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It's known that if $ \left(F,+, *\right) $ is a field, then $ \forall a \in F, 0_F * a = 0_F$.

Question: Is there any intuitive explanation for this? (Not a proof).

Because, after all, $ + $ and $ * $ are different operations.

talopl
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    Just the distributive law. $b\times a=(b+0)\times a = b\times a +0\times a$. – lulu Jun 29 '22 at 00:05
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    And it is true for a ring too, even a rng. – Henry Jun 29 '22 at 00:06
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    @Henry I'm looking for an intuitive explanaton fr this, not a proof – talopl Jun 29 '22 at 00:10
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    The distributive proof is as intuitive as it gets: adding zero before multiplying has no effect, and expanding this means multiplying by zero must lead again to adding nothing – Henry Jun 29 '22 at 00:14
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    I don’t have intuition for this beyond the proof that lulu suggested. (I do have various ways of thinking about multiplication when our field is $\mathbb Q$ or $\mathbb R$, of course.) Since a field is an abstract setting and we have only the axioms to work with, it seems difficult to come up with a more intuitive argument that makes sense in the general setting. If $0 \times a$ were not $0$ then surely we would have modified the field axioms to make this result be true. – littleO Jun 29 '22 at 00:18
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    This is not a duplicate of the linked post. OP asks intuitive explanation, not a proof. –  Jun 29 '22 at 13:15

1 Answers1

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Bill Dubuque explained it in the linked comment: $0$ is that thing which if you add it to anything, the anything does not change. In fact, if there exists something such that adding $x$ to it does not change the something, then $x$ must be $0$. Well, by distributive law adding $0*a$ to any multiple of $a$ will not change it; so $0*a$ must be $0$.

Max
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