MHB How to simplify algebraic expression

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The discussion focuses on simplifying the algebraic expression \frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{2}(\frac{x - 3}{x - 2})^{-\frac{1}{2}} \cdot \frac{1}{(x - 2)^{2}}. Participants confirm that the steps taken to simplify the expression are correct, leading to the final result of \frac{1}{2(x - 3)(x - 2)}. There is a consensus that no errors were made during the simplification process. The discussion emphasizes clarity in each transformation of the expression. Overall, the simplification appears accurate and well-validated by contributors.
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\frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{2}(\frac{x - 3}{x - 2})^{-\frac{1}{2}} \cdot \frac{1}{(x - 2)^{2}}

\frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{2}(\frac{x - 3}{x - 2})^{-\frac{1}{2}} \cdot \frac{1}{(x - 2)^{2}} \\<br /> \frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{2}\frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{(x - 2)^{2}} \\<br /> \frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{\frac{1}{2}}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{(x - 2)^{2}} \\<br /> \frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{2(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{(x - 2)^{2}} \\<br /> \frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{(\frac{x - 3}{x - 2})^{\frac{1}{2}}} \cdot \frac{1}{2(x - 2)^{2}} \\<br /> \frac{1}{(\frac{x - 3}{x - 2})} \cdot \frac{1}{2(x - 2)^{2}} \\<br /> \frac{1}{(x - 3)} \cdot \frac{1}{2(x - 2)} \\<br /> \frac{1}{2(x - 3)(x - 2)}

Which step have I done incorrectly?
 
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I don't see that you have done any step incorrectly! What makes you thing you have?
 
Can't find any errors, seems right.
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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