hey guy could somebody help me to solve this :

is the binary operation * defined for real numbers x,y by x*y=x^2y^2

is it commutative? is it associative?

thank you

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- Oct 15th 2008, 12:01 PMboomshine57thBinary operations?
hey guy could somebody help me to solve this :

is the binary operation * defined for real numbers x,y by x*y=x^2y^2

is it commutative? is it associative?

thank you - Oct 15th 2008, 01:03 PMSoroban
Hello, boomshine57th!

Quote:

Is the binary operation $\displaystyle *$ defined for real numbers $\displaystyle x,y$ by: .$\displaystyle x*y \:=\:x^2y^2$ ?

If $\displaystyle x,y$ are any real numbers, then: .$\displaystyle x*y \:=\:x^2y^2$ is a real number.

The operationdefined for real numbers.*is*

Quote:

Is it commutative?

If the operation is commutative, then: .$\displaystyle x*y \:=\:y*x$

We have: .$\displaystyle \begin{array}{cccc}x*y &=& x^2y^2 \\ y*x &=&y^2x^2 \end{array}$

Since $\displaystyle x^2y^2 = y^2x^2$, the operationcommutative.*is*

Quote:

Is it associative?

If the operation is associative, then: .$\displaystyle x*(y*z) \:=\:(x*y)*z$

We have: .$\displaystyle \begin{array}{ccccccc}

x*(y*z) &=& x*(y^2z^2) &=& x^2(y^2z^2)^2 &=& x^2y^4z^4 \\

(x*y)*z &=& (x^2y^2)*z &=& (x^2y^2)^2 z^2 &=& x^4y^4z^2\end{array}\quad\hdots$ not equal

The operation isassociative.*not*

- Oct 15th 2008, 01:26 PMboomshine57ththank you
thank you soooo much! you are a life saver (Happy)