Simplication: eq1: sqrt[x^2 y^3], eq2: sqrt[x^2 y^2 y], eq3: |xy| sqrt[y], eq4: ...

Dale10101

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\(\displaystyle \begin{array}{l}{\rm{eq1:}}\,\,\,\,\,\sqrt {{x^2}{y^3}} \,\,\,\,\, = \\{\rm{eq2:}}\,\,\,\,\,\sqrt {{x^2}{y^2}y} \,\,\,\,\, = \\{\rm{eq3:}}\,\,\,\,|xy|\sqrt y \,\,\,\,\, = \\{\rm{eq4:}}\,\,\,\,y|x|\sqrt y \end{array}\)

Is the expression 3 or 4 the correct simplification? It seems to me that I haven't seen the last step taken as a normal course yet it seem clear that y >= 0 (else the expression is undefined) and so can be taken out of the absolute values bars?
 
\(\displaystyle \begin{array}{l}{\rm{eq1:}}\,\,\,\,\,\sqrt {{x^2}{y^3}} \,\,\,\,\, = \\{\rm{eq2:}}\,\,\,\,\,\sqrt {{x^2}{y^2}y} \,\,\,\,\, = \\{\rm{eq3:}}\,\,\,\,|xy|\sqrt y \,\,\,\,\, = \\{\rm{eq4:}}\,\,\,\,y|x|\sqrt y \end{array}\)

Is the expression 3 or 4 the correct simplification? It seems to me that I haven't seen the last step taken as a normal course yet it seem clear that y >= 0 (else the expression is undefined) and so can be taken out of the absolute values bars?

How is it clear that \(\displaystyle \displaystyle y \geq 0\)? There's no reason to believe that y can't be complex...

I would write \(\displaystyle \displaystyle \begin{align*} \left| x \right| \, \left| y \right| \,\sqrt{y} \end{align*}\)
 
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How is it clear that \(\displaystyle \displaystyle y \geq 0\)? There's no reason to believe that y can't be complex...

I would write \(\displaystyle \displaystyle \begin{align*} \left| x \right| \, \left| y \right| \,\sqrt{y} \end{align*}\)

If the original function

\(\displaystyle f(x,y) \ = \ \sqrt{x^2y^3}\)

is constrained to real domains and range - then y ≥ 0.
 
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