Discussion
If the operation * is defined for all non-zero x and y by the equation x * y = (xy)2, then (x * y)* z =
*This question is included in Nova Math - Problem Set U: Algebraic Expressions, question #10
| (A) | x2y2z2 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 09/25/2014 13:49
I don't understand why you would add the exponent to z before finishing the inner parentheses area. Doesn't this violate PEMDAS?
Posted: 09/25/2014 15:19
Hi Darren, in this case think of the defined operation x*y like f(x,y). * here is not the multiplication operation. So,
f(x,y) = (xy)^2
f [f(x,y), z] = [(xy)^2 times z]^2
= (xy)^4 times z^2
= x^4 y^4 x^2
To answer your confusion, by adding the exponent to z, they are actually doing the outer parentheses first, since they are also adding the exponent to the other factor in the parentheses, ie, (xy)^2, at the same time as z.
f(x,y) = (xy)^2
f [f(x,y), z] = [(xy)^2 times z]^2
= (xy)^4 times z^2
= x^4 y^4 x^2
To answer your confusion, by adding the exponent to z, they are actually doing the outer parentheses first, since they are also adding the exponent to the other factor in the parentheses, ie, (xy)^2, at the same time as z.
Posted: 09/25/2014 15:34
Hi Darren,
Yes, it does violate PEMDAS. But no harm is done. Sometimes you can violate PEMDAS (though sometimes, of course, you cannot).
On occasion, violating PEMDAS may be more efficient, but not here because distributing the inner exponent first will lead to the same number of steps as distributing the outer exponent first.
Nova Press
Yes, it does violate PEMDAS. But no harm is done. Sometimes you can violate PEMDAS (though sometimes, of course, you cannot).
On occasion, violating PEMDAS may be more efficient, but not here because distributing the inner exponent first will lead to the same number of steps as distributing the outer exponent first.
Nova Press