The product of any number and zero is zero

Question

I don't understand why this proof is valid. The proof just swaps the left and right hand side of the equality. Is this a valid method of proof for equalities? If so, what's the logical intuition behind it?

Theorem: $0⋅a = 0$ for any integer $a$.

Proof: Add zero!

$$\begin{align} 0 + \color{Green}{0⋅a} = 0⋅a & = (0 + 0)⋅a \\ & = 0⋅a + \color{Green}{0⋅a} \end{align}$$ by using the identity property twice and then the distributive law. Use cancellation on the first and last piece to get $0 = 0⋅a$

---

The problem is not algebra. I don't understand why transforming $0\cdot a =0$ to $0=0\cdot a$ proves this. I mean to prove $a=b$, we transform it to $b=a$, why is it a proof method. What logic does it depend on?

Answer 1

The first equality is true since $0$ plus any element is that element

The second equality is true since $0=0+0$ by the same axiom used to get the first equality

The third equality is true since we use the distributive law

Finally, we reduce $0\cdot a$ from both sides (or in other wording: we add the inverse in respect to addition of $0\cdot a$ to both sides) to get $$0\cdot a-0\cdot a=(0\cdot a+0\cdot a)-0\cdot a$$ i.e. $0=0\cdot a$

Answer 2

you want to prove

$$0\cdot a = 0$$

assume, where x is any number

$$0\cdot a = x \tag{1}$$

We know that

$$0\cdot a = (0+0)\cdot a$$

so

$$(0+0)\cdot a = x$$

$$0\cdot a + 0\cdot a = x\tag{2}$$

subtracting $(2)-(1)$

$$(0\cdot a + 0\cdot a) - (0\cdot a) = x-x = 0$$

that is

$$0\cdot a + 0\cdot a = 0\cdot a$$

so

$$0\cdot a = 0\cdot a - 0\cdot a$$

as a result

$$0\cdot a = 0$$

Answer 3

You did not start with $0\cdot a = 0$. What happened is that you started with $0+0\cdot a$, proved that $0+0\cdot a = 0\cdot a + 0\cdot a$ using identities, and ended up with $0 = 0\cdot a$ which is equivalent to saying $0\cdot a = 0$.