y\in F,$$ 2. $$xy \in F.$$ Kind of trivial as you say, but proofs are about covering all of the bases.
P: 11
 Quote by fzero You're on the right track. To be completely rigorous, I'd add the steps to verify that the set F is actually closed under addition and multiplication. That is, for any $$x,y\in F$$ 1. $$x+y\in F,$$ 2. $$xy \in F.$$ Kind of trivial as you say, but proofs are about covering all of the bases.
Just for clarity,

Why is it necessary to claim that the set F is closed under addition and multiplication for any x,y \in F? In other words, what would happen if the set wasn't closed under addition and multiplication, is it possible for the axioms to hold and this to still be true? Or does it mean that it's impossible for the axioms to hold if the set is not closed under addition and multiplication.

HW Helper
PF Gold
P: 2,606

## Prove that F satisfies all field axioms by method of direct verification

 Quote by GOsuchessplayer Just for clarity, Why is it necessary to claim that the set F is closed under addition and multiplication for any x,y \in F? In other words, what would happen if the set wasn't closed under addition and multiplication, is it possible for the axioms to hold and this to still be true? Or does it mean that it's impossible for the axioms to hold if the set is not closed under addition and multiplication.
Closure is usually the first axiom in the definition. I brought it up because it wasn't listed in your problem. If the set wasn't closed under those operations, there'd be no point in verifying any of the other axioms since it couldn't be a field.
P: 11
 Quote by fzero Closure is usually the first axiom in the definition. I brought it up because it wasn't listed in your problem. If the set wasn't closed under those operations, there'd be no point in verifying any of the other axioms since it couldn't be a field.
Alright,

Got it, You've been a big help.

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