Distributive property

[HPedro]

New operation where # is expressed as
a#b=(a+1)(b+1)

Is this distributive property?

 

[tkhunny]

What does this question even mean? Can you demonstrate a single operator that does constitute a "distributive" property?

 

[mmm4444bot]

a#b = (a+1)(b+1)

Is this distributive property?

Nope. :cool:

The multiplication of two numbers -- after first increasing each number by 1 -- is not distribution. It's simply a product of incremented values.


EGs

1#2 = (2)(3) = 6

20#5 = (21)(6) = 126

0#0 = 1

-1#9999999 = 0

 

[soroban]

Hello, HPedro!

New operation where # is expressed as : a#b = (a+1)(b+1)

Is this operation distributive?

I assume you mean "Is multiplication destributive over # ?"

That is, does $\displaystyle a\!\cdot\!(b \# c) \:=\: (a\!\cdot\!b) \# (a\!\cdot\!c)$

. . $\displaystyle a\!\cdot\!(b \# c) \;=\;a\!\cdot\!(b+1)(c+1) \;=\;a(bc + b + c + 1) \;=\;abc + ab + ac + a$

. . $\displaystyle (a\!\cdot\!b) \# (a\!\cdot\!c) \;=\;(a\!\cdot\!b+1)(a\!\cdot\!c+1) \;=\;a^2bc + ab + ac + 1$

They are not equal.
Multiplication is not distributive over the operation.