Prove a(b-c)=ab-ac: Is It Enough?

[Velcroe]

I am currently working my way though Calculus by Tom Apostol. One of the really early proofs ask the reader to prove: a(b-c)=ab-ac. Here is what I did, I let x=b-c which by the definition of subtraction equals x+c=b. Substituting that value into the right hand side I got a((x+c)-c)=a(x+(c-c))=a(x+0)=ax.

I then plugged the exact same value into the right hand side getting a(x+c)-ac=(ax+ac)-ac=ax+(ac-ac)=ax.

Is this sufficient as a proof? In a proof that I looked up the author of the proof instead let part of the left hand side =x and part of the right hand side equal y then showed that x=y. Is that the way I should have approached this problem?

 

[fresh_42]

I am currently working my way though Calculus by Tom Apostol. One of the really early proofs ask the reader to prove: a(b-c)=ab-ac. Here is what I did, I let x=b-c which by the definition of subtraction equals x+c=b. Substituting that value into the right hand side I got a((x+c)-c)=a(x+(c-c))=a(x+0)=ax.

I then plugged the exact same value into the right hand side getting a(x+c)-ac=(ax+ac)-ac=ax+(ac-ac)=ax.

Is this sufficient as a proof? In a proof that I looked up the author of the proof instead let part of the left hand side =x and part of the right hand side equal y then showed that x=y. Is that the way I should have approached this problem?

When you write a(x+c)-ac=(ax+ac)-ac don't you use already what you want to prove? Or is it all about the subtraction in the formula?
You should start with a list of what you are allowed to use (and tell us). There is more than one possible way to establish group operations.

 

[Velcroe]

Sorry let me fix it.
Prove a(b-c)=ab-ac
let x=b-c existence of subtraction (axiom)
x+c=b (part of the subtraction axiom)
a[(x+c)-c] substitution which is never spelled out but used repeatedly in books proofs so I assume its allowed
a[x+(c-c)] associative property axiom
a[x+0] existence of negatives axiom
ax identity axiom

For Right hand side
ab-ac given
a(x+c)-ac substitution as I did above
(ax+ac)-ac distributive axiom
ax+(ac-ac) associative axiom
ax+0 existence of negative axiom
ax identity axiom

This shows both left hand side and right hand side =ax so both sides are equal.

 

[fresh_42]

If you already have the distributivity for addition this is correct. But you could as well simply write
$$a(b-c) = a(b+(-c))= ab + a (-c) = ab + a((-1)c) + ab + (a(-1))c = ab +(-a)c = ab-ac$$
Edit: Maybe $-c = (-1)c$ is cheating here.

 

[Velcroe]

Thanks so much for your help. That makes a lot of sense.