Hi I have a question about the following algebra rule
A + AB = A
My textbook explains this as follows A + AB = A This rule can be proved as such:
Step 1:
Dustributive Law:
A + AB = A*1 = A(1+B) Huh...? Where do they get the one(1) from?
Step 2:
1 + B = 1
Step 3:
: A + 1 = A
Thus A + AB = A
If anyone can clarify this for me it would be greatly appreciated
The 1 would stand for ⊤ or true. To prove the rule they're assuming the right side is true, distributing this information into the left side, and reducing.
Starting with A ∨ (A ∧ B) ↔ A,
A + AB = A
Call A ⊤,
⊤ ∨ (⊤ ∧ B) ↔ ⊤
1 + 1 * B = 1 now it reads "true or true and B equals true" which could just as easily be "with gravy or gravy and something else, you will have gravy"
1 + B = 1 AND having higher precedence...
1 = 1 OR is true if at least one operand is true
A no further reduction possible.
It could just as easily be done using ⊥ (false) instead
⊥ ∨ (⊥ ∧ B) ↔ ⊥
0 + 0 * B = 0 which would read "false or false and B equals false" which could just as easily be "without bananas or bananas and something else, you will not have bananas"
0 + 0 = 0 AND having higher precedence...
0 = 0 OR is false
A no further reduction possible
It might help to construct a truth table and then review the rules for distribution. The 1 appearing in your formula is distributed into the terms to facilitate simplifying the statement.
Since rule 0101 (10) maps to (P ∧ Q) ∨ Q ↔ Q
P Q x
0 0 0
0 1 1
1 0 0
1 1 1
or a Karnaugh map
~Q Q
~P 0 1
P 0 1
see also http://en.wikipedia.org/wiki/Truth_function#Table_of_binary_truth_functions : Proposition Q
So you can always replace X with X*1, and you can always replace 1 with X+1.
