boolean alegbra

[davazzio]

hello
im having trouble applying identitys. for example how to factor A+AB how does this become A(1+B)? It blowes my mind.

 

[davazzio]

with more studing i think i understand to factor out A from both terms i have to get rid of B and i do this by using A+1=1. so in long term it looks like A+A(1+B)
A+A1
A+A=A

OK still a long way to go.

 

[RCinFLA]

Actually not specific to boolean alegbra. A + AB is factored to A(1+B) in any base numbering system.

To check multiple A by both terms in (1+B) = (A*1) + (A*B) = A + AB.

 

[birdman0_o]

If A and B are scaring you then try just x. x + x^2 = x(1 + x)

 

[MrAl]

Hello,


Look up the "distributive law in algebra". It says a*(b+c)=a*b+a*c. As you can see, the 'a' distributes to the 'b' and the 'c'.

 

[BrownOut]

with more studing i think i understand to factor out A from both terms i have to get rid of B and i do this by using A+1=1. so in long term it looks like A+A(1+B)
A+A1
A+A=A

OK still a long way to go.

The identity you want is A*1=A, so then your expression becomes:

A+AB = A*1+A*B; using the above identity.

A*1 + A*B = A(1+B), factoring.

 

[Fitz Gabby]

A Boolean algebra is also called a Boolean lattice. The connection to lattices (special partially ordered sets) is suggested by the parallel between set inclusion, A ⊆ B, and ordering, a ≤ b. Consider the lattice of all subsets of {x,y,z}, ordered by set inclusion. This Boolean lattice is a partially ordered set in which, say, {x} ≤ {x,y}. Any two lattice elements, say p = {x,y} and q = {y,z}, have a least upper bound, here {x,y,z}, and a greatest lower bound, here {y}.
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**broken link removed**

 

[fouadalnoor]

I agree with what RCinFLA said. Factoring is not actually specific to Boolean Algebra, but algebra in general.

If say I have an equation like this:

2x+4x+6 = 0

If you look at the numbers, you can see that they are all multiples of 2. So:

2 (x+2x+3) = 0

They are both the same equation. In the second case we can just multiply it out to get back to the original equation. The trick is to see what is common to all terms and then work from there.

Say we have an equation like this:

ax+abx+a = 0

again, we can see a is common (the common factor) and by taking it out we get this:

a (x+bx+1) = 0

if you now multiply it out again, you get back to the original equation.

Now in your case you had:

A+AB

This is the same as:

AB+A

And we can see that A is in every term (its the common factor remember) so taking it out you get:

A(B+1)

You can multiply it out to get the same equation.

Hope this long winded explanation helps.

 

[crutschow]

A Boolean algebra is also called a Boolean lattice. The connection to lattices (special partially ordered sets) is suggested by the parallel between set inclusion, A ⊆ B, and ordering, a ≤ b. Consider the lattice of all subsets of {x,y,z}, ordered by set inclusion. This Boolean lattice is a partially ordered set in which, say, {x} ≤ {x,y}. Any two lattice elements, say p = {x,y} and q = {y,z}, have a least upper bound, here {x,y,z}, and a greatest lower bound, here {y}.
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**broken link removed**

What? I don't think the op is quite ready for such an advanced mathematical topic. You lost me after "Boolean lattice".

 

[fouadalnoor]

What? I don't think the op is quite ready for such an advanced mathematical topic. You lost me after "Boolean lattice".

Yeah lol, he lost me as well.

 

[carbonzit]

What? I don't think the op is quite ready for such an advanced mathematical topic. You lost me after "Boolean lattice".

I think it's called "showing off".

 

[ARandomOWl]

I think it's called "showing off".

More like "spamming".