Paul notes that if $\mathrm{a}=3$ and $\mathrm{b}=-3$, then $\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{a}}=-1$. He concludes that for any real numbers $\mathrm{a}$ and $\mathrm{b}, \frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{a}}$. Do you agree with his reasoning? If not, why not?

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The Answer of Paul notes that if $\mathrm{a}=3$ and $\mathrm{b}=-3$, then $\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{a}}=-1$. He concludes that for any real numbers $\mathrm{a}$ and $\mathrm{b}, \frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{a}}$. Do you agree with his reasoning? If not, why not?

Paul Notes That If a=3\mathrm{a}=3a=3 And b=−3\mathrm{b}=-3b=−3 , Then ab=ba=−1\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{a}}=-1ba​=ab​=−1 He Concludes That for Any Real Numbers

Paul Notes That If $\mathrm{a}=3$ And $\mathrm{b}=-3$ , Then $\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{a}}=-1$ He Concludes That for Any Real Numbers