# How to prove $a - b = - (b - a)$ using the following laws?

I want to prove: $$a - b = - (b - a)$$

I am only allowed to use the following theorems:

“Associativity of +”: (a + b) + c = a + (b + c)

“Associativity of ·”: (a · b) · c = a · (b · c)

“Symmetry of +”: a + b = b + a

“Symmetry of ·”: a · b = b · a

“Additive identity” “Identity of +”: 0 + a = a

“Multiplicative identity” “Identity of ·”: 1 · a = a

“Distributivity of · over +”: a · (b + c) = a · b + a · c

“Zero of ·”: a · 0 = 0

“Unary minus”: a + (- a) = 0

“Subtraction”: a - b = a + (- b)

• Are you learning groups? – yuanming luo Sep 21 '19 at 0:31

By definition, $$-(b-a)$$ is the unique element (real number?) such that $$-(b-a)+(b-a)=0$$ So, if you are able to prove that $$a-b$$ holds the same property, we can conclude that $$a-b=-(b-a)$$ by the uniqueness. Now, observe that \begin{align} (a-b)+(b-a)&=\big( a+(-b) \big) + \big( b+(-a) \big) & (\textrm{definition of substraction})\\ &= \big( a+(-b+b) \big) +(-a) & (\textrm{by associativity of +}) \\ &= (a+0)+(-a) & (-b+b=0 \textrm{ for that } b)\\ &= a+(-a) & (a+0=a \textrm{ for that } a) \\ &= 0. \end{align} Thus, $$a-b=-(b-a)$$.

I may be misinterpreting the question, but I do believe you apply the Distributive Law. $$-(b-a)=-1\cdot (b+(-a))=-b+-(-a)=\boxed{a-b}.$$ Is this what you want?

• You need before prove that $-a=-1a$ for all $a$. – azif00 Sep 21 '19 at 0:52
• You don't have to prove that, right? Isn't it just a notation? – guest Sep 21 '19 at 1:13
• Yes, we must do it. The notation is for $a-b$ which means $a+(-b)$. – azif00 Sep 21 '19 at 1:20

$$a-b=a+(-b)\\a+(-b)+b-a=0\\a+(-b)=-(b-a)\\a-b=-(b-a)$$

Apply unary minus on both sides:

$$a - b = - (b - a)\\ a - b + b-a= - (b - a)+b-a\\ a-b+b-a=0\\ a-a=0$$

Equality holds, so the proposition holds.

• That could be more persuasive in reverse, starting with $a+(-a)=0$ and ending with the desired result – Henry Sep 21 '19 at 9:52
• I could add a few steps and descriptive language, but I think the TR answer does what is required. This is effectively a hint on an alternative approach at this point. – abiessu Sep 21 '19 at 20:41

First, we have following equation:

\begin{align}(b - a) +(a - b)&=[b+(-a)]+[a+(-b)]\\&=b+\{(-a)+[a+(-b)]\}\\&=b+\{[a+(-a)]+(-b)\}\\&=b+[0+(-b)]\\&=b+(-b)\\&=0\end{align}

Then we can add $$-(b-a)$$ to the both sides of above equation from the left

\begin{align}(-(b-a))+[(b - a) +(a - b)]&=(-(b-a))+0\\ [(-(b-a))+(b - a)] +(a - b)&=-(b-a)\\ 0+(a - b)&=-(b-a)\\ (a - b)&=-(b-a)\end{align}