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I can use an example: the case where we have $x$ such that $f(x)=g(x)$. The quotient is $1$, a non-negligible function. However, we can't conclude that all functions $f(x)/g(x)$ are also negligible.

How can I formally prove this?

tylo
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Akhil Dhir
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1 Answers1

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This is false, take $f(x) = 2^{-2x}$ and $g(x) = 2^{-x}$.

fkraiem
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    This answer just appeared in the "low quality" review, probably because it's just one line. It is a perfectly viable counter example, but some extra information would be nice. – tylo Nov 02 '16 at 13:46
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    @tylo I think the answer is fine as it is, thank you very much. – fkraiem Nov 02 '16 at 14:28
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    @tylo: I agree with fraiem; he could use more words, but it is an effective counterexample – poncho Nov 02 '16 at 14:45
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    Just to add: this counterexample actually comes up in real situations. In particular, when simulating constant-round zero-knowledge proofs (as in the Goldreich Kahan proof system). – Yehuda Lindell Nov 02 '16 at 20:25
  • If two quantities go to zero at different rates their ratio can be made to diverge, just divide by the one going to zero faster. – kodlu Nov 02 '16 at 22:13