An alternative proof (Hopefully not an alternative fact)

[Nerd-ho]

Homework Statement

Hi all, I'm currently studying the amazing Calculus by Spivak. Whenever I study textbooks I always attempt to do all the examples and proofs in the text before looking at the answers.

(Whether this is a good thing or a bad thing I don't know, the examples are similar to the easy to moderate problems in exercise books and are never challenging, but I find it difficult to do it any other way, I'm like the child who refuses to be shown how to undo the knot, I want to work it out for myself.)

One of the proofs I did looked different to that which Spivak provided and I'd like to know if it's still correct.


2. Homework Equations

The theorems used were:
a + (-a) = (-a) + a = 0
a(bc) = (ab)c
a(1) = (1)a = a
a(b + c) = ab + ac
a(0) = 0

Spivak used the first, fourth and fifth theorems and I used the second and third.

The proof was of (-a)b = -(ab)

Spivak's was:
(-a)b + (ab) = (a + (-a))b
(-a)b + (ab) = (0)b
(-a)b + (ab) = 0
(-a)b + (ab) + (-(ab)) = 0 + (-(ab))

Therefore (-a)b = -(ab)

The Attempt at a Solution

My solution was:

(-a)b = (-1a)b
= -1(ab)
= -(1(ab))
= -(ab)

Any feedback would be greatly appreciated.

Regards, Travis.

 

[fresh_42]

Where have you shown, that $-a = -1a$ for the first line? All you have for $-1$ is that $1 + (-1)=0$. Otherwise, $-1a$ isn't defined yet.

 

[Nerd-ho]

Thanks @fresh_42

I see my mistake, I would need to use the theorem in order to prove it's own validity.