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Proof

(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 $ax=b.$
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References

Bibliography

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