Question: If $F$ is a field, and $a, b, c \in F$, then prove that if $a+b = a+c$, then $b=c$ by using the axioms for a field.

Relevant information:
Field Axioms (for $a, b, c \in F$):

$a+b = b+a$ (Commutativity)
$a+(b+c) = (a+b)+c$ (Associativity)
$a+0 = a$ (Identity element exists)
$a+(-a) = 0$ (Inverse exists)

$ab = ba$ (Commutativity)
$a(bc) = (ab)c$ (Associativity)
$a1 = a$ (Identity element exists)
$aa^{-1} = 1$ (Inverse exists)

Distributive Property:
$a(b+c) = ab + ac$

Attempt at solution:
I'm not sure where I can begin. Is it ok to start with adding the inverse of a to both sides, as in the following?
$(a+b)+(-a) = (a+c)+(-a)$ (Justification?)
$(b+a)+(-a) = (c+a)+(-a)$ (Commutativity)
$b+(a+(-a)) = c+(a+(-a))$ (Associativity)
$b+0 = c+0$ (Definition of additive inverse)
$b = c$ (Definition of additive identity)

I'm wondering about my very first step. Specifically, the axioms don't mention anything about doing something to both sides of an equation simultaneously. Is there some other axiom I can use to justify this step?

This is Exercise 1, part b in Section 1 on page 2 of Halmos, Finite Dimensional Vector Spaces (reading book for fun--this is not homework (probably too easy to be a homework problem anyway!)). In part a, I proved that $0+a = a$, in case that is somehow helpful in this problem. Thanks!

  • $\begingroup$ Just a minor note: you may want to write $1/a$ as $a^{-1}$ because you would want to define "/" later using the inverses and it wouldn't be good to have syntax with ambiguous parsing. $\endgroup$
    – user21820
    Jan 25, 2014 at 11:33
  • $\begingroup$ That's a good idea--I made the change. $\endgroup$
    – Mike Bell
    Jan 25, 2014 at 11:44

3 Answers 3


Martín-Blas Pérez Pinilla suggests that "=" can be considered a logical symbol obeying logical axioms. While I agree that it fundamentally is so, I would like to note that it is possible to consider it an equivalence relation obeying 'internal' field axioms, because for example the rational numbers can be taken as equivalence classes of a certain set of pairs of integers, and so it is not quite right to consider the equality between these rationals as a logical equality. Also, Ittay made a mistake where he used an unstated axiom that allows substitution. What you need, either way, is something equivalent to the following for any field $F$:

$a=a$ for any $a \in F$ [reflexivity of =]

$a=b \Rightarrow b=a$ for any $a,b \in F$ [commutativity of =]

$a=b \wedge b=c \Rightarrow a=c$ for any $a,b,c \in F$ [transitivity of =]

(These describe "=" as an equivalence relation on $F$)

$a=b \Rightarrow P(a)=P(b)$ for any $a,b \in F$ and predicate $P$ [substitution]

(This describes substitution, which can be used to replace separate axioms governing how "=" and the field operations interact. Ittay used this in one of his steps.)

These allow us to "do the same thing to both sides", for example:

For any $a,b,c \in F$ such that $a=b$,

  Let $d=a+c$ [closure under +]

  $a+c=a+c$ [transitivity of =; $a+c=d=a+c$]

  $a+c=b+c$ [substitution; where the predicate is given by $P(x) \equiv (a+c=x+c)$]

Note that to prove that something is a field, we will have to prove the substitution axiom, which boils down to proving the following equivalent set of axioms:

$a=b \Rightarrow a+c=b+c$ for any $a,b,c \in F$

$a=b \Rightarrow ac=bc$ for any $a,b,c \in F$

The original problem can then be proven as follows:

For any $a,b,c \in F$ such that $a+b=a+c$,

  $b = 0+b = (-a+a)+b = (-a)+(a+b) = (-a)+(a+c) = (-a+a)+c = 0+c = c$

  • $\begingroup$ Hurkyl had given a comment indicating that we could use substitution of an argument of $+$ considered as a 2-argument function. As I noted, fields constructed through equivalence relations will still require one to prove the substitution rule before using, so it amounts to the same two more basic axioms that I gave. $\endgroup$
    – user21820
    Jan 25, 2014 at 11:48

The first step is justified by the existence of $-a$. No further axiom is required in order to deduce that $(a+b)+(-a)=(a+c)+(-a)$. To see that, just write $a+b=y=a+c$ (after all, it is given that $a+b=a+c$). Now, one of the information defining a field is the function of addition: $(u,v)\mapsto u+v$. Applying it to $u=y$ and $v=(-a)$, yields the element $y+(-a)$. But, since $y=a+b$, $y+(-a)=(a+b)+(-a)$. Similarly, since $y=a+c$, $y+(-a)=(a+c)+(-a)$. By transitivity of equality, it follows that $(a+b)+(-a)=(a+c)+(-a)$.

Note that it would be a lot easier to add $(-a)$ on the left rather than on the right. It will save you using commutativity.

  • 1
    $\begingroup$ The step where you substitute $(a+b)$ into $y$ requires another axiom such as substitution. (See my answer.) With substitution or the alternative, you can just do it in one line. And that very axiom is the most important part of the problem! $\endgroup$
    – user21820
    Jan 25, 2014 at 11:19
  • $\begingroup$ Yes, you can put $(-a)$ on the left. But the axiom says $a + (-a) = 0$, not $(-a) + a = 0$. So you have to use Commutativity anyway! $\endgroup$
    – TonyK
    Jan 25, 2014 at 11:26

"Specifically, the axioms don't mention anything about doing something to both sides of an equation simultaneously." Because is a logical axiom. If you have two equal things and do the same with both things the results are equal. Search "first order logic with identity".

  • $\begingroup$ See also en.wikipedia.org/wiki/…, for when we don't have two equal things but have two things that behave the same way under some operations, and we still want to have 'substitution'. $\endgroup$
    – user21820
    Jan 25, 2014 at 11:29

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