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If an additive function $a(x)$ with the property that $a(xy)=a(x)+a(y)$, when $\gcd(x,y)=1$, then is $a(1)=0$, I think it should be, otherwise I would have for example $a(1)=a(1)+a(1)..., a(1)=c\cdot a(1)$, which implies $1$ is equal to everything?

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To rephrase your argument: we have $a(1) = a(1\cdot 1) = a(1) + a(1)$. Subtracting $a(1)$ from both sides, we get $a(1) = 0$.

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    Yes but a(1)=2a(1), dividing both sides by a(1), gives 1=2, doesn't that imply it must be zero?2012-11-25
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    Right, you're not allowed to divide by $a(1)$ because it is zero.2012-11-25
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    @boby: It does imply $a(1)$ must be zero for the exact reason Ben gave above: $$ a(1) = 2a(1) \iff 2 a(1) - a(1) = 0 \iff a(1) = 0. $$ In general it is a bit sloppy to divide by an unknown, derive a contradiction, and _then_ conclude the unknown must have been zero. It is far easier (and less error-prone) to subtract one side from another and factor.2012-11-25
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    Yes but you assumed a(1) was zero before I divided it by a(1), couldn't there be a different reason for the contracdiction, how do I know its because a(1) is zero.2012-11-25
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    In the argument JavaMan and I are trying to make, we don't make an assumption that $a(1)$ is nonzero and derive a contradiction; we prove directly that $a(1)$ is zero. We see that the definition of additive implies that $a(1) = 2a(1)$, and subtracting $a(1)$ (which is always legal!) gives us that $a(1) = 0$.2012-11-25
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    But why isn't division "legal"?2012-11-25
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    Division is legal only if we are dividing by something nonzero. We can talk about $1/2$ or $1/3$ or $1/(-123)$. We are not allowed to divide by zero, so there is no $1/0$. If we want to talk about $1/a(1)$, then we have to be careful and take into account the possibility that $a(1)$ is zero, in which case we can't divide.2012-11-25
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    In this case $a(1)$ does happen to be zero (which we show without any contradictions), so there is no need to divide by it. It is acceptable (although unusual, as JavaMan remarks) to say that we have $2a(1) = a(1)$, assume $a(1)$ is nonzero, and divide by it to get a contradiction. The reason why the contradiction must come from assuming $a(1) \neq 0$ is that we haven't made any other assumptions.2012-11-25
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    If I don't know for a fact the value of some quanity x, is Not zero, then I can never divide it by some other expression?2012-11-27
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    If you have a quantity $x$ and you don't know that it's not zero, then you can't divide another expression by it. What you can do is to split it into two cases: either $x$ is zero or $x$ is not zero. For example, if you want to solve the equation $x(x+1) = 0$, then one possibility is that $x$ is zero. To find the other solutions, you can assume that $x$ is not zero, divide by it, and get $x+1 = 0$, which gives $x=-1$. So $x$ can be $0$ or $-1$, but you have to split your reasoning into two cases.2012-11-27
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    To expand on @Ben's point, tne problem with dividing by zero to obtain a contradiction is it is not always clear which assumption has been violated. For example, say we have defined $b(x)^2 = b(x) - 1$ for positive x. Dividing both sides by $b(x)^2$, we have $1=1/b(x) - 1/b(x)^2$. The difference of two positive reciprocals is strictly less than 1: contradiction. But it is incorrect to conclude $b(x) \neq 0$ for positive $x$ because the actual assumption being contradicted is that there is any such function with the given properties of $b(x)$.2012-11-27
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    In fact, concluding $a(1) = 0$ also depends on $a(x)$ being a valid definition of a function, although in this case $a(x)=0$ is an obviously valid example of such.2012-11-27