number

• Dec 2nd 2008, 05:13 AM
number
Can someone pls guide me through . Thanks

Q : The binary operation , * , on rational numbers is as follows . For x and y are both rational numbers .

x*y = x+y-xy .

(a) Prove that * is associative
(b) Determine if * is distributive with respect to addition .
• Dec 2nd 2008, 09:48 AM
Soroban

Quote:

The binary operation $*$ on rational numbers is as follows.

For rationals $x\text{ and }y\!:\;\;x*y \:=\: x+y-xy$

(a) Prove that * is associative

We want to show that: . $(a*b)*c \:=\:a*(b*c)$

$(a*b)*c \;=\;(a+b-ab)*c$

. . . . . . $=\;(a + b-ab) + c - (a+b-ab)c$

. . . . . . $=\; a + b - ab + c - ac - bc + abc$

. . . . . . $= \;a + b + c -(ab + bc + ac) + abc$ .[1]

$a*(b*c) \;=\;a*(b + c - bc)$

. . . . . . $= \;a + (b + c - bc) - a(b + c - bc)$

. . . . . . $=\;a + b + c + -bc - ab - ac + abc$

. . . . . . $= \;a + b + c - (ab + bc + ac) + abc$ .[2]

Since [1] = [2], the operation is associative.

Quote:

(b) Determine if * is distributive with respect to addition .
Does $a*(b+c) \:=\:(a*b) + (a*c)$ ?

$a*(b+c) \;=\;a + (b+c) - a(b+c)$

. . . . . . . $= \;a + b + c - ab - ac$ .[3]

$(a*b) + (a*c) \;=\;(a + b - ab) + (a + c - ac)$

. . . . . . . $= \;{\color{red}2}a + b + c - ab - ac$ .[4]

Since [3] $\neq$ [4], the operation is not distributive over addition.