(a)The operation @ and * are defined real numbers as follows :
a*b = a+b+ab
a@b=ab+1
For each operation , determine if @ is distributive on * .
how can this be done ? Thanks for any help .
Here's how to do the first one:
$\displaystyle x*(y+z) = x + (y+z) + x(y+z) = x + y + z + xy + xz $
$\displaystyle x*y + x*z = (x + y + xy) + (x + z + xz) = 2x + y + z + xy + xz $
These are not the same (e.g. let x = 2), so the operation of * as defined here is not distributive.
Take your time and understand this, and if you do you should be able to apply the same idea to the second.
Hello, mathaddict!
The operation @ and * are defined real numbers as follows :
. . $\displaystyle a*b \:= \:a+b+ab$ . (Add the two and add their product)
. . $\displaystyle a\,@\,b \:=\:ab+1$ . (Multiply the two and add 1)
Determine if @ is distributive on *.
If @ is distributive on *, then: .$\displaystyle a\,@\,(b*c) \;=\;(a\,@\,b)*(a\,@\,c) $ . . . Is this true?
On the left:
. . $\displaystyle a\,@\,(b*c) \;=\;a\,@\,(b+c+bc) $
. . . . . . . . .$\displaystyle =\;a(b+c+bc) + 1$
. . . . . . . . .$\displaystyle = \;ab + ac + abc + 1$
On the right:
. . $\displaystyle (a\,@\,b) * (a\,@\,c) \;=\;(ab+1) * (ac+1)$
. . . . . . . . . . . .$\displaystyle = \;(a+b+1) + (a+c+1) + (ab+1)(ac+1) $
. . . . . . . . . . . .$\displaystyle = \;a^2bc + ab + ac + 2a + b + c + 3$
They are not equal . . . @ is not distributive on *.