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

• I
• Velcroe
In summary, to prove a(b-c)=ab-ac, you can let x=b-c and use the properties of subtraction, associativity, existence of negatives, and identity to show that both sides are equal to ax. Alternatively, you can use the distributive property of addition to show that a(b-c) is equal to ab-ac.

#### 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?

Velcroe said:
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.

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.

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.

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

## 1. What does the equation "a(b-c)=ab-ac" mean?

The equation "a(b-c)=ab-ac" is known as the distributive property in algebra. It states that when multiplying a number or variable by a sum or difference in parentheses, the result will be the same as if you distributed the multiplication to each term inside the parentheses. In this case, it means that the product of a and the difference of b and c is equal to the difference of the products of a and b, and a and c.

## 2. How do you prove the equation "a(b-c)=ab-ac" using algebraic manipulation?

To prove the equation "a(b-c)=ab-ac", we can start by expanding the left side using the distributive property. This gives us:
a(b-c) = ab - ac.
Then, we can rearrange the terms on the right side to match the right side of the equation. This gives us:
ab - ac = ab - ac.
Since both sides are now equal, we have proven the equation.

## 3. Can you provide an example to demonstrate the equation "a(b-c)=ab-ac"?

Yes, for example, if we let a = 2, b = 5, and c = 3, the equation becomes:
2(5-3) = 2(5)-2(3)
2(2) = 10-6
4 = 4
This shows that the equation holds true for these values of a, b, and c.

## 4. How is the equation "a(b-c)=ab-ac" useful in solving algebraic problems?

The equation "a(b-c)=ab-ac" is useful in algebraic problems because it allows us to simplify expressions by distributing the multiplication. This can help us solve equations, simplify complex expressions, and factor polynomials.

## 5. Is the equation "a(b-c)=ab-ac" always true?

Yes, the equation "a(b-c)=ab-ac" is always true. This is because it is a fundamental property of algebra, known as the distributive property. It holds true for any values of a, b, and c, as long as they are real numbers or variables.