MHB Negative Exponents: Solve the Problem and Figure it Out

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Negative exponents can be confusing, but they can be simplified by substituting x^-2 with 1/x^2 and y^-2 with 1/y^2. The expression (x^-2)/(x^-2 + y^-2) can be transformed by multiplying the numerator and denominator by x^2y^2. This manipulation leads to the simplified result of (y^2)/(x^2 + y^2). Understanding these substitutions is key to solving problems involving negative exponents effectively.
Circe
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I hope I'm doing this correctly!

I'm having problems understanding negative exponents and how they work...

I have this problem, (x^-2)/(x^-2 + y^-2)

I don't understand how the answer could be, (y^2)/(x^2 + y^2)..

Maybe I'm doing something incorrectly? :/
 
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Circe said:
I hope I'm doing this correctly!

I'm having problems understanding negative exponents and how they work...

I have this problem, (x^-2)/(x^-2 + y^-2)

I don't understand how the answer could be, (y^2)/(x^2 + y^2)..

Maybe I'm doing something incorrectly? :/

welcome on MHB Circe!...

For the solution is sufficient the substitution $\displaystyle x^{- 2} = \frac{1}{x^{2}}$ and $\displaystyle y^{- 2} = \frac{1}{y^{2}}$ ...

Kind regards

$\chi$ $\sigma$
 
Try multiplying the expression by:

$$\frac{x^2y^2}{x^2y^2}$$

And see what you get...:D
 
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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