(related to Proposition: Unique Solvability of $ax=b$)

- By hypothesis, $a\neq 0.$
- First, we show that $x=ba^{-1}$ really
*is*the solution of the equation $ax=b.$- Replacing \(x\) by \(ba^{-1}\) in the equation leads to $a(ba^{-1})=b.$
- The commutativity of multiplying real numbers leads to $a(a^{-1}b)=b.$
- The associativity of multiplying real numbers leads to $(aa^{-1})b=b.$
- The uniqueness of inverse real numbers leads to $1\cdot b=b.$
- The existence of one leads to $b=b.$
- Thus, $x=ba^{-1}$ is the solution of the equation $ax=b.$

- Now, we show this is
*the only*solution of the equation.- Take
*any*real number \(y\) with $ay=b.$ - Multyplying both sides of the equation by \(a^{-1}\) leads to $a^{-1}(ay)=a^{-1}b.$
- By associativity of multiplying real numbers we get $(a^{-1}a)y=a^{-1}b.$
- By the uniqueness of inverse real numbers it follows that $1\cdot y=a^{-1}b.$
- By the existence of one we get $y=a^{-1}b.$
- The commutativity of multiplying real numbers leads to $y=ba^{-1}.$
- Thus, the solution $(ba^{-1})$ is the only solution of the equation $ax=b.$∎

- Take

Parts: 1

**Forster Otto**: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983