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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?

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Make a list of all the properties that you have assumed and are allowed to use. Each step in your deduction can be justified by appealing to these assumptions, hence the proof is valid. – Per Manne Sep 16 '12 at 13:36
I think it is important to understand the question first. You did not write what $a$ means, but I assume that you are working in arbitrary field and the goal is to show $0\cdot a=0$ using nothing but the axioms from the definition of field. – Martin Sleziak Sep 16 '12 at 13:58

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 $0\cdot a = 0$.

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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. $0=0\cdot a$

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you want to prove

$$0\cdot a = 0$$

assume, where x is any number

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

We know that

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


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

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

subtracting $(2)-(1)$

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

that is

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


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

as a result

$$0\cdot a = 0$$

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this is nice tnx – mehdi Sep 16 '12 at 13:59

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