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I'm quite sure about the exactness of my proof, but I'd like to hear (constructive) criticism about my writing.

This is the problem:

Every non-negative integer is coloured white or red, so that:

1) there are at least a white number and a red number;

2) the sum of a white number and a red number is white;

3) the product of a white number and a red number is red. Prove that the product of two red numbers is always a red number, and the sum of two red numbers is always a red number.

And this is my proof: Zero is red: if it were white, then zero times a red number would be zero, which is a white number, contradicting 3).

Case one: zero is red and all the other numbers are white. This is possible since, letting $r$ be a red number, then $r=0$. Therefore $b\cdot r=b\cdot0=0=r$ for any white number $b$, making axiom 3) hold. Also, $b+r=b+0=b$, making 2) hold. Furthermore, the sum of two red numbers would be red: let $r_1, r_2$ be any two red numbers, then $r_1=r_2=0$, therefore $r_1+r_2=0+0=0=r_1=r_2$. The same goes for the product: $r_1 \cdot r_2=0 \cdot 0=0=r_1=r_2$.

Case two: there is at least one non-zero red number. First we prove that the product of two red numbers is red: let $r_1$ and $r_2$ be red numbers, and $b_1$ a white one. Then $r_1r_2+b_1r_1=r_1(r_2+b_1)$; $r_2+b_1$ is white by axiom 2), hence $r_1(r_2+b_1)$ is red by axiom 3), but if $r_1r_2$ were white, then summing it with a red number, $b_1r_1$, would give a white number, therefore $r_1r_2$ is red.

Let $r$ be the smallest non-zero red number. Any multiple of $r$ is also red. Now we prove that only its multiples are red. Suppose that there exists a red number $r_0$ that is not a multiple of $r$, this means that $r_0>r$ and $r_0=rq+m$, where $q$ is the quotient of the divison $r_0:r$ and $m$ the remainder. But since $r_0$ is not a multiple of $r$, $0<m<r$. This means that $m$ is white, because we defined $r$ to be the smallest non-zero red numbers, but then $rq+m$ is the sum of a red number and a white number, which is a white number, contradicting the fact that $r_0$ is red.

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    $\begingroup$ It’s just fine. Note that you could also prove that the product of red numbers is red before splitting the cases. $\endgroup$ – Brian M. Scott Jan 19 '15 at 18:25

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