Can someone help me with proofs?

Question

Prove the theorem using the basic axioms of algebra.

If ac = bc and c=/=0, then a = b.



It's my first time learning proofs. I just need this one example exlpained to me and I'll get it.

Answer 1

Here's a pseudo-proof below.

Assume ac = bc, and c is not equal to 0.
This means there exists a multiplicative inverse 1/c.

Multiply both sides by (1/c) from the RIGHT.

(ac)(1/c) = (bc)(1/c)

By associativity, this becomes

a(c * 1/c) = b(c * 1/c)

One of the axioms is that a number multiplied by its multiplicative inverse is 1.

a(1) = b(1)

By identity, a(1) = a, and b(1) = b, so

a = b

Answer 2

Given ac = bc and c not = 0
Prove a = b

ac= bc [Reason = Given]
c not = 0 [Reason = Given]
ac/c = bc/c [ Reason = Division axiom]
c/c = 1 [ Aquantity divided by itself = 1]
a*1 = b*1 [Substitution Axiom]
therefore a = b [muliplicative identity axiom]

Answer 3

ac=bc subtracting bc from both sides
ac - bc = 0 factor oout the common c
c(a-b) = 0 Now, for teh product to be 0, one of the factors must be 0 and, since c cannot be 0 (from the statement of the problem) we are left with
a-b = 0 and, adding b to both sides
a = b

BTW, proofs are the absolute --heart-- of all of mathematics. Learn them well and things will be much easier ☺

HTH

Doug

Answer 4

ac = bc

ac/c = bc/c { Divided both sides by "c" }

a = b

The reason c can't be zero (0) is because you can't divide by zero.

Answer 5

always hated proofs

Answer 6

no....not really.