Hello, I need some help on how to prove this using the field axioms of the real numbers Given a and b with a =/ 0, there is exactly one x such that ax = b. This x is denoted as b/a. Thanks in advance
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Originally Posted by lds09 Hello, I need some help on how to prove this using the field axioms of the real numbers Given a and b with a =/ 0, there is exactly one x such that ax = b. This x is denoted as b/a. Thanks in advance $\displaystyle ax=b$ then $\displaystyle a^{-1}(ax)=a^{-1}b$ then $\displaystyle (a^{-1}a)x=a^{-1}b$ thus $\displaystyle 1x = a^{-1}b$ thus $\displaystyle x=a^{-1}b$.
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