prove the following rules of addition a/b+c/d=(ad+bc)/bd
Answers
You should simplify / isolate a and b independently, then add the results. a simplified = a(c+d) = ac + ad. b simplified = b(c+d) = bc + bd
Answer:
It depends which are your axioms or known identities.
If you already have field axioms (or have already proved that multiplication distributes addition), then you have right distribution:
a(c+d)=a×(c+d)=(a×c)+(a×d)=ac+ad
And left distribution:
(a+b)c=(a+b)×c=(a×c)+(b×c)=ac+bd .
Just apply left distribution and right distribution and you get:
(a+b)(c+d)=a(c+d)+b(c+d)=(ac+ad)+(bc+bd)=ab+ad+bc+bd
If you begin with Peano's axioms and definitions of addition and multiplication, then you should first prove right and left distribution, using induction.
(a+b)×0=0,a×0=0,b×0=0 , by definition of multiplication.
0=0+0 , by definition of addition, therefore
(a+b)0=a0+b0 .
Let's assume , by induction hypothesis that
(a+b)n=an+bn
Then
(a+b)sucn=(a+b)n+(a+b) , by inductive definition of multiplication.
=(an+bn)+(a+b) , by induction hypothesis.
=an+bn+a+b , let's assume we have already proved addition is associative
=an+a+bn+b , let's assume we have already proved addition is commutative
=a⋅sucn+b⋅sucn , by associative, definition of addition; and q.e.d.
We have proved left distribution. For proving right distribution it is usually easier proving that product is commutative first.
If you get to the real numbers by construction from natural numbers, you need to ensure that after each construction distributive properties are preserved.