I have an exercise in my last assignment for calculus which is the following:
Let $a, b, x \in \mathbb{Q}$ with $a \neq 0$. Use only the field axioms and the properties which we showed in class (and can be found in the course notes) to show that the equation $a \cdot x = b$ holds for $x = b/a = b \cdot a^{−1}$ and that this is the only solution of the equation.
Ok, I have first tried to prove by contradiction (but this is not what they are requiring):
Assume there's another solution $x_i$ different from $x$, such that:
$$a \cdot x_i = b$$
and $x_i \neq \frac{b}{a}$.
This means that there exists a $b_i$ and $a_i$ such that $x_i = \frac{b_i}{a_i}$, and either $b_i \neq b$ or $a_i \neq a$, or both.
We have essentially 3 cases:
- $b_i \neq b \land a_i \neq a$:
$$a \cdot x_i = b$$
$$a \cdot \frac{b_i}{a_i} = b$$
But we cannot simplify that equation, since $\frac{b_i}{a_i} \neq \frac{b}{a}$. So this is not a solution.
- $b_i \neq b \land a_i = a$:
$$a \cdot \frac{b_i}{a_i} = b$$
We can replace $a_i$ with $a$:
$$a \cdot \frac{b_i}{a} = b$$
Since $\frac{1}{a}$ is the multiplicative inverse of $a$, we can simplify them and we obtain $1$:
$$b_i = b$$
But this is a contradiction.
- $b_i = b \land a_i \neq a$
(same as point 2)
This 3 cases contradict the fact that either $b_i \neq b \lor a_i \neq a$. So $b_i = b \land a_i = a$. So whenever we suppose there's another solution $x_i$ that solution is $\frac{b}{a} = x$, the unique solution.
That proof could also be correct (is it correct?), but I need to prove using the field axioms.
What I have tried using the field axioms:
We have the initial equation:
$$a \cdot x = b$$
If I am not wrong, we can multiply both sides by the same number, and the equality remains, even if the number is negative. But, doing this, am I just using the field axioms? If yes, which one?
I will multiply by $a^{-1}$:
$$a \cdot x \cdot a^{-1} = b \cdot a^{-1}$$
Now, by the field axioms associativity, we have:
$$a \cdot x \cdot a^{-1} = a \cdot a^{-1} \cdot x = b \cdot a^{-1}$$
Now, we have $a$ multiplied by its multiplicative inverse $a^{-1}$, which results in $1$. Can we actually do this just using the field axioms?
$$x = b \cdot a^{-1}$$
Which proves what we want to prove, that the only solution for $x$ is $\frac{b}{a}$.
I think we could also start from the solution, and multiplying by $a$, and, doing a similar process as above, we would arrive at saying $x \cdot a = b$.
The pertinent axiom is the fact that in field $\mathbb{F}= (F,+,\cdot)$ the product $\cdot$ is such that $(F,\cdot)$ is a commutative group. This means that every element $a \in F$ has an inverse such that $a\cdot a^{-1}=a^{-1}\cdot a=1$ where $1$ is the neutral element for the product $\cdot$. So, given the equation $a \cdot x=b$ you can mutiply both sides (on the left) for $a^{-1}$ and you find $$ a^{-1}\cdot a \cdot x= a^{-1}\cdot b \Rightarrow x=a^{-1}\cdot b$$ and, from commutativity: $x=b \cdot a^{-1}$
As you have noted here a key pass is to multiply both sides by $a^{-1}$ This is a special case of $ a=b \iff ca=cb$. Is this allways true? In a field YES. You can prove this by axioms noting that: $$ ca=cb \iff ca-cb=cb-cb \iff ca-cb=0 \iff c(a-b)=0 $$ Now, if $c$ is not a zero divisor, this is true only if $(a-b)=0$, i.e. $a=b$. And since in a field there are not zero divisors ( every element has an inverse), you have that $ ca=cb \iff a=b$.
But note that in a different structure, as a ring, the question is not so simple.