1. ## binary operation

(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 .

2. 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.

3. ## Re :

For the second one :

x*(y+z) = x(y+z)+1 = xy+xz+1

x*y+x*z = xy+1+xz+1 = xy+xz+2

Thus , it is not distributive .
Any mistake ?

The operation @ and * are defined real numbers as follows :

. . $\displaystyle a*b \:= \:a+b+ab$ .

. . $\displaystyle a\,@\,b \:=\:ab+1$ .
(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 *.