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2:20 AM
A new tag . I am not really sure what was the intended tag there.
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Q: Showing that $e^{2\pi i /5}\not\in \mathbb{Q}(\sqrt[4]{2},i)$

eraldcoilShow that there are no $a,b,c,d\in \mathbb{Q}(i)$ such that $e^{2\pi i/5}=a+b\sqrt[4]{2}+c\sqrt{2}+d(\sqrt[4]{2})^{-1}$ Previously in this problem, I found all the subfields of $\mathbb{Q}(\sqrt[4]{2},i)$ with the help of the subgroups of $D_4$ and Galois's Theorem as you can see in the followin...

 
 
4 hours later…
5:53 AM
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Q: Showing that $e^{2\pi i /5}\not\in \mathbb{Q}(\sqrt[4]{2},i)$

eraldcoilShow that there are no $a,b,c,d\in \mathbb{Q}(i)$ such that $e^{2\pi i/5}=a+b\sqrt[4]{2}+c\sqrt{2}+d(\sqrt[4]{2})^{-1}$ Previously in this problem, I found all the subfields of $\mathbb{Q}(\sqrt[4]{2},i)$ with the help of the subgroups of $D_4$ and Galois's Theorem as you can see in the followin...

 
 
2 hours later…
8:15 AM
in MO editors' lounge, 9 mins ago, by Martin Sleziak
SEDE query showing (approved) tag synonyms where both tags have some questions: MathOverflow, Mathematics.
in MO editors' lounge, 9 mins ago, by Martin Sleziak
There is no such synonym on MathOverflow Meta and one on Mathematics Meta.
 
 
4 hours later…
12:36 PM
A new tag was created by Devansh Kamra.
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Q: Proving a relation between sum of reciprocal of divisors and $\sigma(n)$

Devansh Kamra Prove that $\sum_{d|n}\dfrac{1}{d}=\dfrac{\sigma(n)}{n}\ \forall\ n\geq 1, n\in \mathbb Z$ My question and my approach is a lot similar to this question but just a bit short, I want to verify my approach. Here it goes: My Approach: Suppose $\sum_{d|n}\dfrac{1}{d}=\dfrac{1}{d_1}+\dfrac{1}{d_2}+...

I'd say , , seem suitable for that question. (With probably the closest to the intended meaning of .)
I edited the tags on that question.
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Q: Proving a relation between sum of reciprocal of divisors and $\sigma(n)$

Devansh Kamra Prove that $\sum_{d|n}\dfrac{1}{d}=\dfrac{\sigma(n)}{n}\ \forall\ n\geq 1, n\in \mathbb Z$ My question and my approach is a lot similar to this question but just a bit short, I want to verify my approach. Here it goes: My Approach: Suppose $\sum_{d|n}\dfrac{1}{d}=\dfrac{1}{d_1}+\dfrac{1}{d_2}+...

 
12:51 PM
Although maybe is a stretch.
 

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