How can I prove that $\sqrt{2}$ is irrational with the ideal $$\{n \in \mathbb{Z} \mid n\times \sqrt{2} \in \mathbb Z\}?$$
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2Just show that there cannot exist a minimum element in the set that can be an integer. – Jan 17 '21 at 16:30
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2Which ring is that an ideal in? – Kenta S Jan 17 '21 at 16:30
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thanks for your answer, math buddy. But you see, my point IS to prove that the ideal is reduced to zéro. Because the Z is principal, so all the ideals have the same structure, e.g. (k)={nk| n \in N}. Get it ? – rorovavro Jan 17 '21 at 18:00
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Yes, see this answer in the linked dupe (and its links). – Bill Dubuque Jan 17 '21 at 19:16
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thanks !!! That's exactly wot I was searching. – rorovavro Jan 18 '21 at 20:52