Proving Theorem i.5 with Calvo Apostol

[cnaeger]

Hello,

I am working my way through Cal vo1 by Apostol and working the problems and was wondering if someone could assist.

The question is on pg 19 to prove theorems i.5-i.15 using axioms and theorems i.1-i.4.

Theorem i.5 states a(b-c) = ab - bc.

I started the proof but am stuck at at the point indicated below.

a(b-c) = a(b + (-c)) = ab + a(-c)

being able to represent a(-c) as -ac is not illustrated until theorem i.12.

I tried representing a(-c) as -(-a)(-c) but that did not go anywhere either due to (-a)(-c) not being illustrated until i.12 either.

Can anyone help me along with this one. This is not a homework assignment as I am a BS in EE with a minor in math. Just working my way through this book to enhance my knowledge.

Thanks,
Chad

 

[Gib Z]

I personally don't have Apostol, though others on this forum most likely do. Either wait for one of them to post, or please post here stating which Theorems and Axioms you are given, ie what you are already allowed to assume.

 

[neutrino]

a(b-c) = a(b + (-c)) = ab + a(-c)

ab+a(-c) + ac = ?

 

[bmm9]

Prove that ac + a(-c) = 0... then a(-c) = -ac.. I suppose you have to prove that additive inverses are unique but you can use the cancellation law for this. Don't stress over this part too much; most proofs aren't like this. If you did most of the set proofs you should be good - generally that is the stuff that is unfamiliar compared to field axioms.

 

[cnaeger]

In response to Gib Z the Axioms and theorems available are:

A.1 x + y = y + x, xy = yx (Comm Laws)
A.2 x + (y + z) = (x + y) + z, x(yz) = (xy)z (Assoc Laws)
A.3 x(y + z) = xy + xz (Dist Law)
A.4 x + 0 = x, 1*x = x (Existence of Identity elements)
A.5 For every real # x there is a real # y such that x + y =0 (Existence of negatives)
A.6 For every real # x not equal to 0 there is a number y such that xy=1 (Reciprocals)

Th.1 If a + b = a + c then b=c (Cancellation law for addition)
Th.2 Given a and b there is exactly one x such that a + x = b. This is denoted by b - a. In particular 0 - a is simply written -a and called the negative of a.
Th.3 b - a = b + (-a)
Th.4 -(-a) = a

Thanks,
Chad

 

[bmm9]

So far, you have:
a(b-c) = a(b + (-c)) = ab + a(-c)

Now you need to show that a(-c) = -ac... what property does -ac have that you suspect a(-c) has? they're both additive inverses of ac, so..
ac - ac = a(c - c) by a.3
and ac + a(-c) = a(c + (-c)) = a(c - c),
so ac - ac = ac + a(-c), and by thm.1, -ac = a(-c),
now just plug this back into what you had (or you could work from that form the entire time, just leaving "ab" in).