I know if you have an integer exponent you flip the fraction ex)x^-2 = 1/x^2
But what if you have it in a binomial? (x+y^-1)^-1

Question

I know if you have an integer exponent you flip the fraction ex)x^-2 = 1/x^2 
But what if you have it in a binomial? (x+y^-1)^-1

amistre64 · Answer

the rules dont change

amistre64 · Answer

y^-1 = 1/y

amistre64 · Answer

(yada)^-1 = 1/yada

anonymous · Answer

But you can't break up a binomial. I think you have to take the y out of the problem before you can do that

amistre64 · Answer

$$(a+b)^{-5}=\frac{1}{\sum_{n=0}^{5}\ C(5,n)\ a^{5-n}\ b^{n}}$$

amistre64 · Answer

what is there to break up?  y is just a "number" so i dont think i understand your .... well, point

anonymous · Answer

(x+y^-1)^-1 = 1/(x+y^-1)
but I don't think you can flip that y from the denominator because its part of a binomial.

anonymous · Answer

I have the answer to the problem but I don't know how you get to it. y/(xy+1)

amistre64 · Answer

you dont flip the y up; "flipping" isnt the rule; it doesnt go up into the numberator; it just flipps itself over and become the reciprocal of y

amistre64 · Answer

$$(x+y^{-1})^{-1}=\frac{1}{x+\frac{1}{y}}=\frac{1(y)}{(x+\frac{1}{y})(y)}=\frac{y}{xy+1}$$

amistre64 · Answer

that is what going on behind the scene

anonymous · Answer

Thank you :)  I see what I was doing wrong. By the way, how do you type a fraction like that?

amistre64 · Answer

magic powers lol ....

*[ \frac{top}{bottom} *]

replace each " * " with a " \ " in order for the parser to recognize it as math code

anonymous · Answer

Ok thanks