Proof of the Possibility of Division

[kripkrip420]

Hi there! I have been reading Apostol's "Calculus: Volume 1" and have been trying to prove a few theorems using specific field axioms(note that this is not a "homework" question since I have not been assigned it, but instead, chosen to attempt it out of curiosity). Although I am not a math major, I am interested in these proofs. The theorem follows as such;

Given a and b and a does not equal zero, there exists one such x that x=b/a. This is called the quotient...etc.

Can someone help me prove the above theorem using the field axioms? I am not sure where to start.

My attempt...

ax=b

choose one y such that ax(y)=1

then,

ax(y)=b(y)=1

and I get stuck there.

Please help me if you can! Thank you in advance!

 

[wisvuze]

use the unique multiplicative inverse law for non zero elements

 

[Stephen Tashi]

I don't have that book, so I'll have to guess at the axioms.

There's no need to think in terms of setting up an equation. The nonzero elements of a field are a group under multiplication. Each element of that group has a unique multiplicative inverse. What is the definition of a/b? I'd define it to mean "a times the multiplicative inverse of b". b inverse is unique so this defines a unique product. The product of two elements of the group exists and is unique. That's one of the properties of a group isn't it?

 

[TylerH]

[MUST READ] This forum is not for homework or any textbook-style questions.

I wish I did, but I do not know how to help with your proof. Try wisvuze's suggestion.

 

[galois26]

Possibility of division: Given a and b with not equal to cero, there is exactly one x such that a.x=b. This x is denoted by b/a and is called the quotient of b and a. In particular, 1/a is also written a^-1 and is calle the reciprocal of a.

My english is bad but i will try to do the proof in an understandable way

proof: because b/a is the solution to the equation a.x=b and this solution is unique. It is enough to prove that b.a^-1 is also solution of this equation. And indeed, if x=b/a^-1 then:

a.x=a.(b.a^-1)=a.(a^-1..b)=(a.a^-1).b=1.1=b so we proved that b/a= b.a^1

End

 

[galois26]

But I have a new question, how to prove that -0=0 with the field axioms? how would you prove that? please help me out of this, i looked this exercise in the book calculus of apostol first chapter perhaps it does ring a bell.

 

[micromass]

But I have a new question, how to prove that -0=0 with the field axioms? how would you prove that? please help me out of this, i looked this exercise in the book calculus of apostol first chapter perhaps it does ring a bell.

Well, -0=0 means that 0 is the additive inverse of 0. So what you must prove is that 0+0=0. Since additive inverses are unique, this would imply -0=0.

 

[TylerH]

I think it could also be proved using the definition of the additive identity.

Hint: Start with $a+0=a$

Spoiler

$$a+0=a \Rightarrow a=a-0 \Rightarrow -a+a=-a+a-0 \Rightarrow (-a+a)=(-a+a)-0 \Rightarrow 0=0-0=-0 \Rightarrow 0=-0$$

 

[galois26]

Thank you very much for your answers micromass and TylerH nice day... :)