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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:

  1. $b_i \neq b \land a_i \neq a$:

$$a \cdot x_i = b$$

$$a \cdot \frac{b_i}{a_i} = b$$

But we cannot simplify that equation, since $\frac{b_i}{a_i} \neq \frac{b}{a}$. So this is not a solution.

  1. $b_i \neq b \land a_i = a$:

$$a \cdot \frac{b_i}{a_i} = b$$

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.

  1. $b_i = b \land a_i \neq a$

(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 $x \cdot a = b$.

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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.

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  • $\begingroup$ Do you have any example where $a=b$ but $ca\neq cb$ does hold? Only that direction is needed. The other direction ($(ca=cb \wedge c\neq 0) \Rightarrow b=a$) is what requires proof and which follows from your line of reasoning. $\endgroup$ – Thomas Mar 5 '15 at 17:18
  • $\begingroup$ put: $C=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, $B=\begin{bmatrix}0&-1\\0&0\end{bmatrix}$, $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$:. You have $C(A-B)=0$ but $C\ne 0$ and $A\ne B$. $\endgroup$ – Emilio Novati Mar 5 '15 at 18:00
  • $\begingroup$ @Thomas. Sorry! I misunderstood your question, and I've not see the typo in my answer where I used $\Rightarrow$ instead $\iff$. I Edit. $\endgroup$ – Emilio Novati Mar 5 '15 at 18:20

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