What is the proof for distribution?

Question

I realize that a(b+c) = ab + ac. It works. But why? I can create a geometric proof... even an algebraic proof... so see how it works for integers. But how can it be verified that it works for all rational and irrational and even transcendental numbers?

Seriously, people. In-depth answers, please, because this is an in-depth question.

Answer 1

We have a Euclidean space, R¹, and this is a vector space. As such there are 8 different things that must be stratified for a vector space to be a vector space. a list is here.

http://mathworld.wolfram.com/VectorSpace.html

distribution of vector sums and scalar sums must be satisfied. Now the easy way out of this is to say that since R¹ is a vector space then the distribution property is true for all real numbers.

This is an axiom, so it is presented without proof. If you can show that it does not hold for elements then you do not have a vector space.

Answer 2

i guess for rational is like
m/n(p/q+r/s) = m/n*(ps/qs+qr/qs)=

mps/nqs+mqr/nqs= m/n*p/q+m/n*r/s.
Why this works? Because we applied properties from the construction of Q.(which is the field of fractions of Z)

For non-rational numbers, we know that any real number is
limit of a sequence of rational numbers
so let a=lim a_n, b=lim b_n, c=lim c_n, where sequences are over rationals.
we have to show
lim a_n(lim b_n+lim c_n) = lima_n lim b_n+lim a_n lim c_n

i guess this boils down to show
lim a_n lim b_n = lim a_n b_n
lim a_n lim c_n = lim a_n lim c_n
and lim b_n+lim c_n = lim (b_n+c_n)
which are done with methods applying definiton of lim.

so you still have to show
lim a_n lim(b_n+c_n)= lim a_n b_n+ lim a_n c_n
Now apply the above relations and you get
lim a_n(b_n+c_n) = lim a_n b_n +a_n c_n.
But we already proved for rationals that distributivity works,
So the limits are identical and the relation is true, qed

Answer 3

it does not occur to me.