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Let $R$ be a ring with $1 \in R$. Multiplication by a positive integer $k$ means multiple addition $ka = a + a + ... + a$. Suppose the positive integer $n$ is such that $n1 = 0$ in $R$. Show that $na = 0$ for every $a \in R$.

It seems to be easy task, but I have not managed to find an appropiate proof of this. For example writing $na=a+a+...+a$, $a1=1a=a \in R$, can I have $na=n(1a)=(n1)a=0a=0$? I believe that this would follow if $n \in R$ (its the multiplication associativity) but that's not the case here. Any clues?

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Your solution is basically correct, but I guess it would help you if you just write it out: $$na = a +\ldots +a = 1\cdot a +\ldots +1\cdot a = (1 +\ldots +1)\cdot a = 0\cdot a = 0.$$ So apart from the properties of 1 and 0, we only use the distributivity property that holds in $R$.

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    You were faster, so you get the credit for the answer! – 2017-02-21
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Your equalities are correct, but with incorrect reasoning, since $$ na = a+a+\ldots +a = (1+1\ldots +1)a = (n1)a =0a =0$$

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    Last equation is probably $a$ instead of $n$. Thanks for your answer. – 2017-02-21
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    Thanks, yeah that is correct. – 2017-02-21